Chapter 4 Represent And Solve Equations And Inequalities
Section 4.2: Apply Property Of Equalities
Page 183 Exercise 1 Answer
The computation to do – add 5 to both sides of the equation.
4 + 8 = 12
4 + 8 + 5 = 12 + 5
17 = 17
The equation is true.
The computation to do – add 3 to the left side of the equation and add 5 to the right side.
4 + 8 = 12
4 + 8 + 3 = 12 + 5
15 ≠ 17
The equation is false.
The computation to do – divide the left side of the equation by 2 and multiply the right side by 2.
4 + 8 = 12
(4 + 8) ÷ 2 = 12 x 2
6 ≠ 24
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The equation is false.
The computation to do – subtract 4 from both sides of the equation.
4 + 8 = 12
4 + 8 – 4 = 12 – 4
8 = 8
The equation is true.
The computations which keep the equation true are the kind that does the same thing to both sides of the equation.
To determine if an equation is true evaluate both sides of the equation. If the result is the same on both sides, the equation is true.
Result
4 + 8 + 5 = 12 + 5 is true.
4 + 8 + 3 = 12 + 6 is false.
(4+8) ÷ 2 = 12 × 2 is false.
4 + 8 − 4 = 12 − 4 is true.
The computations which keep the equation true are the kind that does the same thing to both sides of the equation.
Page 183 Exercise 1 Answer
An equation is given but two numbers are missing. To make sure the equation is true, fill with number such that the result on both sides is the same.
For example, let’s choose numbers so that the result on both sides is 8. To do so, add 1 to the left side, since 7 + 1 = 8, and subtract 2 from the right side, since10 – 2 = 8.
7 + 1 = 10 – 2
8 = 8
Add 10 to both sides of the equation.
7 + 1 + 10 = 10 − 2 + 10
8 + 10 = 8 + 10
18 = 18
Divide both sides by 4.
(7+1) ÷ 4 = (10−2) ÷ 4
8 ÷ 4 = 8 ÷ 4
2 = 2
Notice, when dividing (or multiplying) first put the entire left side in brackets and than divide, the same goes for the right side. Do this because the order of operations is not the same for addition, subtraction, multiplication, and division.
Result
Possible answer: 7 + 1 = 10 − 2
Two other operations that could be done to the completed equation to keep it true are add 10 to both sides of the equation and divide both sides by 4.
Page 184 Exercise 1 Answer
5y = 25
5y – 7 = 25 – 7
Since 7 was subtracted from both sides of the equation the property used is the Subtraction Property of Equality.
Other properties of equality one could apply are the Addition Property of Equality, the Multiplication Property of Equality, and the Division Property of Equality.
For example, the Addition Property of Equality – add 5 to both sides.
5y = 25
5y + 5 = 25 + 5
For example, the Multiplication Property of Equality – multiply both sides by 2.
5y = 25
5y x 2 = 25 x 2
For example, the Division Property of Equality – divide both sides by 5.
5y = 25
5y ÷ 5 = 25 ÷ 5
Result
Subtraction Property of Equality
Page 185 Exercise 2a Answer
4 ⋅ x = 36
(4⋅x) ÷ 4 = 36 ÷ 4
x = 9
The solution to 4 ⋅ x = 36 is given as x = 9. Check this by substituting x by 9 in 4 ⋅ x = 36. The equation is balanced. Thus, both sides can be divided by the same number, that is the scale can be balanced with only one blue x-block.
Notice, if both sides are divided by 4, from the first line we get the third line.
Result
4 . x = 36
(4 . x) ÷4 = 36 ÷ 4
x = 9
Page 185 Exercise 2b Answer
If 25 + d = 36 is true, than 25 + d − 25 = 36 − 20 is false, since from the left side 25 is subtracted, and from the right side only 20 is subtracted. The equation stays balanced if and only if the same number is subtracted from both sides, that is the Subtraction Property of Equality.
Result
If 25 + d = 36 is true, than 25 + d − 25 = 36 − 20 is false.
Page 186 Exercise 1 Answer
If we have an equation, we can use any of the properites of equality, that is we can either add to, subtract from, multiply, or divide both sides of the equation by the same number, and the result will be a new equation that is equivalent to the original one.
