{"id":23,"date":"2019-06-11T13:06:38","date_gmt":"2019-06-11T17:06:38","guid":{"rendered":"https:\/\/pressbooks.library.torontomu.ca\/pohmath\/chapter\/unit-7-equations\/"},"modified":"2024-01-05T09:42:34","modified_gmt":"2024-01-05T14:42:34","slug":"unit-7-equations","status":"publish","type":"chapter","link":"https:\/\/pressbooks.library.torontomu.ca\/pohmath\/chapter\/unit-7-equations\/","title":{"raw":"1.2. Equations","rendered":"1.2. Equations"},"content":{"raw":"
[Latexpage]<\/div>\r\n

Equations<\/h1>\r\nEquation<\/strong>: a mathematical sentence that contains two expressions and is separated by an equal sign (both sides of the equation have the same value).\r\n
Example:\u00a0\u00a0 <\/strong>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a04 + 3 = 7,\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 9x<\/em> \u2013 4 = 5, \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a02y<\/em> \u2013 $\\frac{1}{3}$ = y<\/em><\/div>\r\n \r\n\r\nTo <\/strong>solve<\/strong> an equation<\/strong> we find a particular value for the variable in the equation that makes the equation true (left side = right side).\r\n
\r\n\r\nExample:<\/strong>\u00a0 \u00a0For the equation \u00a0\u00a0x<\/em> + 4 = 5\r\n\r\nonly x<\/em> = 1 can make it true, since \u00a0\u00a01 + 4 = 5 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0(Left side = Right side)\r\n\r\n<\/div>\r\n \r\n\r\nSolution of an equation<\/strong>: the value of the variable in the equation that makes the equation true.\r\n
Example:<\/strong>\u00a0 \u00a0\u00a0For the equation\u00a0 \u00a0x<\/em> + 4 = 5,\u00a0\u00a0 \u00a0\u00a0x<\/em> = 1\u00a0 \u00a0is the solution.<\/div>\r\n \r\n
\r\n

Example 1.2.1<\/p>\r\n\r\n<\/header>\r\n

\r\n\r\nIndicate whether each of the given number is a solution to the given equation.\r\n\r\n\r\n\r\n
1)<\/strong>\u00a0\u00a0\u00a0\u00a0\u00a0 2 :<\/strong> \u00a0\u00a0\u00a0\u00a0\u00a04x<\/em> \u2013 3 = 5<\/td>\r\n4 \u2219 2 \u2013 3 $\\overset{?}{=}$ 5<\/td>\r\n5 $\\overset{\\checkmark}{=}$ 5<\/td>\r\nYes<\/td>\r\nReplace x<\/em> with 2.<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n\r\n\r\n\r\n
2)<\/strong>\u00a0\u00a0\u00a0\u00a0\u00a0 15 :<\/strong>\u00a0 \u00a0 $\\frac{-3}{15}$y<\/em> = -3<\/td>\r\n$\\frac{-3}{15}(15) \\overset{?}{=}$ -3<\/td>\r\n-3 $\\overset{\\checkmark}{=}$ -3<\/td>\r\nYes<\/td>\r\nReplace y<\/em> with 15.<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n\r\n\r\n\r\n
3)<\/strong> \u00a0\u00a0\u00a0\u00a0 $\\bf\\frac{1}{2}$\u00a0:<\/strong>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 8t<\/em> = 3<\/td>\r\n8 ($\\frac{1}{2}$) $\\overset{?}{=}$ 3<\/td>\r\n4 \u2260 3<\/td>\r\nNo<\/td>\r\nCannot replace t<\/em> with $\\frac{1}{2}$.<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/div>\r\n<\/div>\r\n
\r\n

Solving Equations<\/strong><\/h2>\r\n<\/div>\r\n
\r\n\r\nBasic rules for solving one-step equations<\/strong>:<\/span>\r\n\r\n<\/div>\r\n
    \r\n \t
  • Add, subtract, multiply or divide the same quantity to both sides of an equation to obtain a valid equation.<\/li>\r\n \t
  • Remember to always do the same thing to both sides of the equation (balance).<\/li>\r\n<\/ul>\r\nProperties for solving equations:<\/strong>\r\n\r\n\r\n\r\n
    Properties<\/strong><\/td>\r\nEquality<\/strong><\/td>\r\nExample<\/strong><\/td>\r\n<\/tr>\r\n<\/thead>\r\n<\/table>\r\n\r\n\r\n\r\n\r\n\r\n\r\n
    Addition<\/strong> property of equality<\/td>\r\nA =<\/strong> B<\/em>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 A + C =<\/strong> B<\/em> + C<\/em><\/strong><\/td>\r\nSolve $x-6=3$\r\n\r\n$x-\\bcancel{6}+\\bcancel{\\bf6}=3+\\bf6$\u00a0 \u00a0 \u00a0x <\/em>= 9<\/strong><\/td>\r\n<\/tr>\r\n
    Subtraction<\/strong> property of equality<\/td>\r\nA =<\/strong> B<\/em>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 A \u2013<\/em>\u00a0C =<\/em><\/strong> B<\/em> \u2013\u00a0C<\/em><\/strong><\/td>\r\nSolve $y+5=-8$\r\n\r\n$y+\\bcancel{5}-\\bcancel{\\bf5}=-8-\\bf5$\u00a0 \u00a0 \u00a0y<\/em> = -13<\/strong><\/td>\r\n<\/tr>\r\n
    Multiplication<\/strong> property of equality<\/td>\r\nA =<\/strong> B<\/em>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 A \u00b7<\/em>\u00a0C =<\/em><\/strong> B<\/em> \u00b7\u00a0C<\/em><\/strong><\/td>\r\nSolve $\\frac{m}{9}=2$\r\n\r\n$\\bcancel{\\bf9} \\cdot \\frac{m}{\\bcancel{9}}=2 \\cdot \\bf9$\u00a0 \u00a0 \u00a0m = 18<\/strong><\/td>\r\n<\/tr>\r\n
    Division <\/strong>property of equality<\/td>\r\nA =<\/strong> B<\/em>\u00a0 \u00a0 \u00a0 \u00a0 $\\frac{A}{C}=\\frac{B}{C}$\u00a0 \u00a0 \u00a0(C<\/em> \u2260 0)<\/td>\r\nSolve $3n=-15$\r\n\r\n$\\frac{\\bcancel{3}n}{\\bcancel{\\bf3}}=\\frac{-15}{\\bf3}$\u00a0 \u00a0 \u00a0n = -5<\/strong><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n \r\n
    \r\n

    Example 1.2.2<\/p>\r\n\r\n<\/header>\r\n

    \r\n\r\nSolve the following equations.\r\n\r\n\r\n\r\n\r\n\r\n\r\n
    1) <\/strong>$-9 + x = 5$<\/td>\r\n$-\\bcancel{9} + x+ \\bcancel{\\bf9} = 5 + \\bf9$<\/td>\r\nProperty of addition.<\/td>\r\n<\/tr>\r\n
    <\/td>\r\n$x = 14$<\/td>\r\n<\/td>\r\n<\/tr>\r\n
    <\/td>\r\n<\/td>\r\n<\/td>\r\n<\/tr>\r\n
    Check:<\/td>\r\n$-9 + {\\bf14} \\overset{?}{=} 5$\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 $5 \\overset{\\checkmark}{=} 5$<\/td>\r\nReplace x with 14.<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n \r\n\r\n\r\n\r\n\r\n
    2) <\/strong>$t+\\frac{2}{5}=-\\frac{1}{5}$<\/td>\r\n$y+\\frac{2}{5}-\\frac{\\bf2}{\\bf5}=-\\frac{1}{5}-\\frac{\\bf2}{\\bf5}$<\/td>\r\nProperty of subtraction.<\/td>\r\n<\/tr>\r\n
    <\/td>\r\n$y=-\\frac{3}{5}$<\/td>\r\n<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n \r\n\r\n\r\n\r\n\r\n
    3) <\/strong>\u00a0$\\frac{-1}{6}x = 7$<\/td>\r\n${\\bf-6} \\cdot \\frac{-1}{6}x=7(\\bf-6)$<\/td>\r\nProperty of multiplication.<\/td>\r\n<\/tr>\r\n
    <\/td>\r\n$x = -42$<\/td>\r\n<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n \r\n\r\n\r\n\r\n\r\n
    4<\/strong>) <\/strong><\/em>$-5x = 30$<\/td>\r\n$\\frac{-5x}{\\bf-5}=\\frac{30}{\\bf-5}$<\/td>\r\nProperty of division.<\/td>\r\n<\/tr>\r\n
    <\/td>\r\n$x = -6$<\/td>\r\n<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n \r\n\r\n<\/div>\r\n<\/div>\r\n

