Equations with Variables on Both Sides

Some equations contain the same variable on both sides of the equals sign.

For example:

5x + 3 = 2x + 18

To solve this equation, we need to collect the variable terms onto one side and the constants onto the other.

Subtract 2x from both sides:

3x + 3 = 18

Subtract 3:

3x = 15

Divide by 3:

x = 5

The key idea is the same as with simpler equations:

keep the equation balanced while isolating the variable.

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Why Variables Can Appear on Both Sides

An equation compares two expressions.

There is no requirement that the variable appear on only one side.

For example:

7x − 4 = 3x + 20

means that:

7x − 4

and:

3x + 20

have the same value for the solution of x.

Our goal is to determine which value makes that equality true.

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The Main Strategy

For many equations with variables on both sides, use:

Simplify → Collect variables → Collect constants → Isolate → Check

Simplify

• distribute if necessary
• combine like terms

Collect variables

Move the variable terms to one side.

Collect constants

Move constants to the opposite side.

Isolate

Divide or multiply as necessary to get the variable alone.

Check

Substitute the solution into the original equation.

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Collecting Variable Terms

Consider:

6x + 5 = 2x + 21

We can remove 2x from the right by subtracting 2x from both sides:

6x − 2x + 5 = 2x − 2x + 21

Simplify:

4x + 5 = 21

The variable now appears on only one side.

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Finish the Equation

Continue:

4x + 5 = 21

Subtract 5:

4x = 16

Divide by 4:

x = 4

Check:

Left side:

6(4) + 5 = 29

Right side:

2(4) + 21 = 29

Therefore:

29 = 29

So:

x = 4

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Terms Do Not Really "Move"

You may sometimes hear:

"Move 2x to the other side and change its sign."

This shortcut can be useful, but it hides what is actually happening.

From:

6x + 5 = 2x + 21

we do not magically move 2x.

We:

subtract 2x from both sides

That preserves equality.

Understanding this balancing process helps prevent sign errors.

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Choosing Which Variable Term to Remove

Consider:

8x + 3 = 5x + 24

It is convenient to subtract 5x:

3x + 3 = 24

This keeps the coefficient positive.

Then:

3x = 21

x = 7

We could instead subtract 8x, but this would create a negative coefficient.

Both approaches are mathematically valid.

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An Alternative Correct Method

Start with:

8x + 3 = 5x + 24

Subtract 8x:

3 = −3x + 24

Subtract 24:

−21 = −3x

Divide by −3:

x = 7

Same solution.

Therefore:

you can collect variables on either side

as long as you maintain equality correctly.

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Efficient Solving

It is often easier to move the smaller variable term toward the larger variable term.

For:

9x + 2 = 4x + 27

subtract 4x:

5x + 2 = 27

This avoids negative coefficients.

Subtract 2:

5x = 25

Divide by 5:

x = 5

Efficiency matters, but correctness matters more.

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Worked Example

Solve:

7x − 8 = 3x + 20

Subtract 3x:

4x − 8 = 20

Add 8:

4x = 28

Divide by 4:

x = 7

Check:

LHS:

7(7) − 8 = 41

RHS:

3(7) + 20 = 41

Therefore:

41 = 41

Correct.

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Worked Example with Negative Numbers

Solve:

2x + 11 = 5x − 4

Subtract 2x:

11 = 3x − 4

Add 4:

15 = 3x

Divide by 3:

x = 5

Check:

LHS:

2(5) + 11 = 21

RHS:

5(5) − 4 = 21

Correct.

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Negative Coefficients

Solve:

−3x + 7 = 2x − 18

Subtract 2x:

−5x + 7 = −18

Subtract 7:

−5x = −25

Divide by −5:

x = 5

Check:

LHS:

−3(5) + 7 = −8

RHS:

2(5) − 18 = −8

Correct.

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Variables with Several Terms

Sometimes we must combine like terms before collecting the variables.

Solve:

3x + 4 + 2x = 7x − 10

Combine the left-side variable terms:

5x + 4 = 7x − 10

Subtract 5x:

4 = 2x − 10

Add 10:

14 = 2x

Divide by 2:

x = 7

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Simplify First

A common mistake is trying to move terms before simplifying.

Consider:

4x + 3x − 8 = 2x + 17

First combine:

7x − 8 = 2x + 17

Then subtract 2x:

5x − 8 = 17

Add 8:

5x = 25

Divide by 5:

x = 5

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Equations with Parentheses

Parentheses should usually be removed before collecting variable terms.