For example, use th Subtraction Property of Equality to write a new equation if x + 5 = 34 is given.
x + 5 – 4 = 34 – 4
To check that these two equation are really equivalent, find the value of x which is the solution and compare them. if and only if they are the same, the equations are equivalent.
x + 5 = 34
x + 5 – 5 = 34 – 5 (Subtract 5 from both sides)
x = 29
x + 5 – 4 = 34 – 4
x + 1 = 20 (Evaluate)
x + 1 – 1 = 30 – 1 (Subtract 1 from both sides)
x = 29
Result
If we have an equation, we can use any of the properties of equality, that is we can either add to, subtract from, multiply, or divide both sides of the equation by the same number, and the result will be a new equation that is equivalent to the original one.
Page 186 Exercise 2 Answer
Since the equation 7 + 5 = 12 is balanced, if 4 units are removed from one side the same must be done to the other side.
7 + 5 = 12
7 + 5 – 4 = 12 − 4
Result
Subtract 4 from the other side as well.
Page 186 Exercise 3 Answer
Since the equation 32 + 43 = 66 is balanced, if one side is multiplied by 3 than the other side must be multiplied by 3 as well to keep the sides equal.
23 + 43 = 66
66 = 66 (The equation is balanced.)
66 x 3 = 66 x 3
(23 + 43) x 3 = 66 x 3
Result
The other side must be multiplied by 3 as well.
Page 186 Exercise 4 Answer
If x + 5 = 8 is balanced, than adding 9 to one side and (4+5) will keep the equation balanced, since 4 + 5 = 9, s 9 was actually added to both sides.
x + 5 = 8
x + 5 + 9 = 8 + (4+5)
x + 5 + 9 = 8 + 9
Result
The equation will stay equal.
Page 186 Exercise 5 Answer
If 23 + 37 = 60 is true, than 23 + 37 + 9 = 60 + 9 is true as well. 9 is added to both sides so the Addition Property of Equality is used. The equation stays balanced.
23 + 37 = 60
23 + 37 + 9 = 60 + 9
69 = 69
Result
If 23 + 37 = 60 is true, than 23 + 37 + 9 = 60 + 9 is true as well.
Page 186 Exercise 6 Answer
16 + 1 = 16
(16+1) − 1 = 17 − 2
16 ≠ 15
1 is subtracted from the left side and 2 is subtracted from the right side. Since the first equation is balanced, the second is not. The equation stays balanced if and only if the same number is subtracted from both sides, which is not the case here.
Result
If 16 + 1 = 17 is true, than (16+1) − 1 = 17 − 2 is false.
Page 186 Exercise 7 Answer
To apply the Multiplication Property of Equality multiply both sides of the equation by 5.
7n = 28
7n x 5 = 28 × 5
Result
7n × 5 = 28 × 5
Page 186 Exercise 8 Answer
Solve 6m = 9 for m.
6m = 9
m = \(\frac{9}{6}\) (Divide both sides by 6.)
m = \(\frac{3}{2}\)
Tomas changed the equation – he divided the left side by 2 and the right side by 3.
6m ÷ 2 = 9 ÷ 3
3m = 3
m = \(\frac{3}{3}\) (Divide both sides by 3 to find m.)
m = 1 ≠ \(\frac{3}{2}\)
Tomas is not correct. If one side of the equation is divided by 2 and the other by 3, the equation won’t stay balanced. The result is not the same, thus the equations are not equivalent.
Result
Tomas is not correct.
Page 187 Exercise 9 Answer
5m + 4 = 19
5m + 4 – 3 = 19 − 3
The same number, 3, was subtracted from both sides. That is the equation is balanced.
The property of equality used is Subtraction Property of Equality.
Result
Subtraction Property of Equality.
Page 187 Exercise 10 Answer
3t = 20
3t ÷ 2 = 20 ÷ 2
Both sides are divided by the same number, 2. That is the equation is balanced.
The property of equality used is Division Property of Equality.
Result
Division Property of Equality.
Page 187 Exercise 11 Answer
\(\frac{n}{6}\) = 9
(\(\frac{n}{6}\)) x 5 = 9 × 5
Both sides are multiplied by the same number, 5. That is the equation is balanced.
