Algebra · Variables and Equations · Lesson 4

Solving Linear Equations in One Variable

You have solved equations with counters, balances and tiles. Now you will do the same moves with symbols only, which is faster and works for any numbers — negatives, fractions and letters on both sides. This is the skill you will use in BECE questions and whenever you need to find an unknown price, age or quantity.

NaCCA B7.2.3.1.4 ⏱ about 50 min

By the end you will be able to

✔Use inverse operations and the balance idea to solve equations such as $3 x + 5 = 20$ .
✔Solve equations with the unknown on both sides, such as $7 x + 3 = 2 x + 18$ .
✔Solve equations whose answers are negative or fractions, and equations containing fractions of $x$ .
✔Check a solution by substituting it into the original equation.
✔Form and solve equations from word problems.

Key words

Solve: Find the value of the unknown that makes the equation true.
Inverse operation: The operation that undoes another: + and − undo each other; × and ÷ undo each other.
Like terms: Terms with the same letter part, e.g. $7 x$ and $2 x$ , or 3 and 18. Only like terms can be added or subtracted.
Substitute: Replace the letter with a number, e.g. put $x = 5$ into $3 x + 5$ to get 20.

Curriculum reference

Content standard: B7.2.3.1 Demonstrate an understanding of linear equations of the form x + a = b (where a and b are integers) by modelling problems as a linear equation and solving the problems concretely, pictorially, and symbolically.

Indicator: Solve linear equations in one variable.

Starter

Abena's MoMo balance

Abena checks her mobile money wallet in Takoradi. Since morning she has received GH₵35 from her brother and paid GH₵12 for data, and her balance is now GH₵80. How much did she have this morning? We can write $x + 35 - 12 = 80$ , so $x + 23 = 80$ and $x = 57$ . She had GH₵57. In this lesson you will learn to solve equations like this — and harder ones — quickly and accurately, using symbols alone.

1. The balance method in symbols

When you removed 5 counters from both sides of a balance, you were really subtracting 5 from both sides of the equation. When you shared counters among 3 bags, you were dividing both sides by 3. Now we write those moves in symbols.

Inverse operations
If the equation has…	undo it by…	Example
+ 5	subtracting 5 from both sides	$x + 5 = 9 \Rightarrow x = 4$
− 5	adding 5 to both sides	$x - 5 = 9 \Rightarrow x = 14$
× 3 (as in 3x)	dividing both sides by 3	$3 x = 12 \Rightarrow x = 4$
÷ 3 (as in $\frac{x}{3}$ )	multiplying both sides by 3	$\frac{x}{3} = 4 \Rightarrow x = 12$

In an equation such as $3 x + 5 = 20$ , the $x$ was first multiplied by 3, then 5 was added. We undo in reverse order: first remove the + 5, then the × 3.

▶ Notice the order: the added number is removed first, then we divide. The scale stays level at every step.

2. Letters on both sides

In $7 x + 3 = 2 x + 18$ there are $x$ terms on both sides. First collect all the $x$ terms on one side by subtracting the smaller $x$ term from both sides. Then collect the plain numbers on the other side, and finally divide.

▶ We remove the smaller x term (2x) so the x terms stay positive. Then the numbers, then divide.

1Subtract the smaller $x$ term from both sides, so the $x$ terms are on one side only.
2Add or subtract numbers on both sides so the plain numbers are on the other side.
3Divide both sides by the number in front of $x$ .
4Check: substitute your answer into both sides of the original equation; they must be equal.

3. Negatives, fractions and subtractions

Answers are not always whole positive numbers. In $5 x - 3 = x - 1$ , subtract $x$ from both sides to get $4 x - 3 = - 1$ . Add 3 to both sides: $4 x = 2$ . Divide by 4: $x = \frac{2}{4} = \frac{1}{2}$ . A fraction is a perfectly good answer.

When $x$ is subtracted, as in $20 - 3 x = x + 12$ , add the $x$ term to both sides to make it positive: $20 = 4 x + 12$ . Subtract 12: $8 = 4 x$ . Divide by 4: $2 = x$ , which is the same as $x = 2$ . It does not matter which side $x$ ends up on.