Consider:

3(x + 2) = 2x + 10

Distribute:

3x + 6 = 2x + 10

Subtract 2x:

x + 6 = 10

Subtract 6:

x = 4

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Parentheses on Both Sides

Solve:

2(x + 5) = 3(x + 1)

Distribute:

2x + 10 = 3x + 3

Subtract 2x:

10 = x + 3

Subtract 3:

7 = x

Therefore:

x = 7

Check:

LHS:

2(7 + 5) = 24

RHS:

3(7 + 1) = 24

Correct.

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Negative Distribution

Solve:

−2(x − 4) = 3x − 7

Distribute:

−2x + 8 = 3x − 7

Subtract 3x:

−5x + 8 = −7

Subtract 8:

−5x = −15

Divide by −5:

x = 3

Check:

LHS:

−2(3 − 4) = 2

RHS:

3(3) − 7 = 2

Correct.

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A More Complex Equation

Solve:

3(2x + 1) − 4 = 2(x + 6)

Distribute:

6x + 3 − 4 = 2x + 12

Combine:

6x − 1 = 2x + 12

Subtract 2x:

4x − 1 = 12

Add 1:

4x = 13

Divide by 4:

x = 13/4

or:

x = 3.25

Not every equation produces an integer solution.

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Three Possible Outcomes

Equations with variables on both sides can produce three important outcomes:

one solution

no solution

infinitely many solutions

Understanding why these occur is an important part of algebra.

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One Solution

Consider:

5x + 4 = 2x + 19

Subtract 2x:

3x + 4 = 19

Subtract 4:

3x = 15

Divide by 3:

x = 5

The variable remains during the solving process and eventually produces one specific value.

Therefore the equation has:

one solution

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No Solution

Consider:

4x + 3 = 4x + 9

Subtract 4x from both sides:

3 = 9

This statement is false.

No value of x can make:

3 = 9

true.

Therefore:

the equation has no solution

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Why No Solution Occurs

Consider:

4x + 3 = 4x + 9

Both sides contain exactly the same variable term:

4x

But one side is always 6 greater than the other.

For any x:

Left side:

4x + 3

Right side:

4x + 9

They can never be equal.

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Infinitely Many Solutions

Consider:

3x + 6 = 3(x + 2)

Distribute the right side:

3x + 6 = 3x + 6

Subtract 3x:

6 = 6

This statement is always true.

Therefore:

every permissible value of x is a solution

The equation has:

infinitely many solutions

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Why Infinitely Many Solutions Occur

Consider:

2(x + 4) = 2x + 8

The left side simplifies to:

2x + 8

So the equation becomes:

2x + 8 = 2x + 8

Both sides are actually the same expression.

Therefore they are equal for every value of x.

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Recognizing the Outcome

After simplifying and collecting terms:

If you get:

x = a number

there is:

one solution

If the variable disappears and you get a false statement:

5 = 12

there is:

no solution

If the variable disappears and you get a true statement:

8 = 8

there are:

infinitely many solutions

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Worked Example: Determine the Number of Solutions

Solve:

5x − 2 = 5x − 2

Subtract 5x:

−2 = −2

This is always true.

Therefore:

infinitely many solutions

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Worked Example: No Solution

Solve:

7x + 4 = 7x − 3

Subtract 7x:

4 = −3

False.

Therefore:

no solution

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Worked Example: One Solution

Solve:

7x + 4 = 5x + 18

Subtract 5x:

2x + 4 = 18

Subtract 4:

2x = 14

Divide by 2:

x = 7

Therefore:

one solution: x = 7

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The Graph Connection

A linear equation with variables on both sides can also be interpreted as asking:

When do two linear expressions have the same value?

For example:

2x + 3 = x + 7

can be represented by:

y = 2x + 3

and:

y = x + 7

The solution occurs where the two lines intersect.

If they intersect once:

one solution

If they are parallel:

no solution

If they are the same line:

infinitely many solutions

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One Intersection

Consider:

2x + 3 = x + 7

Solve:

x + 3 = 7

x = 4

The corresponding lines intersect once.

Therefore the equation has:

one solution

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Parallel Lines and No Solution

Consider:

2x + 3 = 2x + 8

Both expressions have the same coefficient of x but different constants.

Graphically:

y = 2x + 3

and:

y = 2x + 8

have the same slope but different y-intercepts.

They are parallel.

They never intersect.

Therefore:

no solution

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Same Line and Infinite Solutions

Consider:

2(x + 3) = 2x + 6

Both sides simplify to:

2x + 6

Graphically, both expressions represent the same line.

Every point on one is also on the other.

Therefore:

infinitely many solutions

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Fractions with Variables on Both Sides

Solve:

x/2 + 3 = x/4 + 8

One approach is to clear the fractions.

The lowest common denominator is 4.

Multiply every term by 4:

2x + 12 = x + 32

Subtract x:

x + 12 = 32

Subtract 12:

x = 20

Check:

LHS:

20/2 + 3 = 13

RHS:

20/4 + 8 = 13

Correct.