The property of equality used is Multiplication Property of Equality.
Result
Multiplication Property of Equality.
Page 187 Exercise 12 Answer
5b – 6 = 14
5b – 6 + 2 = 14 + 2
The same number, 2, is added to both sides. That is the equation is balanced.
The property of equality used is Addition Property of Equality.
Result
Addition Property of Equality.
Page 187 Exercise 13 Answer
r + 9 = 42
r + 9 – 9 = 42 + 9
r + 9 – 9 ≠ 42 + 9
If equality r + 9 = 42 is true, than equality r + 9 − 9 = 42 + 9 is false since the equality stays true if and only if the same number is either added or subtracted from both sides. In this case, to one side 9 was added and from the other side 9 was subtracted. The equation is not balanced.
Result
If r + 9 = 42, than r + 9 − 9 ≠ 42 + 9.
Page 187 Exercise 14 Answer
6s = 24
6s ÷ 6 = 24 ÷ 6
Since both sides were divided by the same number, that is by 6, the equation stays balanced and is true.
Result
If 6s = 24, than 6s ÷ 6 = 24 ÷ 6.
Page 187 Exercise 15 Answer
On the left side of the scale is 12 and on the right 2 + 7 + 3 + 16, since the scale is not even, it is an inequality:
12 ≠ 2 + 7 + 3 + 16.
To make this two sides equal, evaluate the right side.
2 + 7 + 3 + 16 = 28
The right side is equal to 28. To make the scale balance we need to make the left side equal to 28 as well.
Since 28 − 12 = 16, that is 28 can be written as 12 + 16, add 16 to the left side to make the scale balance.
12 + 16 = 2 + 7 + 3 + 16
Result
12 + 16 = 2 + 7 + 3 + 16
Page 187 Exercise 16 Answer
It is given that the scale balanced with 3 green blocks on one side and 1 blue x-block on the other side so we know that:
3 = x
The scale now has 15 green blocks on the left side. Since 3 ⋅ 5 = 15, then the left side of the equation was multiplied by 5. To make the completed equation true, the right side must also be multiplied by 5:
15 = 5 ⋅ x
Result
15 = 5 ⋅ x
Page 187 Exercise 17 Answer
Since 8x = 24 is true, and your friend divides the right side by 4, to make the equation true again, divide the left side by 4 as well.
8x = 24
8x ≠ 24 ÷ 4
8x ÷ 4 = 24 ÷ 4
Result
Divide the left side by 4.
Page 187 Exercise 18 Answer
The scale originally had 1 blue x-block and 20 green blocks on the left side and 40 green blocks on the right side. The balance was then represented by the equation: x + 20 = 40
If the scale now has 1 blue x-block and 3 green blocks on the left side, then 17 green blocks were removed from the left side since 20 − 17 = 3.
The scale and equation are only balanced if the same thing is done to both sides so you must then remove 17 green blocks from the right side to make the scale balanced:
x + 20 − 17 = 40 − 17
Result
17 green blocks
Page 188 Exercise 19 Answer
(y+6) ÷ 3 = 15
The second equation is not equivalent to the first because the left side was divided by 3 and the right side was not. That is, the Division Property of Equality is not used.
To make these two equations equivalent Bobbie must divide the right side by 3 as well.
Result
Bobbie must divide the right side of the equation by 3 as well.
Page 188 Exercise 20 Answer
5 + 5 = 10
5 + 5 + n = 11 + n
Since the same change was made to both sides of the equation, the equations are equivalent.
John used the Addition Property of Equality – he added the same number, that is n, to both sides of the equation.
Result
The equations are equivalent.
Page 188 Exercise 21 Answer
4 + 3 − 1 = 7 − 1
Since the scientist took one unit of mass from each side of a pan, and in the equation this is represented by subtracting one from both sides of the equation, the pan stays balanced. The equation stays balanced since the same change was made to both sides, that is he used the Subtraction Property of Equality.
Result
Subtraction Property of Equality.
