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All courses Math Foundations · MF.EQ.6 33 of 60
Solve two-step equations and justify each inverse operation.

Solve a two-step equation with a constant and coefficient

Problem
Solve 3x + 5 = 20.
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Hint

Look at the left side: the +5 is attached to 3x. Undo that first by subtracting 5 from both sides.

Use inverse operations in reverse order: remove the outside addition before undoing the multiplication by 3, or x will not be isolated.

Solution walkthrough

01

Identify the outer operation first

\[3x~+~5~=~20\]

The variable term is not just 3x; it has +5 attached to it. Since +5 is the outside operation on the left side, undo that first to start isolating x.

02

Remove the 5 by subtracting 5 from both sides

\[3x~+~5~-~5~=~20~-~5~→~3x~=~15\]

Subtracting 5 is the inverse of adding 5. We remove the constant first because dividing before removing the +5 would not isolate x; the 5 is still attached to the variable expression.

03

Undo the multiplication by 3

\[3x~÷~3~=~15~÷~3~→~x~=~5\]

Now x is being multiplied by 3, so divide both sides by 3. This isolates x and gives the solution.

04

State and verify the solution

\[x~=~5\]

Substituting 5 back into the original equation gives 20, so the solution checks. The value of the variable is 5.

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Another way

  1. Check by substitution after solving: replace x with 5 in 3x + 5 = 20 and confirm both sides are 20.

!

Common mistake

Dividing 20 by 3 right away and writing x + 5 = 20/3. That is wrong because only 3x could be divided by 3 cleanly; the +5 must be removed first since it is still attached to the variable side.