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Decimals with Variables on Both Sides

Solve:

0.5x + 4 = 0.2x + 10

Subtract 0.2x:

0.3x + 4 = 10

Subtract 4:

0.3x = 6

Divide by 0.3:

x = 20

Check:

LHS:

0.5(20) + 4 = 14

RHS:

0.2(20) + 10 = 14

Correct.

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Clearing Decimals First

We could also solve:

0.5x + 4 = 0.2x + 10

by multiplying every term by 10:

5x + 40 = 2x + 100

Subtract 2x:

3x + 40 = 100

Subtract 40:

3x = 60

x = 20

Both methods are valid.

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Variables in Numerators

Solve:

(x + 2)/3 = (x − 2)/2

Multiply every term by 6:

2(x + 2) = 3(x − 2)

Distribute:

2x + 4 = 3x − 6

Subtract 2x:

4 = x − 6

Add 6:

x = 10

Check:

LHS:

(10 + 2)/3 = 4

RHS:

(10 − 2)/2 = 4

Correct.

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Verification by Substitution

After finding a solution, substitute it into the original equation.

Suppose:

4x + 5 = 2x + 17

Solving gives:

x = 6

Check the left-hand side:

4(6) + 5

= 24 + 5

= 29

Check the right-hand side:

2(6) + 17

= 12 + 17

= 29

Therefore:

LHS = RHS

and x = 6 is verified.

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Why Verification Matters

Equations with variables on both sides involve several opportunities for mistakes:

• incorrect subtraction
• lost negative signs
• incorrect distribution
• combining unlike terms
• arithmetic errors

Substitution tests whether the final answer actually satisfies the original equation.

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Worked Example: Catching an Error

A student claims that:

5x + 4 = 3x + 18

has solution:

x = 5

Check:

LHS:

5(5) + 4 = 29

RHS:

3(5) + 18 = 33

Since:

29 ≠ 33

x = 5 is incorrect.

Solve correctly:

5x + 4 = 3x + 18

Subtract 3x:

2x + 4 = 18

Subtract 4:

2x = 14

Divide by 2:

x = 7

Check:

39 = 39

Correct.

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Word Problem: Two Pricing Plans

Company A charges:

$15 + $4 per unit

Company B charges:

$27 + $2 per unit

When will the costs be equal?

Let x be the number of units.

4x + 15 = 2x + 27

Subtract 2x:

2x + 15 = 27

Subtract 15:

2x = 12

Divide by 2:

x = 6

The plans cost the same at:

6 units

Check:

Company A:

4(6) + 15 = $39

Company B:

2(6) + 27 = $39

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Word Problem: Two Moving Objects

One object is:

10 m

ahead and moves at:

3 m/s

Another starts at 0 m and moves at:

5 m/s

Their positions can be represented as:

First object:

3t + 10

Second object:

5t

They meet when:

3t + 10 = 5t

Subtract 3t:

10 = 2t

Therefore:

t = 5 s

After 5 seconds, both are at:

25 m

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Equations in Geometry

Suppose two sides of a shape must have equal lengths.

One side is:

3x + 5

The other is:

5x − 9

Set them equal:

3x + 5 = 5x − 9

Subtract 3x:

5 = 2x − 9

Add 9:

14 = 2x

Therefore:

x = 7

Equations with variables on both sides frequently appear in geometry.

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Equations in Science

Suppose two objects have position equations:

d₁ = 4t + 6

d₂ = 7t − 9

To find when they have the same position:

4t + 6 = 7t − 9

Subtract 4t:

6 = 3t − 9

Add 9:

15 = 3t

Therefore:

t = 5

This shows how variables-on-both-sides equations can model physical situations.

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A Systematic Method

For complicated equations, follow this order:

Remove parentheses

↓

Combine like terms on each side

↓

Collect variable terms on one side

↓

Collect constants on the other side

↓

Divide by the variable's coefficient

↓

Check the solution

↓

Classify the equation

This process keeps the algebra organized.

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Full Worked Example

Solve:

4(2x − 3) + 5 = 3(x + 4) + 8

Distribute:

8x − 12 + 5 = 3x + 12 + 8

Combine:

8x − 7 = 3x + 20

Subtract 3x:

5x − 7 = 20

Add 7:

5x = 27

Divide by 5:

x = 27/5

or:

x = 5.4

Check using the original equation.

LHS:

4(2(5.4) − 3) + 5

= 4(10.8 − 3) + 5

= 4(7.8) + 5

= 36.2

RHS:

3(5.4 + 4) + 8

= 3(9.4) + 8

= 36.2

Correct.