    <\/h1>\r\nMulti-step equation<\/strong>: an equation that requires more than one step to solve it.\r\n\r\nProcedure for solving multi-step equations:<\/strong>\r\n\r\n\r\n\r\n
    \r\n
      \r\n \t
    • Clear the fractions or decimals if necessary.<\/li>\r\n<\/ul>\r\n
        \r\n \t
      • Simplify and remove parentheses if necessary.<\/li>\r\n<\/ul>\r\n
          \r\n \t
        • Combine like terms on each side of the equation.<\/li>\r\n<\/ul>\r\n
            \r\n \t
          • Collect the variable terms on one side of the equation and the constants on the other side.<\/li>\r\n<\/ul>\r\n
              \r\n \t
            • Isolate the variable (to get the\u00a0variable alone on one side\u00a0of the equation).<\/li>\r\n<\/ul>\r\n
                \r\n \t
              • Check the solution with the original equation.<\/li>\r\n<\/ul>\r\n<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n \r\n\r\n\r\n\r\n
                Steps<\/strong><\/td>\r\nExample<\/strong><\/td>\r\n<\/tr>\r\n<\/thead>\r\n<\/table>\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
                <\/td>\r\nSolve $\\bf\\frac{1}{5}(y+10)=3y-\\frac{9}{5}y$<\/td>\r\n<\/td>\r\n<\/tr>\r\n
                \r\n
                  \r\n \t
                • Eliminate the denominators if the equation has fractions.<\/li>\r\n<\/ul>\r\n<\/td>\r\n
                		$\\bcancel{5}\\cdot \\frac{1}{\\bcancel5}(y+10)={\\bf5}(3y)-{\\bf\\bcancel5}(\\frac{9}{\\bcancel5}y)$<\/td>\r\nMultiply each term by 5.<\/td>\r\n<\/tr>\r\n
                \r\n
                  \r\n \t
                • Remove parentheses.<\/li>\r\n<\/ul>\r\n<\/td>\r\n
                		$y+10=15y-9y$<\/td>\r\n<\/td>\r\n<\/tr>\r\n
                \r\n
                  \r\n \t
                • Combine like terms.<\/li>\r\n<\/ul>\r\n<\/td>\r\n
                		$y+10=6y$<\/td>\r\n<\/td>\r\n<\/tr>\r\n
                \r\n
                  \r\n \t
                • Collect variable terms on one side and the constants on the other side.<\/li>\r\n<\/ul>\r\n<\/td>\r\n
                		$y+\\bcancel{10}-\\bcancel{\\bf10}=6y-\\bf10$\r\n\r\n$y=6y-\\bf10$\r\n\r\n$y-{\\bf6y}=\\bcancel{6y}-10-\\bcancel{\\bf6y}$<\/td>\r\n\r\n

                Subtract 10 from both sides.<\/p>\r\n \r\n

                Subtract 6y<\/em> from both sides.<\/p>\r\n<\/td>\r\n<\/tr>\r\n

                \r\n
                  \r\n \t
                • Isolate the variable.<\/li>\r\n<\/ul>\r\n<\/td>\r\n
                		$-5y=-10$\r\n\r\n$y=\\frac{-10}{-5}$\r\n\r\n$y=2$<\/td>\r\nDivide both sides by -5.<\/td>\r\n<\/tr>\r\n
                \r\n
                  \r\n \t
                • Check with the original equation.<\/li>\r\n<\/ul>\r\n<\/td>\r\n
                		$\\frac{1}{5}(2+10)=3 \\cdot 2-\\frac{9}{5} \\cdot 2$\r\n\r\n$\\bcancel{5}\\frac{1}{\\bcancel5}(2+10)=5 \\cdot 3 \\cdot 2-\\bcancel{5} \\cdot \\frac{9}{\\bcancel5} \\cdot 2$\r\n\r\n$(2+10)=30-18$\r\n\r\n$12=12$<\/td>\r\nReplace y<\/em> with 2.\r\n\r\nMultiply each term by 5.\r\n\r\n \r\n\r\nLS = RS\u00a0 (correct)<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n \r\n\r\nEquations involving decimals: <\/strong>Multiply every term of both sides of the equation by a multiple of 10 (10, 100, 1000, etc.) to clear the decimals (based on the number with the largest number of decimal places in the equation).\r\n\r\n\r\n\r\n
                Steps<\/strong><\/td>\r\nExample<\/strong><\/td>\r\n<\/tr>\r\n<\/thead>\r\n<\/table>\r\n\r\n\r\n\r\n\r\n\r\n\r\n
                <\/td>\r\nSolve\u00a0 \u00a0 $\\bf0.34x -\u00a00.12 = -4.26x$<\/td>\r\n<\/td>\r\n<\/tr>\r\n
                \r\n
                  \r\n \t
                • Multiply each term by 100 to clear the decimal<\/li>\r\n<\/ul>\r\n<\/td>\r\n
                		${\\bf100}(0.34x) - {\\bf100}(0.12) = {\\bf100}(-4.26x)$<\/td>\r\nThe largest number of decimal place is two.<\/td>\r\n<\/tr>\r\n
                \r\n
                  \r\n \t
                • Collect the variable terms on one side of the equation and the constants on the other side.<\/li>\r\n<\/ul>\r\n<\/td>\r\n
                		$34x-12 = -426x$\r\n\r\n$34x + 426x = 12$\r\n\r\n$460x = 12$<\/td>\r\n \r\n

                Add 12 to both sides.<\/p>\r\n

                Add 426x<\/em> to both sides.<\/p>\r\n<\/td>\r\n<\/tr>\r\n

                \r\n
                  \r\n \t
                • Isolate the variable.<\/li>\r\n<\/ul>\r\n<\/td>\r\n
                		$x \\approx 0.026$<\/td>\r\n<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n \r\n
                \r\n

                Example 1.2.3<\/p>\r\n\r\n<\/header>\r\n

                \r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
                Solve\u00a0\u00a0 0.4y <\/em>+ 0.08 = 0.016<\/strong><\/td>\r\nThe largest number of decimal place is three.<\/td>\r\n<\/tr>\r\n
                1000<\/strong>(0.4y<\/em>) + 1000<\/strong>(0.08) = 1000<\/strong>(0.016)<\/td>\r\nMultiply each term by 1000.<\/td>\r\n<\/tr>\r\n
                400y <\/em>+ 80 = 16<\/td>\r\nCombine like terms.<\/td>\r\n<\/tr>\r\n
                400y<\/em> = -64<\/td>\r\nDivide both sides by 400.<\/td>\r\n<\/tr>\r\n
                y<\/em> = - 0.16<\/td>\r\n<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n \r\n\r\n<\/div>\r\n<\/div>\r\n \r\n\r\nEquations involving fractions:<\/strong>\r\n\r\n\r\n\r\n
                Steps<\/strong><\/td>\r\nExample<\/strong><\/td>\r\n<\/tr>\r\n<\/thead>\r\n<\/table>\r\n\r\n\r\n\r\n\r\n\r\n\r\n
                <\/td>\r\nSolve\u00a0 \u00a0 $\\bf\\frac{t}{3}+\\frac{3}{4}=-\\frac{t}{2}-\\frac{1}{3}$<\/td>\r\n<\/td>\r\n<\/tr>\r\n
                \r\n
                  \r\n \t
                • Multiply each term by the lowest common denominator (LCD).<\/li>\r\n<\/ul>\r\n<\/td>\r\n
                		${\\bf12} \\cdot \\frac{t}{3}+{\\bf12} \\cdot \\frac{3}{4}={\\bf12}(-\\frac{t}{2})-{\\bf12} \\cdot \\frac{1}{3}$<\/td>\r\n\r\n

