Algebra 1 · Grades 8, 9

Solving Equations with Variables on Both Sides

Quick answer

When an equation has variables on both sides, move all variable terms to one side and all constants to the other, then solve as usual. For 5x + 3 = 2x + 18, subtract 2x from both sides to get 3x + 3 = 18, then solve to get x = 5. If the variables cancel and leave a false statement there is no solution; if they leave a true statement, every number is a solution.

What you'll learn

  • Collect variable terms on one side of an equation
  • Recognise equations with no solution and with infinitely many solutions
  • Choose which side to collect variables on to avoid negatives

What changes when variables appear on both sides

Up to now the variable lived on one side and the numbers on the other. Now both sides have variable terms:

5x+3=2x+185x + 3 = 2x + 18

You cannot start undoing operations yet, because there is no single xx to isolate — there are two xx-terms in different places. One extra move fixes that: get all the variable terms onto one side.

Why you can subtract a variable term from both sides

The balance rule has not changed. “Do the same thing to both sides” was never restricted to numbers — it works for any expression, including one containing a variable.

Subtracting 2x2x from both sides of 5x+3=2x+185x + 3 = 2x + 18 removes the xx-term from the right and shrinks the one on the left:

5x+3−2x=2x+18−2x5x + 3 - 2x = 2x + 18 - 2x 3x+3=183x + 3 = 18

Whatever number xx turns out to be, 2x2x is some specific quantity, and removing the same quantity from both pans keeps the scale level. You are now back to a two-step equation.

How to solve

  1. Simplify each side: distribute and combine like terms.
  2. Move all variable terms to one side by adding or subtracting a variable term.
  3. Move all constants to the other side.
  4. Divide by the coefficient.
  5. Check in the original equation.

Worked examples

The three possible outcomes

{x=a numberexactly one solutiontrue statement, e.g. 6=6infinitely many solutionsfalse statement, e.g. 5=9no solution\begin{cases} x = \text{a number} & \text{exactly one solution} \\[2pt] \text{true statement, e.g. } 6 = 6 & \text{infinitely many solutions} \\[2pt] \text{false statement, e.g. } 5 = 9 & \text{no solution} \end{cases}

Common mistakes

Practice problems

  1. Solve 7x−4=3x+127x - 4 = 3x + 12.

    Hint

    Subtract the smaller variable term, 3x3x, from both sides.

    Answer

    x=4x = 4

    Full solution

    Subtract 3x3x: 4x−4=124x - 4 = 12. Add 44: 4x=164x = 16. Divide by 44: x=4x = 4.

    Check: 7(4)−4=247(4) - 4 = 24 and 3(4)+12=243(4) + 12 = 24 ✓

  2. Solve 9y=4y+259y = 4y + 25.

    Answer

    y=5y = 5

    Full solution

    Subtract 4y4y: 5y=255y = 25. Divide by 55: y=5y = 5.

    Check: 9(5)=459(5) = 45 and 4(5)+25=454(5) + 25 = 45 ✓

  3. Solve 2x+11=6x−52x + 11 = 6x - 5.

    Answer

    x=4x = 4

    Full solution

    Subtract 2x2x: 11=4x−511 = 4x - 5. Add 55: 16=4x16 = 4x. Divide by 44: x=4x = 4.

    Check: 2(4)+11=192(4) + 11 = 19 and 6(4)−5=196(4) - 5 = 19 ✓

  4. Solve 3x+7=3x+73x + 7 = 3x + 7.

    Hint

    Look carefully at both sides before doing any work.

    Answer

    Infinitely many solutions

    Full solution

    Subtract 3x3x from both sides: 7=77 = 7, which is true.

    The two sides are identical expressions, so every real number is a solution. This is an identity.

  5. Solve 6x−1=6x+86x - 1 = 6x + 8.

    Answer

    No solution

    Full solution

    Subtract 6x6x: −1=8-1 = 8, which is false.

    No value of xx can make this true, so there is no solution.

  6. Solve 4(x−2)=2x+64(x - 2) = 2x + 6.

    Answer

    x=7x = 7

    Full solution

    Distribute: 4x−8=2x+64x - 8 = 2x + 6. Subtract 2x2x: 2x−8=62x - 8 = 6.

    Add 88: 2x=142x = 14. Divide by 22: x=7x = 7.

    Check: 4(5)=204(5) = 20 and 2(7)+6=202(7) + 6 = 20 ✓

  7. Solve 5−2x=3x−155 - 2x = 3x - 15.

    Hint

    Add 2x2x to both sides to keep the variable positive.

    Answer

    x=4x = 4

    Full solution

    Add 2x2x: 5=5x−155 = 5x - 15. Add 1515: 20=5x20 = 5x. Divide by 55: x=4x = 4.

    Check: 5−2(4)=−35 - 2(4) = -3 and 3(4)−15=−33(4) - 15 = -3 ✓

  8. Two phone plans: Plan A costs $20 plus $5 per GB. Plan B costs $35 plus $2 per GB. At how many GB do they cost the same?

    Hint

    Set the two cost expressions equal to each other.

    Answer

    55 GB, at $45 each

    Full solution

    Let gg be gigabytes: 5g+20=2g+355g + 20 = 2g + 35.

    Subtract 2g2g: 3g+20=353g + 20 = 35. Subtract 2020: 3g=153g = 15. Divide by 33: g=5g = 5.

    Check: Plan A =5(5)+20=45= 5(5) + 20 = 45; Plan B =2(5)+35=45= 2(5) + 35 = 45 ✓

Frequently asked questions

Which side should I move the variables to?

Move them to whichever side has the larger coefficient, so the variable you keep stays positive. For 2x + 8 = 6x, subtracting 2x gives 8 = 4x rather than a negative coefficient. Both choices give the same answer, but one has less room for sign errors.

What does it mean when the variable disappears completely?

It means the equation is either always true or never true. If what is left is true, like 7 = 7, every number is a solution. If what is left is false, like 7 = 3, no number works and there is no solution.

What to learn next

Key terms in this lesson

Equation
An equation is a statement that two expressions are equal, joined by an equals sign. Solving one means finding every value of the variable that makes the statement true.
Solution set
The solution set is every value that makes an equation or inequality true. An equation often has one solution, an inequality usually has infinitely many, and some have none at all.

Standards alignment

This lesson covers the following Common Core State Standards for Mathematics.

  • CCSS.MATH.CONTENT.8.EE.C.7aExpressions and EquationsGive examples of linear equations in one variable with one solution, infinitely many solutions, or no solutions. Show which of these possibilities is the case by successively transforming the given equation into simpler forms, until an equivalent equation of the form x = a, a = a, or a = b results (where a and b are different numbers).
  • CCSS.MATH.CONTENT.8.EE.C.7bExpressions and EquationsSolve linear equations with rational number coefficients, including equations whose solutions require expanding expressions using the distributive property and collecting like terms.
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