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Inequalities with Variables on Both Sides

When an inequality like 5x+3>2x+155x + 3 > 2x + 15 has variable terms on both sides, the first job is to gather them onto one side. Add or subtract a variable term from both sides until only one side has an xx — then it's a two-step inequality you already know how to finish.

Here's the part students like: adding or subtracting variable terms never flips the inequality sign. The only move that ever flips it is multiplying or dividing both sides by a negative number, and with one smart choice at the start, you can usually avoid that entirely.

Gather the variables on one side

Pick one side to hold the variable and subtract the other side's variable term from both sides. In 5x+3>2x+155x + 3 > 2x + 15, subtracting 2x2x from both sides leaves 3x+3>153x + 3 > 15 — one variable side, and the symbol never moved.

From there it's the standard finish: subtract 33 from both sides to get 3x>123x > 12, then divide by 33 to get x>4x > 4.

Choose the side that keeps the coefficient positive

You can collect the variables on either side — the solution comes out the same. But one choice is cleaner: move the smaller variable term over to the larger one, so the remaining coefficient is positive and no flip is ever needed.

Take 2x75x+82x - 7 \leq 5x + 8. Subtracting 5x5x from both sides gives 3x78-3x - 7 \leq 8, which works but forces a divide-by-negative-and-flip later. Subtracting 2x2x instead gives 73x+8-7 \leq 3x + 8 — positive coefficient, no flip coming. Then 153x-15 \leq 3x, so 5x-5 \leq x, which is the same statement as x5x \geq -5.

If you do end up dividing by a negative, that's fine — just remember the flip. Both paths land on the same answer.

Reading an answer like 5x-5 \leq x

When the variable lands on the right, flip your reading, not the math: 5x-5 \leq x says 5-5 is less than or equal to xx, which means xx is greater than or equal to 5-5, or x5x \geq -5. Notice the symbol still opens toward the xx in both versions — the open end of the symbol always faces the larger side.

Worked examples

Example 1: collect on the left

Solve 7x95x+397x - 9 \geq -5x + 39.

Start with the inequality7x95x+397x - 9 \geq -5x + 39
Add 5x5x to both sides12x93912x - 9 \geq 39
Add 99 to both sides12x4812x \geq 48
Divide both sides by 1212 — positive, no flipx4x \geq 4

Answer: x4x \geq 4

Example 2: collect on the right to avoid a flip

Solve 2x75x+82x - 7 \leq 5x + 8.

Start with the inequality2x75x+82x - 7 \leq 5x + 8
Subtract 2x2x from both sides73x+8-7 \leq 3x + 8
Subtract 88 from both sides153x-15 \leq 3x
Divide both sides by 335x-5 \leq x
Rewrite with xx firstx5x \geq -5

Answer: x5x \geq -5

Example 3: variable term with a negative sign

Solve 9x>4x69 - x > 4x - 6.

Start with the inequality9x>4x69 - x > 4x - 6
Add xx to both sides9>5x69 > 5x - 6
Add 66 to both sides15>5x15 > 5x
Divide both sides by 553>x3 > x
Rewrite with xx firstx<3x < 3

Answer: x<3x < 3

Try one yourself

Common questions

Does moving a variable term across the inequality flip the sign?

No. Moving a variable term is adding or subtracting the same thing from both sides, and adding or subtracting never flips the symbol. Only multiplying or dividing both sides by a negative number does.

Which side should I collect the variables on?

Either side gives the same solution, but collecting them where the coefficient stays positive is cleaner. Move the smaller variable term to the side with the larger one and you'll never be forced to divide by a negative.

What if the variables cancel completely?

Then the answer is about the leftover number statement. If you reach something always true, like 3<73 < 7, every real number is a solution. If you reach something false, like 5<25 < 2, there is no solution.

How do I turn 3>x3 > x into an xx-first answer?

Swap the sides and reverse the symbol: 3>x3 > x becomes x<3x < 3. The relationship hasn't changed — the open end of the symbol still faces the 33, the larger side — you've just written the same fact with xx first.

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