                <\/p>\r\n<\/td>\r\n<\/tr>\r\n

                \r\n
                  \r\n \t
                • Collect the variable terms on one side of the equation and the constants on the other side.<\/li>\r\n<\/ul>\r\n<\/td>\r\n
                		$4t+9=-6t-4$\r\n\r\n$10t=-13$<\/td>\r\n\r\n

                Add 6t<\/em> to both sides.<\/p>\r\n

                Subtract 9 from both sides.<\/p>\r\n<\/td>\r\n<\/tr>\r\n

                \r\n
                  \r\n \t
                • Isolate the variable.<\/li>\r\n<\/ul>\r\n<\/td>\r\n
                		$t=\\frac{-13}{10}=-1\\frac{3}{10}$<\/td>\r\nDivide both sides by 10.<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n \r\n
                \r\n

                Word Problems<\/strong><\/h2>\r\n<\/div>\r\nIdentifying keywords:<\/strong>\r\n
                  \r\n \t
                • When trying to figure out the correct operation ( + , \u2013 , \u00d7 <\/strong>, \u00f7 , etc.) in a word problem it is important to pay attention to keywords (clues to what the problem is asking).<\/li>\r\n \t
                • Identifying keywords and pulling out relevant information that appear in the word problem are effective ways for solving mathematical\u00a0word problems.<\/li>\r\n<\/ul>\r\nKey or clue words in word problems:<\/strong>\r\n\r\n\r\n\r\n
                  Addition\u00a0 (+)<\/strong><\/td>\r\nSubtraction (<\/strong>-)<\/strong><\/td>\r\nMultiplication (\u00d7)<\/strong><\/td>\r\nDivision \u00a0(\u00f7)<\/strong><\/td>\r\nEquals to\u00a0 (=)<\/strong><\/td>\r\n<\/tr>\r\n<\/thead>\r\n<\/table>\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
                  add<\/td>\r\nsubtract<\/td>\r\ntimes<\/td>\r\ndivided by<\/td>\r\nequals<\/td>\r\n<\/tr>\r\n
                  sum (of)<\/td>\r\ndifference<\/td>\r\nproduct<\/td>\r\nquotient<\/td>\r\nis<\/td>\r\n<\/tr>\r\n
                  plus<\/td>\r\ntake away<\/td>\r\nmultiplied by<\/td>\r\nover<\/td>\r\nwas<\/td>\r\n<\/tr>\r\n
                  total (of)<\/td>\r\nminus<\/td>\r\ndouble<\/td>\r\nsplit up<\/td>\r\nare<\/td>\r\n<\/tr>\r\n
                  altogether<\/td>\r\nless (than)<\/td>\r\ntwice<\/td>\r\nfit into<\/td>\r\nwere<\/td>\r\n<\/tr>\r\n
                  increased by<\/td>\r\ndecreased by<\/td>\r\ntriple<\/td>\r\nper<\/td>\r\namounts to<\/td>\r\n<\/tr>\r\n
                  gain (of)<\/td>\r\nloss (of)<\/td>\r\nof<\/td>\r\neach<\/td>\r\ntotals<\/td>\r\n<\/tr>\r\n
                  combined<\/td>\r\n(amount) left<\/td>\r\nhow much (total)<\/td>\r\ngoes into<\/td>\r\nresults in<\/td>\r\n<\/tr>\r\n
                  in all<\/td>\r\nsavings<\/td>\r\nhow many<\/td>\r\nas much as<\/td>\r\nthe same as<\/td>\r\n<\/tr>\r\n
                  greater than<\/td>\r\nwithdraw<\/td>\r\n<\/td>\r\nout of<\/td>\r\ngives<\/td>\r\n<\/tr>\r\n
                  complete<\/td>\r\nreduced by<\/td>\r\n<\/td>\r\nratio\/rate<\/td>\r\nyields<\/td>\r\n<\/tr>\r\n
                  together<\/td>\r\nfewer (than)<\/td>\r\n<\/td>\r\npercent<\/td>\r\n<\/td>\r\n<\/tr>\r\n
                  more (than)<\/td>\r\nhow much more<\/td>\r\n<\/td>\r\nshare<\/td>\r\n<\/td>\r\n<\/tr>\r\n
                  additional<\/td>\r\nhow long<\/td>\r\n<\/td>\r\naverage<\/td>\r\n<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n
                  <\/div>\r\n
                  \r\n

                  Example 1.2.4<\/p>\r\n\r\n<\/header>\r\n

                  \r\n

                  1) <\/strong>Edward drove from Prince George to Williams Lake (235 km), then to Cache Creek (203 km) and finally to Vancouver (390 km). How many kilometres in total<\/em><\/strong> did Edward drive?<\/p>\r\n\r\n\r\n\r\n\r\n
                  235km + 203 km + 390 km = 828 km<\/td>\r\nThe key word: total (+)<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n

                  2) <\/strong>Emma had \\$150 in her purse on Friday. She bought a pizza for \\$15, and a pair of shoes for \\$35. How much money does she have left<\/em><\/strong>?<\/p>\r\n\r\n\r\n\r\n\r\n
                  \\$150 \u2013 15 \u2013<\/strong> 35 = \\$100<\/td>\r\nThe key word: left (\u2013)<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n

                  3) <\/strong>Lucy received \\$950 per month of rent from Mark for the months September to November. How much<\/em><\/strong> rent in total<\/em><\/strong> did she receive?<\/p>\r\n\r\n\r\n\r\n\r\n
                  \\$950 $\\cdot$ 3 = \\$2850<\/td>\r\nThe key word: how much total (\u00d7)<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n

                  4) <\/strong>Julia is going to buy a \\$7500 used car from her uncle.\u00a0 She promises to pay \\$500 per<\/em><\/strong> month. In how many months can she pay it off?<\/p>\r\n\r\n\r\n\r\n\r\n
                  \\$7500 \u00f7<\/strong> \\$500 = 15 months<\/td>\r\nThe key word: per (\u00f7)<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/div>\r\n<\/div>\r\nSteps for solving word problems:\u00a0<\/strong>\r\n\r\n\r\n\r\n
                  -\u00a0\u00a0\u00a0\u00a0\u00a0 Organize<\/strong> the facts<\/strong> given from the problem (create a table<\/strong> or diagram<\/strong> if it will make the problem clearer)\r\n\r\n-\u00a0\u00a0\u00a0\u00a0\u00a0 Identify and label the unknown quantity (let x <\/em>= unknown<\/strong>).\r\n\r\n-\u00a0\u00a0\u00a0\u00a0\u00a0 Convert words into mathematical symbols, and determine the operation<\/strong> \u2013 write an equation<\/strong> (looking for \u2018key\u2019 or \u2018clue\u2019 words).\r\n\r\n-\u00a0\u00a0\u00a0\u00a0\u00a0 Estimate <\/strong>and solve<\/strong> the equation and find the solution(s).\r\n\r\n-\u00a0\u00a0\u00a0\u00a0\u00a0 Check<\/strong> and state the answer<\/strong>. (Check the solution to the equation and check it back into the problem \u2013 is it logical?)<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n \r\n
                  \r\n