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Common Mistakes

Mistake: Changing the sign of a term without performing an operation on both sides.

Instead of saying:

"Move 3x and change its sign,"

understand that you are:

subtracting 3x from both sides.

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Mistake: Combining terms across the equals sign.

In:

5x + 2 = 3x + 10

you cannot combine 5x and 3x immediately because they are on different sides.

First perform the same operation on both sides.

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Mistake: Combining unlike terms.

4x + 7

cannot become:

11x

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Mistake: Forgetting to distribute.

3(x + 5)

equals:

3x + 15

not:

3x + 5

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Mistake: Losing a negative sign during distribution.

−2(x − 4)

equals:

−2x + 8

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Mistake: Assuming every equation has exactly one solution.

Variables-on-both-sides equations may have:

• one solution
• no solution
• infinitely many solutions

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Mistake: Thinking 5 = 5 means x = 5.

If the variable disappears and the result is:

5 = 5

the equation is true for every permissible value.

Therefore there are:

infinitely many solutions

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Mistake: Thinking 5 = 8 means x has been calculated incorrectly.

If the algebra is correct and the variable disappears to produce a false statement, the equation has:

no solution

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Key Terms

Variable on both sides — equation in which the same variable occurs on both sides of the equals sign.

Variable term — term containing a variable.

Constant — number without a variable.

Coefficient — number multiplying a variable.

Like terms — terms containing the same variables raised to the same powers.

Collect terms — combine or move terms so similar terms appear together.

Inverse operations — operations that undo one another.

Balance — maintaining equality by performing equivalent operations on both sides.

Isolate — get the variable alone on one side.

Distributive property — property represented by a(b + c) = ab + ac.

Substitution — replacing a variable with a particular value.

Verification — checking that a solution satisfies the original equation.

One solution — exactly one value satisfies the equation.

No solution — no value satisfies the equation.

Infinitely many solutions — every permissible value satisfies the equation.

Identity — equation that is true for every permissible value of its variable.

Equivalent equations — equations with the same solution set.

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Key Takeaways

• Variables can appear on both sides of an equation.
• The goal is still to isolate the variable.
• Variable terms should usually be collected onto one side.
• Constants should usually be collected onto the other side.
• Performing the same operation on both sides maintains equality.
• Terms do not literally jump across the equals sign and change signs.
• You can collect the variable on either side of the equation.
• Choosing the side that produces a positive coefficient can make calculations easier.
• Simplify each side before collecting terms.
• Use the distributive property to remove parentheses.
• Combine like terms carefully.
• Negative coefficients and constants follow the same algebraic rules.
• Fractions and decimals do not change the fundamental solving process.
• Clearing fractions or decimals can sometimes simplify an equation.
• Equations with variables on both sides can have one solution, no solution, or infinitely many solutions.
• A false statement after the variable disappears indicates no solution.
• A true statement after the variable disappears indicates infinitely many solutions.
• One solution corresponds graphically to two lines intersecting once.
• No solution corresponds to distinct parallel lines.
• Infinitely many solutions corresponds to the same line represented in two equivalent forms.
• Substitution into the original equation is an effective way to verify a solution.
• These equations can model comparisons such as pricing plans, motion, geometry, and scientific relationships.

Check Your Understanding

1. What does it mean to collect variable terms onto one side of an equation?
2. Why must the same operation be performed on both sides?
3. Solve: 5x + 3 = 2x + 18
4. Solve: 8x − 7 = 3x + 23
5. Solve: 2x + 14 = 6x − 10
6. Solve: −3x + 8 = 2x − 17
7. Solve: 3(x + 4) = 2x + 17
8. Solve: 2(x + 5) = 4(x − 2)
9. Solve: 5x + 3x − 4 = 2x + 20
10. Solve: x/2 + 4 = x/4 + 9
11. Solve: 0.6x + 5 = 0.2x + 13
12. Determine whether 4x + 7 = 4x + 12 has one solution, no solution, or infinitely many solutions.
13. Determine whether 5x − 3 = 5x − 3 has one solution, no solution, or infinitely many solutions.
14. Determine whether 3(x + 2) = 3x + 6 has one solution, no solution, or infinitely many solutions.
15. Explain why 6 = 11 appearing after simplification means that an equation has no solution.
16. Explain why 8 = 8 appearing after simplification can mean that an equation has infinitely many solutions.
17. Verify by substitution that x = 4 solves 5x + 2 = 3x + 10.
18. Two companies charge 5x + 10 and 3x + 24 dollars. Find the value of x at which their costs are equal.
19. Explain the connection between one solution, no solution, and infinitely many solutions and the intersections of two straight lines.
20. Describe a systematic method for solving an equation containing parentheses, like terms, and variables on both sides.