1.2. Equations
Equations
Equation: a mathematical sentence that contains two expressions and is separated by an equal sign (both sides of the equation have the same value).
To solve an equation we find a particular value for the variable in the equation that makes the equation true (left side = right side).
Example: For the equation x + 4 = 5
only x = 1 can make it true, since 1 + 4 = 5 (Left side = Right side)
Solution of an equation: the value of the variable in the equation that makes the equation true.
Example 1.2.1
Indicate whether each of the given number is a solution to the given equation.
| 1) 2 : 4x – 3 = 5 | 4 ∙ 2 – 3 5 | 5 5 | Yes | Replace x with 2. |
| 2) 15 : y = -3 | -3 | -3 -3 | Yes | Replace y with 15. |
| 3) : 8t = 3 | 8 () 3 | 4 ≠ 3 | No | Cannot replace t with . |
Solving Equations
Basic rules for solving one-step equations:
- Add, subtract, multiply or divide the same quantity to both sides of an equation to obtain a valid equation.
- Remember to always do the same thing to both sides of the equation (balance).
Properties for solving equations:
| Properties | Equality | Example |
| Addition property of equality | A = B A + C = B + C | Solve
x = 9 |
| Subtraction property of equality | A = B A – C = B – C | Solve
y = -13 |
| Multiplication property of equality | A = B A · C = B · C | Solve
m = 18 |
| Division property of equality | A = B (C ≠ 0) | Solve
n = –5 |
Example 1.2.2
Solve the following equations.
| 1) | Property of addition. | |
| Check: | Replace x with 14. |
| 2) | Property of subtraction. | |
| 3) | Property of multiplication. | |
| 4) | Property of division. | |
Multi-step equation: an equation that requires more than one step to solve it.
Procedure for solving multi-step equations:
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| Steps | Example |
| Solve | ||
|
Multiply each term by 5. | |
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Subtract 10 from both sides.
Subtract 6y from both sides. |
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Divide both sides by -5. | |
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Replace y with 2.
Multiply each term by 5.
LS = RS (correct) |
Equations involving decimals: 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).
| Steps | Example |
| Solve | ||
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The largest number of decimal place is two. | |
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Add 12 to both sides. Add 426x to both sides. |
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Example 1.2.3
| Solve 0.4y + 0.08 = 0.016 | The largest number of decimal place is three. |
| 1000(0.4y) + 1000(0.08) = 1000(0.016) | Multiply each term by 1000. |
| 400y + 80 = 16 | Combine like terms. |
| 400y = -64 | Divide both sides by 400. |
| y = – 0.16 |
Equations involving fractions:
| Steps | Example |
| Solve | ||
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Add 6t to both sides. Subtract 9 from both sides. |
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Divide both sides by 10. |
Word Problems
Identifying keywords:
- When trying to figure out the correct operation ( + , – , × , ÷ , etc.) in a word problem it is important to pay attention to keywords (clues to what the problem is asking).
- Identifying keywords and pulling out relevant information that appear in the word problem are effective ways for solving mathematical word problems.
Key or clue words in word problems:
| Addition (+) | Subtraction (–) | Multiplication (×) | Division (÷) | Equals to (=) |
| add | subtract | times | divided by | equals |
| sum (of) | difference | product | quotient | is |
| plus | take away | multiplied by | over | was |
| total (of) | minus | double | split up | are |
| altogether | less (than) | twice | fit into | were |
| increased by | decreased by | triple | per | amounts to |
| gain (of) | loss (of) | of | each | totals |
| combined | (amount) left | how much (total) | goes into | results in |
| in all | savings | how many | as much as | the same as |
| greater than | withdraw | out of | gives | |
| complete | reduced by | ratio/rate | yields | |
| together | fewer (than) | percent | ||
| more (than) | how much more | share | ||
| additional | how long | average |
Example 1.2.4
1) 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 did Edward drive?
| 235km + 203 km + 390 km = 828 km | The key word: total (+) |
2) 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?
| $150 – 15 – 35 = $100 | The key word: left (–) |
3) Lucy received $950 per month of rent from Mark for the months September to November. How much rent in total did she receive?
| $950 3 = $2850 | The key word: how much total (×) |
4) Julia is going to buy a $7500 used car from her uncle. She promises to pay $500 per month. In how many months can she pay it off?
| $7500 ÷ $500 = 15 months | The key word: per (÷) |
Steps for solving word problems:
| – Organize the facts given from the problem (create a table or diagram if it will make the problem clearer)
– Identify and label the unknown quantity (let x = unknown). – Convert words into mathematical symbols, and determine the operation – write an equation (looking for ‘key’ or ‘clue’ words). – Estimate and solve the equation and find the solution(s). – Check and state the answer. (Check the solution to the equation and check it back into the problem – is it logical?) |
Example 1.2.5
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?
- Organize the facts (make a table):
| 5 socks | $4.35 each |
| Sales tax | $2.15 |
| Money left | $5.15 |
- Determine the unknown: How much did William start with? (x = ?)
- Convert words into math symbols, and determine the operation (find keywords):
|
$4.35 × 5 |
|
($4.35 × 5) + $2.15 |
|
x = [($4.35 × 5) + $2.15] + $5.15 |
- Estimate and solve the unknown:
|
x = [($4 × 5) + $2] + $5 |
| = $27 | |
|
x = [($4.35 × 5) + $2.15] + $5.15 |
| = $29.05 |
- Check: 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.
| $29.05 – [($4.35 × 5) + $2.15] $5.15 | |
| $29.05 – $23.9 $5.15 | Correct! |
More examples:
Example 1.2.6
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?
- Organize the facts:
| James had | 96 toys |
| The total number of toys sold | 13 + 32 + 21 + 14 + 7 |
| The toys not sold | 96 – the total number of toys sold |
|
Let x = percentage of the toys were not sold |
|
13 + 32 + 21 + 14 + 7 = 87 |
|
96 – 87 = 9 |
|
x = = 0.094 = 9.4% |
Example 1.2.7
The 60-litre gas tank in Robert’s 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?
- Let x = litres of fuel are required to get to Vancouver.
- The 60-litre gas tank in Robert’s car is 1/2 full:
| 60 L × = 30 L | Robert has 30 litres gas in his car. |
- Robert’s car averages 7 litres per 100 km, and Vancouver is about 390 km from Kelowna.
| Proportion: | |
| (x)(100km) = (7 L) (390 km) | Cross multiply and solve for x. |
| x = = 27.3 L | Robert needs 27.3 litres gas to get to Vancouver. |
- 30L > 27.3L. Therefore, yes, Robert can make his trip.
Practice questions
1. Solve the following equations:
a. x – 7 = 12
b. y + =
c. =
d. 14t + 5 = 8
e. 7(x – 3) + 3x – 5 = 2(5 – 4x)
f. (y + 12) = 4y – y
g. 0.5t + 0.05 = 0.025
h.
2. Write an expression for each of the following:
a. Susan has $375 in her checking account. If she makes a deposit of y dollars, how much in total will she have in her account?
b. Mark weighs 175 pounds. If he loses y pounds, how much will he weigh?
c. A piece of wire 45 metres long was cut in two pieces and one piece is w metres long. How long is the other piece?
d. Emily made 4 dozen muffins. If it cost her x dollars, what was her cost per dozen muffins? What was her cost per muffin?