Page 188 Exercise 22 Answer
Lexi used a property of equality to write an equation equivalent to n − 3 = 4. She could have used either of these four properites: Addition Property of Equality, Subtraction, Multiplication, or Division.
For example, if she used Multiplication Property of Equality, she multiplied both sides of the equation by the same number.
n − 3 = 4
(n-3) x 5 = 4 × 5
Notice, the left side must be in brackets.
The equations are equivalent because the same change was made to both sides of the first equation, and the result is the second equation. Also, one could check by finding n for both equation. The equation are equivalent if and only if the n is the same for both of them.
Result
(n−3) × 5 = 4 × 5
Page 188 Exercise 23 Answer
Since a quarter can be thought of, either as \(\frac{25}{100}\), that is as 25 cents, or more literally as \(\frac{1}{4}\), Emil’s money can be represented as a mixed number \(1 \frac{1}{4}\). Jade’s money, her 5 quarters, can similarly be represented as 5 x \(\frac{1}{4}\).
\(1 \frac{1}{4}=\frac{4}{4}+\frac{1}{4}\)= \(\frac{5}{4}\)
\(5 \times \frac{1}{4}=\frac{5}{1} \times \frac{1}{4}\)= \(\frac{5 \times 1}{1 \times 4}\)
= \(\frac{5}{4}\)
If Emil gives Jade $1 and she gives him four quarters, which is equal to one, they each still have the same amount of money.
\(1 \frac{1}{4}-1=\frac{1}{4}\)(Emil gives Jade $1.)
\(\frac{1}{4}+4 \times \frac{1}{4}=\frac{1}{4}+\frac{4}{4}\) (Jade gives Emil four quarters.)
= \(\frac{5}{4}\)
\(5 \times \frac{1}{4}+1=\frac{5}{4}+\frac{4}{4}\) (Emil give Jade $1.)
= \(\frac{9}{4}\)
\(\frac{9}{4}-4 \times \frac{1}{4}=\frac{9}{4}-\frac{4}{4}\) (Jade gives Emil four quarters.)
= \(\frac{5}{4}\)
Result
After the exchange, they each still have the same amount of money.
Page 188 Exercise 24 Answer
To change 7w = 49 into 7w ÷ 7 = 49 ÷ 7 both side were divided by 7.
The property used to find these equivalent equations is Division Property of Equality.
Result
Division Property of Equality.
Page 188 Exercise 25 Answer
The equation 12b = 24 is given. The find the quantity that equals 4b divide both sides by three.
12b = 24
12b ÷ 3 = 24 ÷ 3
4b = 8
Result
Divide both sides by 3.
Page 188 Exercise 26 Answer
For the equation to be equivalent to n + 4 = 11 whatever is done to one side of the equation, must be done to the other side of the equation as well.
(n+4) × 2 = 11
Since only one side is multiplied by 2, this equation is not equivalent to n + 4 = 11.
(n+4) × 2 = 11 ÷ 2
Since one side is multiplied by 2 and the other is divided by 2, this equation is not equivalent to n + 4 = 11.
(n+4) × 2 = 11 × 4
Since one side is multiplied by 2 and the other by 4, this equation is not equivalent to n + 4 = 11.
(n+4) × 2 = 11 × 2
Since both sides are multiplied by 2, the equation marked D, is equivalent to n + 4 = 11.
Result
D
Page 188 Exercise 27 Answer
8p ÷ 8 = 12 ÷ 8
To get 8p ÷ 8 = 12 ÷ 8 from 8p = 12 divide both sides by 8. Since the same thing is done to both sides, the equation stays balanced. Thus, these two are equivalent.
8p ÷ 8 = 12 ÷ 12
Since one side is divided by 8 and the other by 12 the equation doesn’t stay balanced, that is the solution will not be the same, thus these two equations are not equivalent.
8p + 4 = 12 + 4
To get 8p + 4 = 12 + 4 from 8p = 12 add 4 to both sides. Since the same thing is done to both sides, the equation stays balanced. Thus, these two are equivalent.
8p − 2 = 12 − 2
To get 8p − 2 = 12 − 2 from 8p = 12 subtracted 2 from both sides. Since the same thing is done to both sides, the equation stays balanced. Thus, these two are equivalent.
Result
B