                  Example 1.2.5<\/p>\r\n\r\n<\/header>\r\n

                  \r\n\r\nWilliam bought 5 pairs of socks for \\$4.35 each. The cashier charged him an additional \\$2.15 in sales tax. He left the store with a measly \\$5.15. How much money did William start with?\r\n
                    \r\n \t
                  • Organize the facts (make a table):<\/li>\r\n<\/ul>\r\n\r\n\r\n\r\n\r\n\r\n
                    5 socks<\/td>\r\n\\$4.35 each<\/td>\r\n<\/tr>\r\n
                    Sales tax<\/td>\r\n\\$2.15<\/td>\r\n<\/tr>\r\n
                    Money left<\/td>\r\n\\$5.15<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n
                      \r\n \t
                    • Determine the unknown: How much did William start with?\u00a0 (x <\/em>= ?)<\/strong><\/li>\r\n \t
                    • Convert words into math symbols, and determine the operation (find keywords<\/em><\/strong>):<\/li>\r\n<\/ul>\r\n\r\n\r\n\r\n\r\n\r\n
                      \r\n
                        \r\n \t
                      • The total<\/em><\/strong> cost without the sales tax:<\/span><\/li>\r\n<\/ul>\r\n<\/td>\r\n
                      		\r\n

                      \\$4.35 \u00d7<\/strong> 5<\/p>\r\n<\/td>\r\n<\/tr>\r\n

                      \r\n
                        \r\n \t
                      • With an additional<\/em><\/strong> \\$2.15 sales tax:<\/span><\/li>\r\n<\/ul>\r\n<\/td>\r\n
                      		(\\$4.35 \u00d7 5) + \\$2.15<\/td>\r\n<\/tr>\r\n
                      \r\n
                        \r\n \t
                      • Estimate and solve the unknown:<\/li>\r\n<\/ul>\r\n<\/td>\r\n
                      		x <\/em>= [(\\$4.35 \u00d7 5) + \\$2.15] + \\$5.15<\/strong><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n
                        \r\n \t
                      • Estimate and solve the unknown:<\/li>\r\n<\/ul>\r\n\r\n\r\n\r\n\r\n\r\n\r\n
                        \r\n
                          \r\n \t
                        • Estimate:<\/li>\r\n<\/ul>\r\n<\/td>\r\n
                        		x<\/em> = [(\\$4 \u00d7 5) + \\$2] + \\$5<\/td>\r\n<\/tr>\r\n
                        <\/td>\r\n= \\$27<\/td>\r\n<\/tr>\r\n
                        \r\n
                          \r\n \t
                        • Actual solution:<\/li>\r\n<\/ul>\r\n<\/td>\r\n
                        		x<\/em> = [(\\$4.35 \u00d7 5) + \\$2.15] + \\$5.15<\/td>\r\n<\/tr>\r\n
                        <\/td>\r\n= \\$29.05<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n
                          \r\n \t
                        • \u00a0Check: If William started with \\$29.05, and subtract 5 socks for \\$4.35 each and sales tax in \\$2.15 to see if it equals \\$5.15.<\/li>\r\n \t
                        • <\/li>\r\n<\/ul>\r\n\r\n\r\n\r\n\r\n
                          \\$29.05 - [(\\$4.35 \u00d7 5) + \\$2.15] $\\overset{?}{=}$ \\$5.15<\/td>\r\n<\/td>\r\n<\/tr>\r\n
                          \\$29.05 \u2013 \\$23.9 $\\overset{\\checkmark}{=}$ \\$5.15<\/td>\r\nCorrect!<\/strong><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n \r\n\r\n<\/div>\r\n<\/div>\r\n \r\n\r\nMore examples:<\/strong>\r\n
                          \r\n

                          Example 1.2.6<\/p>\r\n\r\n<\/header>\r\n

                          \r\n\r\nJames had 96 toys. He sold 13 on first day, 32 on second day, 21 on third day, 14 on fourth day and 7 on the last day. What percentage of the toys were not sold?\r\n
                            \r\n \t
                          • Organize the facts:<\/li>\r\n<\/ul>\r\n\r\n\r\n\r\n\r\n\r\n
                            James had<\/td>\r\n96 toys<\/td>\r\n<\/tr>\r\n
                            The total number of toys sold<\/td>\r\n13 + 32 + 21 + 14 + 7<\/td>\r\n<\/tr>\r\n
                            The toys not sold<\/td>\r\n96 \u2013 the total number of toys sold<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n\r\n\r\n\r\n\r\n\r\n\r\n
                            \r\n
                              \r\n \t
                            • Determine the unknown:<\/li>\r\n<\/ul>\r\n<\/td>\r\n
                            		Let x<\/em> = percentage of the toys were not sold<\/td>\r\n<\/tr>\r\n
                            \r\n
                              \r\n \t
                            • The total number of toys sold:<\/li>\r\n<\/ul>\r\n<\/td>\r\n
                            		13 + 32 + 21 + 14 + 7 = 87<\/td>\r\n<\/tr>\r\n
                            \r\n
                              \r\n \t
                            • The toys not sold:<\/li>\r\n<\/ul>\r\n<\/td>\r\n
                            		96 \u2013 87 = 9<\/td>\r\n<\/tr>\r\n
                            \r\n
                              \r\n \t
                            • Percentage of the toys were not sold:<\/li>\r\n<\/ul>\r\n<\/td>\r\n
                            		x<\/em> = $\\frac{Toys\\,not\\,sold}{Total\\,number\\,of\\,toys}$ = $\\frac{9}{96} \\approx$ 0.094 = 9.4%<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/div>\r\n<\/div>\r\n \r\n
                            \r\n

                            Example 1.2.7<\/p>\r\n\r\n<\/header>\r\n

                            \r\n\r\nThe 60-litre gas tank in Robert\u2019s car is 1\/2 full. Kelowna is about 390 km from Vancouver and his car averages 7 litres per 100 km. Can Robert make his trip to Vancouver?\r\n
                              \r\n \t
                            • Let x<\/em> = litres of fuel are required to get to Vancouver.<\/li>\r\n \t
                            • The 60-litre gas tank in Robert\u2019s car is 1\/2 full:<\/li>\r\n<\/ul>\r\n\r\n\r\n\r\n
                              60 L \u00d7 $\\frac{1}{2}$ = 30 L<\/td>\r\nRobert has 30 litres gas in his car.<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n
                                \r\n \t
                              • Robert\u2019s car averages 7 litres per 100 km, and Vancouver is about 390 km from Kelowna.<\/li>\r\n<\/ul>\r\n\r\n\r\n\r\n\r\n\r\n
                                $\\frac{7L}{100 km} = \\frac{x}{390 km}$<\/td>\r\nProportion: $\\frac{a}{b} = \\frac{c}{d}$<\/td>\r\n<\/tr>\r\n
                                (x<\/em>)(100km) = (7 L) (390 km)<\/td>\r\nCross multiply and solve for x<\/em>.<\/td>\r\n<\/tr>\r\n
                                x\u00a0<\/em>= $\\frac{(7 L)(390 km)}{100 km}$ = 27.3 L<\/td>\r\nRobert needs 27.3 litres gas to get to Vancouver.<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n
                                  \r\n \t
                                • 30L > 27.3L. Therefore, yes, Robert can make his trip.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<\/div>\r\n
                                  \r\n

                                  Practice questions<\/strong><\/h2>\r\n<\/div>\r\n1.<\/strong> Solve the following equations:\r\n

                                  a. <\/strong>x<\/em> \u2013\u00a07 = 12<\/p>\r\n

                                  b. <\/strong>y<\/em> + $\\frac{3}{8}$ = $\\frac{5}{8}$<\/p>\r\n

                                  c. <\/strong>$\\frac{3x}{2}$ = $\\frac{9}{16}$<\/p>\r\n

                                  d. <\/strong>14t<\/em> + 5 = 8<\/p>\r\n

                                  e. <\/strong>7(x <\/em>\u2013 3) + 3x<\/em> \u2013 5 = 2(5 \u2013 4x<\/em>)<\/p>\r\n