When the equation contains a fraction of x, multiply to clear it. For $\frac{x}{3} + 2 = 6$ : subtract 2 to get $\frac{x}{3} = 4$ , then multiply both sides by 3 to get $x = 12$ . For $\frac{2}{3} x = 4$ : multiply both sides by 3 to get $2 x = 12$ , then divide by 2 to get $x = 6$ .

Equation	Key move	Solution
$x - 8 = - 3$	add 8 to both sides	$x = 5$
$5 x - 3 = x - 1$	subtract $x$ , add 3, divide by 4	$x = \frac{1}{2}$
$20 - 3 x = x + 12$	add $3 x$ , subtract 12, divide by 4	$x = 2$
$\frac{x}{3} + 2 = 6$	subtract 2, multiply by 3	$x = 12$

4. Avoiding mistakes and solving word problems

For word problems, use the steps below. They work for money at the market, ages, sharing and shapes.

1Read carefully and decide what the unknown is. Write "Let $x$ = …".
2Write the other quantities in terms of $x$ .
3Find the fact that makes two things equal, and write the equation.
4Solve the equation.
5Answer the question in words with units, and check it against the story.

Worked examples

Try each one yourself first, then open the solution and compare every step.

Example 1

A one-sided equation

Solve

6 x + 4 = 28

Show the solution

Step 1Subtract 4 from both sides: $6 x = 24$ .
Step 2Divide both sides by 6: $x = 4$ .
Step 3Check: 6 × 4 + 4 = 24 + 4 = 28 ✓

Answer: $x = 4$

Example 2

Unknowns on both sides

Solve

4 x + 1 = 3 x + 7

Show the solution

Step 1Subtract $3 x$ from both sides: $x + 1 = 7$ .
Step 2Subtract 1 from both sides: $x = 6$ .
Step 3Check: left 4 × 6 + 1 = 25; right 3 × 6 + 7 = 25 ✓

Answer: $x = 6$

Example 3

A fraction answer

Solve

5 x - 3 = x - 1

Show the solution

Step 1Subtract $x$ from both sides: $4 x - 3 = - 1$ .
Step 2Add 3 to both sides: $4 x = 2$ .
Step 3Divide by 4: $x = \frac{2}{4} = \frac{1}{2}$ .
Step 4Check: left $5 \times \frac{1}{2} - 3 = 2 \frac{1}{2} - 3 = - \frac{1}{2}$ ; right $\frac{1}{2} - 1 = - \frac{1}{2}$ ✓

Answer: $x = \frac{1}{2}$

Example 4

When x is subtracted

Solve

20 - 3 x = x + 12

▶ Adding 3x to both sides makes the x term positive. The unknown ends up on the right, and that is fine: 2 = x means x = 2.

Show the solution

Step 1Add $3 x$ to both sides: $20 = 4 x + 12$ .
Step 2Subtract 12 from both sides: $8 = 4 x$ .
Step 3Divide by 4: $2 = x$ , so $x = 2$ .
Step 4Check: left 20 − 3 × 2 = 14; right 2 + 12 = 14 ✓

Answer: $x = 2$

Example 5

A sharing problem

Kofi and Ama share GH₵85. Ama gets GH₵15 more than Kofi. How much does each person get?

Show the solution

Step 1Let Kofi's share be GH₵ $x$ . Ama's share is GH₵ $(x + 15)$ .
Step 2Equation: $x + (x + 15) = 85$ , so $2 x + 15 = 85$ .
Step 3Subtract 15: $2 x = 70$ . Divide by 2: $x = 35$ .
Step 4Ama: 35 + 15 = 50. Check: 35 + 50 = 85 ✓

Answer: Kofi gets GH₵35 and Ama gets GH₵50

Summary

Solve by doing the same to both sides, using inverse operations: + ↔ −, × ↔ ÷.
Undo in reverse order: remove added or subtracted numbers first, then divide (or multiply to clear a fraction).
With $x$ on both sides, subtract the smaller $x$ term from both sides first.
Answers can be negative or fractions; $2 = x$ means the same as $x = 2$ .
Always check by substituting into both sides of the original equation.

Abena's MoMo balance

1. The balance method in symbols

2. Letters on both sides

3. Negatives, fractions and subtractions

4. Avoiding mistakes and solving word problems

Worked examples

A one-sided equation

Unknowns on both sides

A fraction answer

When x is subtracted

A sharing problem

Summary

Practice: 15 questions