By LearningExpress Editors
Updated on Oct 3, 2011
Review these concepts at Solving Multistep Algebraic Equations Study Guide.
Solving Multistep Algebraic Equations Practice Questions
Problems
Find the value of the variable in each equation.
- 5u + 3 = 48
- 3f = 4 + f
- –2p + 3 = 13
- 6r + 4 = –r – 24
- 4(q – 3) = 16
- 12b + 21 = –2b – 21
- –3(t – 11) = 4t – 2
- 9(4p + 12) = 15p + 3
Solutions
1. | The equation 5u + 3 = 48 shows multiplication and addition. We will need to use their opposites, division and subtraction, to find the value of u. |
Subtract first: | |
5u + 3 – 3 = 48 – 3 | |
5u = 45 | |
Because u is multiplied by 5, divide both sides of the equation by 5: | |
u = 9 | |
2. | The equation 3f = 4 + f shows multiplication and addition. We will need to use their opposites, division and subtraction, to find the value of f. Subtracting 4 from both sides of the equation will not reduce the number of terms, but subtracting f from both will: |
3f – f = 4 + f – f | |
2f = 4 | |
Because f is multiplied by 2, divide both sides of the equation by 2: | |
f = 2 | |
3. | The equation –2 = 6 shows division and subtraction. We will need to use their opposites, multiplication and addition, to find the value ofa. If we add first, we will have smaller numbers with which to work: |
– 2 + 2 = 6 + 2 | |
= 8 | |
Because a is divided by 7, multiply both sides of the equation by 7: | |
a = 56 | |
4. | The equation –2p + 3 = 13 shows multiplication and addition. We will need to use their opposites, division and subtraction, to find the value of p. Subtract first: |
–2p + 3 – 3 = 13 – 3 | |
–2p = 10 | |
Because p is multiplied by –2, divide both sides of the equation by –2: | |
p = –5 | |
5. | The equation shows division, addition, and multiplication. Because we have like terms on the left side of the equation, we can simplify by adding: |
Because c is multiplied by, divide both sides of the equation by : | |
c = –4 | |
6. | The equation 6r + 4 = –r – 24 shows multiplication, addition, and subtraction. Both sides of the equation contain an r term and both sides contain constants. We must get the variable alone on one side and a constant alone on the other. We can remove r from the right side by adding r to both sides of the equation: |
6r + r + 4 = –r + r – 24 | |
7r + 4 = –24 | |
Now the equation shows addition and multiplication, so we must use subtraction and division to solve it. Subtract 4 from both sides of the equation: | |
7r + 4 – 4 = –24 – 4 | |
7r = –28 | |
Because r is multiplied by 7, divide both sides of the equation by 7: | |
r = –4 | |
7. | The equation 4(q – 3) = 16 has a constant multiplying an expression. The first step to solving this equation is to simplify the left side of the equation using the distributive law. Multiply q and –3 by 4: |
4(q – 3) = 4q – 12 | |
The equation is now: | |
4q – 12 = 16 | |
The equation shows subtraction and multiplication, so we must use addition and division to solve it. Add 12 to both sides of the equation: | |
4q – 12 + 12 = 16 + 12 | |
4q = 28 | |
Because q is multiplied by 4, divide both sides of the equation by 4: | |
q = 7 | |
8. | The equation has a constant multiplying an expression. The first step to solving this equation is to simplify using the distributive law. Multiply and 1 by 10: |
The equation is now: | |
5v + 10 = 6v + 4 | |
We must get the variable alone on one side and a constant alone on the other. We can remove v from the left side by subtracting 5v from both sides of the equation: | |
5v – 5v + 10 = 6v – 5v + 4 | |
10 = v + 4 | |
To get v alone on the right side of the equation, subtract 4 from both sides: | |
10 – 4 = v + 4 – 4 | |
6 = v | |
9. | The equation shows division, addition, and subtraction. The variable e appears on both sides of the equation, and a constant is on both sides of the equation. Start by subtracting from both sides. This will leave us with just one e term: |
Because 8 is subtracted from , add 8 to both sides of the equation: | |
Finally, because e is multiplied by divide both sides of the equation by | |
24 = e | |
Instead of dividing by , we could have also multiplied by its reciprocal, Both operations get e alone on one side of the equation and give us an answer of 24. | |
10. | The equation 12b + 21 = –2b – 21 shows multiplication, addition, and subtraction. Both sides of the equation contain a b term and both sides contain constants. We must get the variable alone on one side and a constant alone on the other. We can remove b from the right side by adding 2b to both sides of the equation: |
12b + 2b + 21 = –2b + 2b – 21 | |
14b + 21 = –21 | |
Now, the equation shows addition and multiplication, so we must use subtraction and division to solve it. Subtract 21 from both sides of the equation: | |
14b + 21 – 21 = –21 – 21 | |
14b = –42 | |
Because b is multiplied by 14, divide both sides of the equation by 14: | |
b = –3 | |
11. | The equation shows multiplication and addition. We will need to use their opposites, division and subtraction, to find the value of z. Subtract 8 from both sides of the equation: |
We can divide both sides of the equation by which is the same as multiplying by its reciprocal . Let's multiply: | |
z = 9 | |
12. | The equation shows multiplication, subtraction, division, and addition. Both sides of the equation contain an x term and both sides contain constants. We must get the variable alone on one side and a constant alone on the other. We can remove x from the right side by subtracting from both sides of the equation: |
Now, the equation shows multiplication and subtraction, so we must use addition and division to solve it. Add 20 to both sides of the equation: | |
Divide both sides of the equation by or multiply by its reciprocal to get x alone on the left side: | |
x = 6 | |
13. | The equation –3(t – 11) = 4t – 2 has a constant multiplying an expression. The first step to solving this equation is to use the distributive law. Multiply t and –11 by –3: |
–3(t – 11) = –3t + 33 | |
The equation is now: | |
–3t + 33 = 4t – 2 | |
We must get the variable alone on one side and a constant alone on the other. We can remove t from the left side by adding 3t to both sides of the equation: | |
–3t + 3t + 33 = 4t + 3t – 2 | |
33 = 7t – 2 | |
Next, remove –2 from the right side of the equation by adding 2 to both sides: | |
33 + 2 = 7t – 2 + 2 | |
35 = 7t | |
Because t is multiplied by 7, divide both sides of the equation by 7: | |
5 = t | |
14. | The equation has a constant multiplying an expression. The first step to solving this equation is to use the distributive law. Multiply x and 10 by |
The equation is now: | |
We must get the variable alone on one side and a constant alone on the other. We can remove x from the right side by adding x to both sides of the equation: | |
Then, subtract 2 from both sides: | |
Multiply both sides of the equation by to get x alone on the left side: | |
x = 20 | |
15. | The equation 9(4p + 12) = 15p + 3 has a constant multiplying an expression. The first step to solving this equation is to use the distributive law. Multiply 4p and 12 by 9: |
9(4p + 12) = 36p + 108 | |
The equation is now: | |
36p + 108 = 15p + 3 | |
We must get the variable alone on one side and a constant alone on the other. We can remove p from the right side by subtracting 15p from both sides of the equation: | |
36p – 15p + 108 = 15p – 15p + 3 | |
21p + 108 = 3 | |
Next, remove 108 from the left side of the equation by subtracting 108 from both sides: | |
21p + 108 – 108 = 3 – 108 | |
21p = –105 | |
Because p is multiplied by 21, divide both sides of the equation by 21: | |
p = –5 |
From Algebra in 15 Minutues A Day. Copyright © 2009 by LearningExpress, LLC. All Rights Reserved.
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