                                  f. <\/strong>$\\frac{1}{7}$(y<\/em> + 12) = 4y<\/em> \u2013 $\\frac{3}{7}$y<\/em><\/p>\r\n

                                  g. <\/strong>0.5t <\/em>+ 0.05 = 0.025<\/p>\r\n

                                  h.<\/strong>\u00a0$\\frac{x}{4}+\\frac{2}{5}=-\\frac{x}{2}-\\frac{1}{5}$<\/p>\r\n2.<\/strong> Write an expression for each of the following:\r\n

                                  a. <\/strong>Susan has \\$375 in her checking account. If she makes a deposit of y<\/em> dollars, how much in total will she have in her account?<\/p>\r\n

                                  b. <\/strong>Mark weighs 175 pounds. If he loses y<\/em> pounds, how much will he weigh?<\/p>\r\n

                                  c. <\/strong>A piece of wire 45 metres long was cut in two pieces and one piece is w<\/em> metres long. How long is the other piece?<\/p>\r\n

                                  d. <\/strong>Emily made 4 dozen muffins. If it cost her x<\/em> dollars, what was her cost per dozen muffins? What was her cost per muffin?<\/p>\r\n ","rendered":"

                                  <\/div>\n

                                  Equations<\/h1>\n

                                  Equation<\/strong>: a mathematical sentence that contains two expressions and is separated by an equal sign (both sides of the equation have the same value).<\/p>\n

                                  Example:\u00a0\u00a0 <\/strong>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a04 + 3 = 7,\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 9x<\/em> \u2013 4 = 5, \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a02y<\/em> \u2013 \"\frac{1}{3}\" = y<\/em><\/div>\n

                                   <\/p>\n

                                  To <\/strong>solve<\/strong> an equation<\/strong> we find a particular value for the variable in the equation that makes the equation true (left side = right side).<\/p>\n

                                  \n

                                  Example:<\/strong>\u00a0 \u00a0For the equation \u00a0\u00a0x<\/em> + 4 = 5<\/p>\n

                                  only x<\/em> = 1 can make it true, since \u00a0\u00a01 + 4 = 5 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0(Left side = Right side)<\/p>\n<\/div>\n

                                   <\/p>\n

                                  Solution of an equation<\/strong>: the value of the variable in the equation that makes the equation true.<\/p>\n

                                  Example:<\/strong>\u00a0 \u00a0\u00a0For the equation\u00a0 \u00a0x<\/em> + 4 = 5,\u00a0\u00a0 \u00a0\u00a0x<\/em> = 1\u00a0 \u00a0is the solution.<\/div>\n

                                   <\/p>\n

                                  \n
                                  \n

                                  Example 1.2.1<\/p>\n<\/header>\n

                                  \n

                                  Indicate whether each of the given number is a solution to the given equation.<\/p>\n\n\n\n
                                  1)<\/strong>\u00a0\u00a0\u00a0\u00a0\u00a0 2 :<\/strong> \u00a0\u00a0\u00a0\u00a0\u00a04x<\/em> \u2013 3 = 5<\/td>\n4 \u2219 2 \u2013 3 \"\overset{?}{=}\" 5<\/td>\n5 \"\overset{\checkmark}{=}\" 5<\/td>\nYes<\/td>\nReplace x<\/em> with 2.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n\n\n\n
                                  2)<\/strong>\u00a0\u00a0\u00a0\u00a0\u00a0 15 :<\/strong>\u00a0 \u00a0 \"\frac{-3}{15}\"y<\/em> = -3<\/td>\n\"\frac{-3}{15}(15) \overset{?}{=}\" -3<\/td>\n-3 \"\overset{\checkmark}{=}\" -3<\/td>\nYes<\/td>\nReplace y<\/em> with 15.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n\n\n\n
                                  3)<\/strong> \u00a0\u00a0\u00a0\u00a0 \"\bf\frac{1}{2}\"\u00a0:<\/strong>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 8t<\/em> = 3<\/td>\n8 (\"\frac{1}{2}\") \"\overset{?}{=}\" 3<\/td>\n4 \u2260 3<\/td>\nNo<\/td>\nCannot replace t<\/em> with \"\frac{1}{2}\".<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<\/div>\n
                                  \n

                                  Solving Equations<\/strong><\/h2>\n<\/div>\n
                                  \n

                                  Basic rules for solving one-step equations<\/strong>:<\/span><\/p>\n<\/div>\n

                                    \n
                                  • Add, subtract, multiply or divide the same quantity to both sides of an equation to obtain a valid equation.<\/li>\n
                                  • Remember to always do the same thing to both sides of the equation (balance).<\/li>\n<\/ul>\n

                                    Properties for solving equations:<\/strong><\/p>\n\n\n\n
                                    Properties<\/strong><\/td>\nEquality<\/strong><\/td>\nExample<\/strong><\/td>\n<\/tr>\n<\/thead>\n<\/table>\n\n\n\n\n\n\n
                                    Addition<\/strong> property of equality<\/td>\nA =<\/strong> B<\/em>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 A + C =<\/strong> B<\/em> + C<\/em><\/strong><\/td>\nSolve \"x-6=3\"<\/p>\n

                                    \"x-\bcancel{6}+\bcancel{\bf6}=3+\bf6\"\u00a0 \u00a0 \u00a0x <\/em>= 9<\/strong><\/td>\n<\/tr>\n

                                    Subtraction<\/strong> property of equality<\/td>\nA =<\/strong> B<\/em>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 A \u2013<\/em>\u00a0C =<\/em><\/strong> B<\/em> \u2013\u00a0C<\/em><\/strong><\/td>\nSolve \"y+5=-8\"<\/p>\n

                                    \"y+\bcancel{5}-\bcancel{\bf5}=-8-\bf5\"\u00a0 \u00a0 \u00a0y<\/em> = -13<\/strong><\/td>\n<\/tr>\n

                                    Multiplication<\/strong> property of equality<\/td>\nA =<\/strong> B<\/em>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 A \u00b7<\/em>\u00a0C =<\/em><\/strong> B<\/em> \u00b7\u00a0C<\/em><\/strong><\/td>\nSolve \"\frac{m}{9}=2\"<\/p>\n

                                    \"\bcancel{\bf9} \cdot \frac{m}{\bcancel{9}}=2 \cdot \bf9\"\u00a0 \u00a0 \u00a0m = 18<\/strong><\/td>\n<\/tr>\n

                                    Division <\/strong>property of equality<\/td>\nA =<\/strong> B<\/em>\u00a0 \u00a0 \u00a0 \u00a0 \"\frac{A}{C}=\frac{B}{C}\"\u00a0 \u00a0 \u00a0(C<\/em> \u2260 0)<\/td>\nSolve \"3n=-15\"<\/p>\n

                                    \"\frac{\bcancel{3}n}{\bcancel{\bf3}}=\frac{-15}{\bf3}\"\u00a0 \u00a0 \u00a0n = –5<\/strong><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

                                     <\/p>\n

                                    \n
                                    \n

                                    Example 1.2.2<\/p>\n<\/header>\n

                                    \n

                                    Solve the following equations.<\/p>\n\n\n\n\n\n\n
                                    1) <\/strong>\"-9 + x = 5\"<\/td>\n\"-\bcancel{9} + x+ \bcancel{\bf9} = 5 + \bf9\"<\/td>\nProperty of addition.<\/td>\n<\/tr>\n
                                    <\/td>\n\"x = 14\"<\/td>\n<\/td>\n<\/tr>\n
                                    <\/td>\n<\/td>\n<\/td>\n<\/tr>\n
                                    Check:<\/td>\n\"-9 + {\bf14} \overset{?}{=} 5\"\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \"5 \overset{\checkmark}{=} 5\"<\/td>\nReplace x with 14.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

