Balancing Algebraic Equations – Practice Problems
If you are looking for balancing algebraic equations practice problems in math, please go to the next section below. If you need to see the examples first, please scroll down to the second half of the page.
Instructions:
Balance the following equations as indicated in each question. Express your answer as a simplified fraction, if possible.
1) If A + B – C = D, then A = ?
2) If (W + X) ÷ Y = Z, then X = ?
3) If 5a + 3b = c, then a = ?
4) \(\text {If x = } \frac{yv}{w} \text {, then w = ?}\)
5) \(\text {If } \frac{JKL}{M} \text { + N = P, then K = ?}\)
Balancing Equations Practice – Answers
Answer 1
1) The correct answer is: A = D + C – B
If A + B – C = D, then A = ?
A + B – C = D
A + B – C + C = D + C
A + B – C + C = D + C
A + B – B = D + C – B
A + B – B = D + C – B
A = D + C – B
Answer 2
2) The correct answer is: X = ZY – W
If (W + X) ÷ Y = Z, then X = ?
(W + X) ÷ Y = Z
(W + X) ÷ Y × Y = Z × Y
(W + X) ÷ Y × Y = Z × Y
W + X = ZY
W – W + X = ZY – W
W – W + X = ZY – W
X = ZY – W
Answer 3
3) The correct answer is: \(\text {a = } \frac{c – 3b}{5}\)
If 5a + 3b = c, then a = ?
5a + 3b = c
5a + 3b – 3b = c – 3b
5a + 3b – 3b = c – 3b
5a = c – 3b
5 × a = c – 3b
5 ÷ 5 × a = c – 3b ÷ 5
5 ÷ 5 × a = c – 3b ÷ 5
a = c – 3b ÷ 5
\(\text {a = } \frac{c – 3b}{5}\)
Answer 4
4) The correct answer is: \(\text {w = } \frac{yv}{x} \)
\(\text {If x = } \frac{yv}{w} \text {, then w = ?}\)
\(\text {x = } \frac{yv}{w}\)
\(\text {x = yv} \div w\)
\(\text {x} \times \text {w = yv} \div \text {w} \times \text {w}\)
\(\text {xw = yv}\)
\(\text {xw} \div \text {x = yv} \div \text {x}\)
\(\text {w = } \frac{yv}{x} \)
Answer 5
5) The correct answer is: \(\text { K = } \frac{MP – MN}{JL}\)
\(\text {If } \frac{JKL}{M} \text { + N = P, then K = ?}\)
\(\frac{JKL}{M} \text { + N = P}\)
\(\frac{JKL}{M} \text { + N} – N = P – N\)
\(\frac{JKL}{M} \text { = P} – N\)
\(\text {JKL} \div \text {M} = P – N\)
\(\text {JKL} \div \text { M} \times \text {M} = (P – N) \times \text {M}\)
\(\text {JKL} \text { = } (P – N) \times \text {M}\)
\(\text {JL} \times \text { K} = (P – N) \times \text {M}\)
\(\text {JL} \div \text {JL} \times \text { K} = (P – N) \times \text {M} \div \text {JL}\)
\(\text { K} = (P – N) \times \text {M} \div \text {JL}\)
\(\text { K} = (MP – MN) \div \text {JL}\)
\(\text { K = } \frac{MP – MN}{JL}\)
What is “Balancing Equations” in Math?
“Balancing equations” means that you can perform any operation on one side of an algebraic equation if you perform the exact same operation on the other side of the equation. You will need to balance equations in order to “undo” operations to solve the problem.
Balancing Equations with Binary Operations
The phrase “binary operations” also means that whatever operation you do to one side of an equation, you need to do the exact same operation on the other side of the equation.
Balancing equations with binary operations refers to addition, subtraction, multiplication, and division.
After you balance the equation, you can cancel out by drawing a line through the inverse operations as shown in the examples below.
Examples:
Balancing Algebraic Equations with Addition & Subtraction
Look at the examples below on how to balance equations by using addition and subtraction.
Example 1: Solve by adding
The problem below has the minus sign. So to solve we need to do the opposite operation; in other words, we have to add back to solve.
Solve for x by adding 10 to each side.
x – 10 = 56
x – 10 + 10 = 56 + 10
x – 10 + 10 = 56 + 10
x = 66
Example 2:
The problem below has the plus sign. So to solve we need to do the opposite operation; so, we have to subtract to solve.
Solve for y by subtracting 15 from each side.
y + 15 = 45
y + 15 – 15 = 45 – 15
y + 15 – 15 = 45 – 15
y = 30
Balancing Equations Practice- Multiplying Each Side by the Same Number
Look at the example below on how to balance equations by using multiplication
Example 3: Solve by Multiplying
The problem below has the divide sign. Remember that to solve, we always need to do the opposite operation; so in this case, we have to multiply to solve the problem.
Solve for a by multiplying each side by 2
a ÷ 2 = 11
a ÷ 2 × 2 = 11 × 2
a ÷ 2 × 2 = 11 × 2
a = 22
Balancing Equations Practice – Dividing Each Side by the Same Number
The problems below has the minus sign. So to solve we need to do the opposite operation; in other words, we have to add back to solve.Look at the example below on how to balance equations by using division.
Example 4:
The problem below has the times sign. So to solve we need to divide to undo the multiplication.
Your first step is to divide each side by 6.
5b × 6 = 90
5b × 6 ÷ 6 = 90 ÷ 6
5b × 6 ÷ 6 = 90 ÷ 6
5b = 90 ÷ 6
Now we have to get rid of the 5 in front of the b, so now we have to divide each side by 5.
5b = 90 ÷ 6
5 × b = 90 ÷ 6
5 ÷ 5 × b = 90 ÷ 6 ÷ 5
5 ÷ 5 × b = 90 ÷ 6 ÷ 5
b = 90 ÷ 6 ÷ 5
b = 3
Balancing Equations on the Maths Exam
For problems on your exam, you will usually have to perform more than one operation in order to balance the equation.
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