Solve (x + 5)/(x − 3) > 0
Rational inequality, worked out line by line the way a teacher would write it.
Answer
| Solution | x < −5 or x > 3 |
| Interval notation | (−∞, −5) ∪ (3, ∞) |
Step-by-step solution
5 steps-
1 Given\frac{x + 5}{x - 3} > 0
Solve for x.
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2 Critical points: where the expression is 0 or undefined
A polynomial or a fraction can only change sign where it equals 0 or where it is undefined. Between two neighbouring critical points the sign stays the same, so one test value decides each whole interval.
\text{zeros: } x = -5\qquad \text{undefined at: } x = 3The sign can only change at these points, so test one value in each interval between them.
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3 Sign chart: the sign of each factor in each interval
Each row shows the sign of one factor in each interval, and the bottom row multiplies them: an even number of minus signs gives +, an odd number gives −.
\begin{array}{c|ccccc} & \left(-\infty,\, -5\right) & -5 & \left(-5,\, 3\right) & 3 & \left(3,\, \infty\right) \\ \hline x + 5 & - & 0 & + & + & + \\ \hline x - 3\;\scriptstyle(\text{denominator}) & - & - & - & 0 & + \\ \hline \text{whole} & + & 0 & - & \text{\,undef.} & + \end{array}Test values: x = -6, x = -4, x = 4.
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4 Keep the intervals where the expression is positive
Keep the intervals whose sign matches the inequality: > 0 means positive, < 0 negative. With ≥ or ≤ the zeros count as well, but points where the expression is undefined never do.
x < -5 \;\text{ or }\; x > 3The critical points are left out: there the expression is 0 or undefined, not positive.
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5 Solutionx < -5 \;\text{ or }\; x > 3\qquad \left( -\infty, -5 \right) \cup \left( 3, \infty \right)
Check with test values
| x | Left side | Right side | Holds? |
|---|---|---|---|
| −6 | 1/9 | 0 | ✓ true |
| 4 | 9 | 0 | ✓ true |
| −4 | −1/7 | 0 | ✗ false |
Change a number or try your own problem: the solver works it out the same way, with a graph and hint mode.