                                     <\/p>\n\n\n\n\n
                                    2) <\/strong>\"t+\frac{2}{5}=-\frac{1}{5}\"<\/td>\n\"y+\frac{2}{5}-\frac{\bf2}{\bf5}=-\frac{1}{5}-\frac{\bf2}{\bf5}\"<\/td>\nProperty of subtraction.<\/td>\n<\/tr>\n
                                    <\/td>\n\"y=-\frac{3}{5}\"<\/td>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

                                     <\/p>\n\n\n\n\n
                                    3) <\/strong>\u00a0\"\frac{-1}{6}x = 7\"<\/td>\n\"{\bf-6} \cdot \frac{-1}{6}x=7(\bf-6)\"<\/td>\nProperty of multiplication.<\/td>\n<\/tr>\n
                                    <\/td>\n\"x = -42\"<\/td>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

                                     <\/p>\n\n\n\n\n
                                    4<\/strong>) <\/strong><\/em>\"-5x = 30\"<\/td>\n\"\frac{-5x}{\bf-5}=\frac{30}{\bf-5}\"<\/td>\nProperty of division.<\/td>\n<\/tr>\n
                                    <\/td>\n\"x = -6\"<\/td>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

                                     <\/p>\n<\/div>\n<\/div>\n

                                    <\/h1>\n

                                    Multi-step equation<\/strong>: an equation that requires more than one step to solve it.<\/p>\n

                                    Procedure for solving multi-step equations:<\/strong><\/p>\n\n\n\n
                                    \n
                                      \n
                                    • Clear the fractions or decimals if necessary.<\/li>\n<\/ul>\n
                                        \n
                                      • Simplify and remove parentheses if necessary.<\/li>\n<\/ul>\n
                                          \n
                                        • Combine like terms on each side of the equation.<\/li>\n<\/ul>\n
                                            \n
                                          • Collect the variable terms on one side of the equation and the constants on the other side.<\/li>\n<\/ul>\n
                                              \n
                                            • Isolate the variable (to get the\u00a0variable alone on one side\u00a0of the equation).<\/li>\n<\/ul>\n
                                                \n
                                              • Check the solution with the original equation.<\/li>\n<\/ul>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

                                                 <\/p>\n\n\n\n
                                                Steps<\/strong><\/td>\nExample<\/strong><\/td>\n<\/tr>\n<\/thead>\n<\/table>\n\n\n\n\n\n\n\n\n\n
                                                <\/td>\nSolve \"\bf\frac{1}{5}(y+10)=3y-\frac{9}{5}y\"<\/td>\n<\/td>\n<\/tr>\n
                                                \n
                                                  \n
                                                • Eliminate the denominators if the equation has fractions.<\/li>\n<\/ul>\n<\/td>\n
                                                		\"\bcancel{5}\cdot \frac{1}{\bcancel5}(y+10)={\bf5}(3y)-{\bf\bcancel5}(\frac{9}{\bcancel5}y)\"<\/td>\nMultiply each term by 5.<\/td>\n<\/tr>\n
                                                \n
                                                  \n
                                                • Remove parentheses.<\/li>\n<\/ul>\n<\/td>\n
                                                		\"y+10=15y-9y\"<\/td>\n<\/td>\n<\/tr>\n
                                                \n
                                                  \n
                                                • Combine like terms.<\/li>\n<\/ul>\n<\/td>\n
                                                		\"y+10=6y\"<\/td>\n<\/td>\n<\/tr>\n
                                                \n
                                                  \n
                                                • Collect variable terms on one side and the constants on the other side.<\/li>\n<\/ul>\n<\/td>\n
                                                		\"y+\bcancel{10}-\bcancel{\bf10}=6y-\bf10\"<\/p>\n

                                                \"y=6y-\bf10\"<\/p>\n

                                                \"y-{\bf6y}=\bcancel{6y}-10-\bcancel{\bf6y}\"<\/td>\n

                                                		\n

                                                Subtract 10 from both sides.<\/p>\n

                                                 <\/p>\n

                                                Subtract 6y<\/em> from both sides.<\/p>\n<\/td>\n<\/tr>\n

                                                \n
                                                  \n
                                                • Isolate the variable.<\/li>\n<\/ul>\n<\/td>\n
                                                		\"-5y=-10\"<\/p>\n

                                                \"y=\frac{-10}{-5}\"<\/p>\n

                                                \"y=2\"<\/td>\n

                                                		Divide both sides by -5.<\/td>\n<\/tr>\n
                                                \n
                                                  \n
                                                • Check with the original equation.<\/li>\n<\/ul>\n<\/td>\n
                                                		\"\frac{1}{5}(2+10)=3 \cdot 2-\frac{9}{5} \cdot 2\"<\/p>\n

                                                \"\bcancel{5}\frac{1}{\bcancel5}(2+10)=5 \cdot 3 \cdot 2-\bcancel{5} \cdot \frac{9}{\bcancel5} \cdot 2\"<\/p>\n

                                                \"(2+10)=30-18\"<\/p>\n

                                                \"12=12\"<\/td>\n

                                                		Replace y<\/em> with 2.<\/p>\n

                                                Multiply each term by 5.<\/p>\n

                                                 <\/p>\n

                                                LS = RS\u00a0 (correct)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

                                                 <\/p>\n

                                                Equations involving decimals: <\/strong>Multiply every term of both sides of the equation by a multiple of 10 (10, 100, 1000, etc.) to clear the decimals (based on the number with the largest number of decimal places in the equation).<\/p>\n\n\n\n
                                                Steps<\/strong><\/td>\nExample<\/strong><\/td>\n<\/tr>\n<\/thead>\n<\/table>\n\n\n\n\n\n\n
                                                <\/td>\nSolve\u00a0 \u00a0 \"\bf0.34x - 0.12 = -4.26x\"<\/td>\n<\/td>\n<\/tr>\n
                                                \n
                                                  \n
                                                • Multiply each term by 100 to clear the decimal<\/li>\n<\/ul>\n<\/td>\n
                                                		\"{\bf100}(0.34x) - {\bf100}(0.12) = {\bf100}(-4.26x)\"<\/td>\nThe largest number of decimal place is two.<\/td>\n<\/tr>\n
                                                \n
                                                  \n
                                                • Collect the variable terms on one side of the equation and the constants on the other side.<\/li>\n<\/ul>\n<\/td>\n
                                                		\"34x-12 = -426x\"<\/p>\n

                                                \"34x + 426x = 12\"<\/p>\n

                                                \"460x = 12\"<\/td>\n

                                                		 <\/p>\n

                                                Add 12 to both sides.<\/p>\n

                                                Add 426x<\/em> to both sides.<\/p>\n<\/td>\n<\/tr>\n

                                                \n
                                                  \n
                                                • Isolate the variable.<\/li>\n<\/ul>\n<\/td>\n
                                                		\"x \approx 0.026\"<\/td>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

                                                 <\/p>\n

                                                \n
                                                \n

                                                Example 1.2.3<\/p>\n<\/header>\n

                                                \n\n\n\n\n\n\n\n
                                                Solve\u00a0\u00a0 0.4y <\/em>+ 0.08 = 0.016<\/strong><\/td>\nThe largest number of decimal place is three.<\/td>\n<\/tr>\n
                                                1000<\/strong>(0.4y<\/em>) + 1000<\/strong>(0.08) = 1000<\/strong>(0.016)<\/td>\nMultiply each term by 1000.<\/td>\n<\/tr>\n
                                                400y <\/em>+ 80 = 16<\/td>\nCombine like terms.<\/td>\n<\/tr>\n
                                                400y<\/em> = -64<\/td>\nDivide both sides by 400.<\/td>\n<\/tr>\n
                                                y<\/em> = – 0.16<\/td>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

