Solve (x + 2)/(x + 1) > 0
Rational inequality, worked out line by line the way a teacher would write it.
Answer
| Solution | x < −2 or x > −1 |
| Interval notation | (−∞, −2) ∪ (−1, ∞) |
Step-by-step solution
5 steps-
1 Given\frac{x + 2}{x + 1} > 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 = -2\qquad \text{undefined at: } x = -1The 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,\, -2\right) & -2 & \left(-2,\, -1\right) & -1 & \left(-1,\, \infty\right) \\ \hline x + 2 & - & 0 & + & + & + \\ \hline x + 1\;\scriptstyle(\text{denominator}) & - & - & - & 0 & + \\ \hline \text{whole} & + & 0 & - & \text{\,undef.} & + \end{array}Test values: x = -3, x = -3/2, x = 0.
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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 < -2 \;\text{ or }\; x > -1The critical points are left out: there the expression is 0 or undefined, not positive.
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5 Solutionx < -2 \;\text{ or }\; x > -1\qquad \left( -\infty, -2 \right) \cup \left( -1, \infty \right)
Check with test values
| x | Left side | Right side | Holds? |
|---|---|---|---|
| −3 | 1/2 | 0 | ✓ true |
| 0 | 2 | 0 | ✓ true |
| −3/2 | −1 | 0 | ✗ false |
Change a number or try your own problem: the solver works it out the same way, with a graph and hint mode.