How to solve linear inequalities with negative numbers

The statement x < 4 includes every number smaller than 4, such as 3, 0 and −10. It excludes 4 itself. The statement x ≤ 4 also includes 4. Similarly, > means strictly greater than, while ≥ includes equality.

Adding or subtracting the same quantity on both sides preserves the direction. From x − 5 < 2, add 5 to both sides to get x < 7. The negative term did not require a sign reversal because the operation was addition.

Multiplying or dividing both sides by a positive number also preserves direction. From 3x ≤ 12, divide by positive 3 to get x ≤ 4. State the operation explicitly if you tend to lose signs while rearranging.

Start with a comparison you know: 2 < 5. Multiply both numbers by −1. Their new values are −2 and −5, and −2 is greater than −5. The true comparison is now −2 > −5. Multiplication by a negative number reverses their order on the number line.

Consider 7 − 3x ≤ 19. Subtract 7 from both sides: −3x ≤ 12. Now divide both sides by −3 and reverse ≤ to ≥. The solution is x ≥ −4.

Check the boundary first: at x = −4, the original left side is 7 − 3(−4) = 19, which satisfies ≤ 19. Check a value inside the proposed range, such as x = 0: 7 ≤ 19 is true. Then try a value outside, such as x = −5: 22 ≤ 19 is false. These checks help catch a reversed range; the algebra establishes the complete solution.

You can avoid negative division in this example. Add 3x to both sides to get 7 ≤ 19 + 3x, then subtract 19: −12 ≤ 3x. Divide by positive 3 to get −4 ≤ x. This means the same thing as x ≥ −4. When swapping the left and right sides of a comparison, rewrite the symbol so it still points toward the smaller side.

Solve 2(x − 3) > 5x + 6. Expand to get 2x − 6 > 5x + 6. Subtract 5x from both sides: −3x − 6 > 6. Add 6: −3x > 12. Dividing by −3 reverses > to <, giving x < −4.

This inequality is strict, so x = −4 is excluded. Substitution gives −14 > −14, which is false. At x = −5, the comparison is −16 > −19, which is true. Negative numbers farther left on the number line can still produce the larger expression value when the coefficient of x is negative.

Do not multiply or divide by an expression containing x unless you have established its sign. The expression x − 2 is negative for some values and positive for others. Clearing such a denominator as though it were always positive can discard or introduce solutions. The examples here use division by known nonzero constants.

The statement −2 < 3x + 1 ≤ 10 requires both comparisons to hold. Subtract 1 from all three parts: −3 < 3x ≤ 9. Divide every part by positive 3: −1 < x ≤ 3. The left boundary is excluded and the right boundary is included.

With a negative coefficient, every comparison reverses. From −2 < −3x + 1 ≤ 10, subtract 1 to get −3 < −3x ≤ 9. Dividing all parts by −3 gives 1 > x ≥ −3. Rewrite it in increasing order as −3 ≤ x < 1.

On a number line, use an open circle for an excluded boundary and a filled circle for an included boundary. For −3 ≤ x < 1, fill the circle at −3, leave the circle at 1 open and mark the segment between them. For x ≥ −4, fill −4 and mark the ray to the right.

Sometimes the variable cancels. In 2x + 3 < 2x + 5, subtracting 2x leaves 3 < 5, which is always true, so every real x is a solution. In 2x + 3 > 2x + 5, the result is 3 > 5, which is false, so there are no solutions. Do not turn either statement into x = 0.

Use this working record: original inequality; operation applied to both sides or all three parts; whether the multiplier or divisor is negative; final range; included or excluded boundaries; one value inside and one outside checked in the original.

Try 4 − 2x > 10 before reading the answer. Subtract 4 to get −2x > 6. Divide by −2 and reverse the sign: x < −3. At x = −4, the original gives 12 > 10, which is true. At x = −2 it gives 8 > 10, which is false. The boundary x = −3 gives 10 > 10 and is excluded.