                                                 <\/p>\n<\/div>\n<\/div>\n

                                                 <\/p>\n

                                                Equations involving fractions:<\/strong><\/p>\n\n\n\n
                                                Steps<\/strong><\/td>\nExample<\/strong><\/td>\n<\/tr>\n<\/thead>\n<\/table>\n\n\n\n\n\n\n
                                                <\/td>\nSolve\u00a0 \u00a0 \"\bf\frac{t}{3}+\frac{3}{4}=-\frac{t}{2}-\frac{1}{3}\"<\/td>\n<\/td>\n<\/tr>\n
                                                \n
                                                  \n
                                                • Multiply each term by the lowest common denominator (LCD).<\/li>\n<\/ul>\n<\/td>\n
                                                		\"{\bf12} \cdot \frac{t}{3}+{\bf12} \cdot \frac{3}{4}={\bf12}(-\frac{t}{2})-{\bf12} \cdot \frac{1}{3}\"<\/td>\n\n

                                                \n<\/td>\n<\/tr>\n

                                                \n
                                                  \n
                                                • Collect the variable terms on one side of the equation and the constants on the other side.<\/li>\n<\/ul>\n<\/td>\n
                                                		\"4t+9=-6t-4\"<\/p>\n

                                                \"10t=-13\"<\/td>\n

                                                		\n

                                                Add 6t<\/em> to both sides.<\/p>\n

                                                Subtract 9 from both sides.<\/p>\n<\/td>\n<\/tr>\n

                                                \n
                                                  \n
                                                • Isolate the variable.<\/li>\n<\/ul>\n<\/td>\n
                                                		\"t=\frac{-13}{10}=-1\frac{3}{10}\"<\/td>\nDivide both sides by 10.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

                                                 <\/p>\n

                                                \n

                                                Word Problems<\/strong><\/h2>\n<\/div>\n

                                                Identifying keywords:<\/strong><\/p>\n

                                                  \n
                                                • When trying to figure out the correct operation ( + , \u2013 , \u00d7 <\/strong>, \u00f7 , etc.) in a word problem it is important to pay attention to keywords (clues to what the problem is asking).<\/li>\n
                                                • Identifying keywords and pulling out relevant information that appear in the word problem are effective ways for solving mathematical\u00a0word problems.<\/li>\n<\/ul>\n

                                                  Key or clue words in word problems:<\/strong><\/p>\n\n\n\n
                                                  Addition\u00a0 (+)<\/strong><\/td>\nSubtraction (<\/strong>–)<\/strong><\/td>\nMultiplication (\u00d7)<\/strong><\/td>\nDivision \u00a0(\u00f7)<\/strong><\/td>\nEquals to\u00a0 (=)<\/strong><\/td>\n<\/tr>\n<\/thead>\n<\/table>\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
                                                  add<\/td>\nsubtract<\/td>\ntimes<\/td>\ndivided by<\/td>\nequals<\/td>\n<\/tr>\n
                                                  sum (of)<\/td>\ndifference<\/td>\nproduct<\/td>\nquotient<\/td>\nis<\/td>\n<\/tr>\n
                                                  plus<\/td>\ntake away<\/td>\nmultiplied by<\/td>\nover<\/td>\nwas<\/td>\n<\/tr>\n
                                                  total (of)<\/td>\nminus<\/td>\ndouble<\/td>\nsplit up<\/td>\nare<\/td>\n<\/tr>\n
                                                  altogether<\/td>\nless (than)<\/td>\ntwice<\/td>\nfit into<\/td>\nwere<\/td>\n<\/tr>\n
                                                  increased by<\/td>\ndecreased by<\/td>\ntriple<\/td>\nper<\/td>\namounts to<\/td>\n<\/tr>\n
                                                  gain (of)<\/td>\nloss (of)<\/td>\nof<\/td>\neach<\/td>\ntotals<\/td>\n<\/tr>\n
                                                  combined<\/td>\n(amount) left<\/td>\nhow much (total)<\/td>\ngoes into<\/td>\nresults in<\/td>\n<\/tr>\n
                                                  in all<\/td>\nsavings<\/td>\nhow many<\/td>\nas much as<\/td>\nthe same as<\/td>\n<\/tr>\n
                                                  greater than<\/td>\nwithdraw<\/td>\n<\/td>\nout of<\/td>\ngives<\/td>\n<\/tr>\n
                                                  complete<\/td>\nreduced by<\/td>\n<\/td>\nratio\/rate<\/td>\nyields<\/td>\n<\/tr>\n
                                                  together<\/td>\nfewer (than)<\/td>\n<\/td>\npercent<\/td>\n<\/td>\n<\/tr>\n
                                                  more (than)<\/td>\nhow much more<\/td>\n<\/td>\nshare<\/td>\n<\/td>\n<\/tr>\n
                                                  additional<\/td>\nhow long<\/td>\n<\/td>\naverage<\/td>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n
                                                  <\/div>\n
                                                  \n
                                                  \n

                                                  Example 1.2.4<\/p>\n<\/header>\n

                                                  \n

                                                  1) <\/strong>Edward drove from Prince George to Williams Lake (235 km), then to Cache Creek (203 km) and finally to Vancouver (390 km). How many kilometres in total<\/em><\/strong> did Edward drive?<\/p>\n\n\n\n
                                                  235km + 203 km + 390 km = 828 km<\/td>\nThe key word: total (+)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

                                                  2) <\/strong>Emma had $150 in her purse on Friday. She bought a pizza for $15, and a pair of shoes for $35. How much money does she have left<\/em><\/strong>?<\/p>\n\n\n\n
                                                  $150 \u2013 15 \u2013<\/strong> 35 = $100<\/td>\nThe key word: left (\u2013)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

                                                  3) <\/strong>Lucy received $950 per month of rent from Mark for the months September to November. How much<\/em><\/strong> rent in total<\/em><\/strong> did she receive?<\/p>\n\n\n\n
                                                  $950 \"\cdot\" 3 = $2850<\/td>\nThe key word: how much total (\u00d7)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

                                                  4) <\/strong>Julia is going to buy a $7500 used car from her uncle.\u00a0 She promises to pay $500 per<\/em><\/strong> month. In how many months can she pay it off?<\/p>\n\n\n\n
                                                  $7500 \u00f7<\/strong> $500 = 15 months<\/td>\nThe key word: per (\u00f7)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<\/div>\n

                                                  Steps for solving word problems:\u00a0<\/strong><\/p>\n\n\n\n
                                                  –\u00a0\u00a0\u00a0\u00a0\u00a0 Organize<\/strong> the facts<\/strong> given from the problem (create a table<\/strong> or diagram<\/strong> if it will make the problem clearer)<\/p>\n

                                                  –\u00a0\u00a0\u00a0\u00a0\u00a0 Identify and label the unknown quantity (let x <\/em>= unknown<\/strong>).<\/p>\n

                                                  –\u00a0\u00a0\u00a0\u00a0\u00a0 Convert words into mathematical symbols, and determine the operation<\/strong> \u2013 write an equation<\/strong> (looking for \u2018key\u2019 or \u2018clue\u2019 words).<\/p>\n

                                                  –\u00a0\u00a0\u00a0\u00a0\u00a0 Estimate <\/strong>and solve<\/strong> the equation and find the solution(s).<\/p>\n

                                                  –\u00a0\u00a0\u00a0\u00a0\u00a0 Check<\/strong> and state the answer<\/strong>. (Check the solution to the equation and check it back into the problem \u2013 is it logical?)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

                                                   <\/p>\n

                                                  \n
                                                  \n

                                                  Example 1.2.5<\/p>\n<\/header>\n

                                                  \n

                                                  William bought 5 pairs of socks for $4.35 each. The cashier charged him an additional $2.15 in sales tax. He left the store with a measly $5.15. How much money did William start with?<\/p>\n

                                                    \n
                                                  • Organize the facts (make a table):<\/li>\n<\/ul>\n\n\n\n\n\n
                                                    5 socks<\/td>\n$4.35 each<\/td>\n<\/tr>\n
                                                    Sales tax<\/td>\n$2.15<\/td>\n<\/tr>\n
                                                    Money left<\/td>\n$5.15<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n
                                                      \n
                                                    • Determine the unknown: How much did William start with?\u00a0 (x <\/em>= ?)<\/strong><\/li>\n
                                                    • Convert words into math symbols, and determine the operation (find keywords<\/em><\/strong>):<\/li>\n<\/ul>\n\n\n\n\n\n
                                                      \n
                                                        \n
                                                      • The total<\/em><\/strong> cost without the sales tax:<\/span><\/li>\n<\/ul>\n<\/td>\n
                                                      		\n

