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Solving two-step linear equations using multiplication and division: x/a + b = c

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Solving Two-Step Equations with Division: x/a + b = c

This lesson explains how to solve two-step linear equations of the form x/a + b = c. You will learn why the order of operations matters when undoing addition and division, walk through fully worked examples, and see common mistakes to avoid.

What Does \(x/a + b = c\) Mean?

A two-step linear equation in the form \(x/a + b = c\) combines two operations acting on the variable \(x\): first \(x\) is divided by a number \(a\), and then a number \(b\) is added to the result. To solve it, you need to "undo" both operations, but in the opposite order in which they were applied. This is the same idea used in two-step equations solved with addition and subtraction, just with division instead of multiplication as the second step.

Before tackling these two-step equations, it helps to be comfortable with simpler one-step equations like \(x/a = b\), covered in modeling and solving one-step linear equations. It also helps to be confident dividing positive and negative numbers, since many answers involve dividing integers.

To solve \(x/a + b = c\) for \(x\), follow these steps in order:

  1. Step 1: Isolate the division term by subtracting \(b\) from both sides (or adding, if \(b\) is negative). This leaves you with \(x/a = c - b\).
  2. Step 2: Multiply both sides by \(a\) to cancel the division. This gives \(x = a \times (c - b)\).
  3. Step 3: Simplify and check your answer by substituting it back into the original equation.

Notice the order: subtraction (or addition) happens first, then multiplication. This is the reverse of the order of operations, because you're peeling the equation apart from the outside in — undoing the "last thing done" first.

Solve for \(x\): \(\frac{x}{4} + 3 = 7\)

Step 1: Subtract 3 from both sides.

\(\frac{x}{4} + 3 - 3 = 7 - 3\)

\(\frac{x}{4} = 4\)

Step 2: Multiply both sides by 4.

\(4 \times \frac{x}{4} = 4 \times 4\)

\(x = 16\)

Step 3: Check. Substitute \(x = 16\) back into the original equation: \(\frac{16}{4} + 3 = 4 + 3 = 7\). It checks out.

The figure below shows the line \(y = \frac{x}{4} + 3\). The solution to \(\frac{x}{4} + 3 = 7\) is the \(x\)-value where this line reaches a height of 7, which is exactly \(x = 16\).

Graph of y equals x divided by 4 plus 3, with the solution at x = 16 marked Plot of y = x/4 + 3 for x in [-8, 32] 0 10 20 30 0 2 4 6 8 10 12 x y Solution: x = 16
Graph of \(y = \frac{x}{4} + 3\), with the solution point marked at \(x = 16\).

Solve for \(x\): \(\frac{x}{-3} - 5 = 1\)

Step 1: Add 5 to both sides.

\(\frac{x}{-3} - 5 + 5 = 1 + 5\)

\(\frac{x}{-3} = 6\)

Step 2: Multiply both sides by \(-3\).

\((-3) \times \frac{x}{-3} = (-3) \times 6\)

\(x = -18\)

Step 3: Check. Substitute \(x = -18\): \(\frac{-18}{-3} - 5 = 6 - 5 = 1\). Correct.

This example is a good reminder to be extra careful with signs. Dividing a negative by a negative gives a positive, and multiplying both sides by a negative number does not flip an equation's equal sign the way it flips an inequality's sign — it only changes the value.

  • Multiplying before isolating the division term. You must deal with the added or subtracted number \(b\) first, before multiplying by \(a\).
  • Forgetting to apply the operation to both sides. Whatever you subtract, add, or multiply on one side of the equation must be done to the other side too, to keep it balanced.
  • Sign errors with negative values of \(a\) or \(b\). Take an extra moment when \(a\) or \(b\) is negative, and double-check with the earlier work on dividing integers if needed.
  • Skipping the check. Plugging your solution back into the original equation is the fastest way to catch an arithmetic slip.

Once you're comfortable solving \(x/a + b = c\), you're ready to compare it with two-step equations that use multiplication instead of division, such as \(ax + b = c\), and with equations built using the distributive property. If a problem involves multiple variables and you need to solve for one of them in terms of the others, that skill is covered separately under solving literal equations.

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