                                                      $4.35 \u00d7<\/strong> 5<\/p>\n<\/td>\n<\/tr>\n

                                                      \n
                                                        \n
                                                      • With an additional<\/em><\/strong> $2.15 sales tax:<\/span><\/li>\n<\/ul>\n<\/td>\n
                                                      		($4.35 \u00d7 5) + $2.15<\/td>\n<\/tr>\n
                                                      \n
                                                        \n
                                                      • Estimate and solve the unknown:<\/li>\n<\/ul>\n<\/td>\n
                                                      		x <\/em>= [($4.35 \u00d7 5) + $2.15] + $5.15<\/strong><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n
                                                        \n
                                                      • Estimate and solve the unknown:<\/li>\n<\/ul>\n\n\n\n\n\n\n
                                                        \n
                                                          \n
                                                        • Estimate:<\/li>\n<\/ul>\n<\/td>\n
                                                        		x<\/em> = [($4 \u00d7 5) + $2] + $5<\/td>\n<\/tr>\n
                                                        <\/td>\n= $27<\/td>\n<\/tr>\n
                                                        \n
                                                          \n
                                                        • Actual solution:<\/li>\n<\/ul>\n<\/td>\n
                                                        		x<\/em> = [($4.35 \u00d7 5) + $2.15] + $5.15<\/td>\n<\/tr>\n
                                                        <\/td>\n= $29.05<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n
                                                          \n
                                                        • \u00a0Check: If William started with $29.05, and subtract 5 socks for $4.35 each and sales tax in $2.15 to see if it equals $5.15.<\/li>\n
                                                        • <\/li>\n<\/ul>\n\n\n\n\n
                                                          $29.05 – [($4.35 \u00d7 5) + $2.15] \"\overset{?}{=}\" $5.15<\/td>\n<\/td>\n<\/tr>\n
                                                          $29.05 \u2013 $23.9 \"\overset{\checkmark}{=}\" $5.15<\/td>\nCorrect!<\/strong><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

                                                           <\/p>\n<\/div>\n<\/div>\n

                                                           <\/p>\n

                                                          More examples:<\/strong><\/p>\n

                                                          \n
                                                          \n

                                                          Example 1.2.6<\/p>\n<\/header>\n

                                                          \n

                                                          James had 96 toys. He sold 13 on first day, 32 on second day, 21 on third day, 14 on fourth day and 7 on the last day. What percentage of the toys were not sold?<\/p>\n

                                                            \n
                                                          • Organize the facts:<\/li>\n<\/ul>\n\n\n\n\n\n
                                                            James had<\/td>\n96 toys<\/td>\n<\/tr>\n
                                                            The total number of toys sold<\/td>\n13 + 32 + 21 + 14 + 7<\/td>\n<\/tr>\n
                                                            The toys not sold<\/td>\n96 \u2013 the total number of toys sold<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n\n\n\n\n\n\n
                                                            \n
                                                              \n
                                                            • Determine the unknown:<\/li>\n<\/ul>\n<\/td>\n
                                                            		Let x<\/em> = percentage of the toys were not sold<\/td>\n<\/tr>\n
                                                            \n
                                                              \n
                                                            • The total number of toys sold:<\/li>\n<\/ul>\n<\/td>\n
                                                            		13 + 32 + 21 + 14 + 7 = 87<\/td>\n<\/tr>\n
                                                            \n
                                                              \n
                                                            • The toys not sold:<\/li>\n<\/ul>\n<\/td>\n
                                                            		96 \u2013 87 = 9<\/td>\n<\/tr>\n
                                                            \n
                                                              \n
                                                            • Percentage of the toys were not sold:<\/li>\n<\/ul>\n<\/td>\n
                                                            		x<\/em> = \"\frac{Toys\,not\,sold}{Total\,number\,of\,toys}\" = \"\frac{9}{96} \approx\" 0.094 = 9.4%<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<\/div>\n

                                                             <\/p>\n

                                                            \n
                                                            \n

                                                            Example 1.2.7<\/p>\n<\/header>\n

                                                            \n

                                                            The 60-litre gas tank in Robert\u2019s car is 1\/2 full. Kelowna is about 390 km from Vancouver and his car averages 7 litres per 100 km. Can Robert make his trip to Vancouver?<\/p>\n

                                                              \n
                                                            • Let x<\/em> = litres of fuel are required to get to Vancouver.<\/li>\n
                                                            • The 60-litre gas tank in Robert\u2019s car is 1\/2 full:<\/li>\n<\/ul>\n\n\n\n
                                                              60 L \u00d7 \"\frac{1}{2}\" = 30 L<\/td>\nRobert has 30 litres gas in his car.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n
                                                                \n
                                                              • Robert\u2019s car averages 7 litres per 100 km, and Vancouver is about 390 km from Kelowna.<\/li>\n<\/ul>\n\n\n\n\n\n
                                                                \"\frac{7L}{100 km} = \frac{x}{390 km}\"<\/td>\nProportion: \"\frac{a}{b} = \frac{c}{d}\"<\/td>\n<\/tr>\n
                                                                (x<\/em>)(100km) = (7 L) (390 km)<\/td>\nCross multiply and solve for x<\/em>.<\/td>\n<\/tr>\n
                                                                x\u00a0<\/em>= \"\frac{(7 L)(390 km)}{100 km}\" = 27.3 L<\/td>\nRobert needs 27.3 litres gas to get to Vancouver.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n
                                                                  \n
                                                                • 30L > 27.3L. Therefore, yes, Robert can make his trip.<\/li>\n<\/ul>\n<\/div>\n<\/div>\n
                                                                  \n

                                                                  Practice questions<\/strong><\/h2>\n<\/div>\n

                                                                  1.<\/strong> Solve the following equations:<\/p>\n

                                                                  a. <\/strong>x<\/em> \u2013\u00a07 = 12<\/p>\n

                                                                  b. <\/strong>y<\/em> + \"\frac{3}{8}\" = \"\frac{5}{8}\"<\/p>\n

                                                                  c. <\/strong>\"\frac{3x}{2}\" = \"\frac{9}{16}\"<\/p>\n

                                                                  d. <\/strong>14t<\/em> + 5 = 8<\/p>\n

                                                                  e. <\/strong>7(x <\/em>\u2013 3) + 3x<\/em> \u2013 5 = 2(5 \u2013 4x<\/em>)<\/p>\n

                                                                  f. <\/strong>\"\frac{1}{7}\"(y<\/em> + 12) = 4y<\/em> \u2013 \"\frac{3}{7}\"y<\/em><\/p>\n

                                                                  g. <\/strong>0.5t <\/em>+ 0.05 = 0.025<\/p>\n

                                                                  h.<\/strong>\u00a0\"\frac{x}{4}+\frac{2}{5}=-\frac{x}{2}-\frac{1}{5}\"<\/p>\n

                                                                  2.<\/strong> Write an expression for each of the following:<\/p>\n

                                                                  a. <\/strong>Susan has $375 in her checking account. If she makes a deposit of y<\/em> dollars, how much in total will she have in her account?<\/p>\n

                                                                  b. <\/strong>Mark weighs 175 pounds. If he loses y<\/em> pounds, how much will he weigh?<\/p>\n

                                                                  c. <\/strong>A piece of wire 45 metres long was cut in two pieces and one piece is w<\/em> metres long. How long is the other piece?<\/p>\n

                                                                  d. <\/strong>Emily made 4 dozen muffins. If it cost her x<\/em> dollars, what was her cost per dozen muffins? What was her cost per muffin?<\/p>\n

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