Solve (x + 5)/(x + 3) < 0
Rational inequality, worked out line by line the way a teacher would write it.
Answer
| Solution | −5 < 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 = -2.
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4 Keep the intervals where the expression is negative
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.
-5 < x < -3The critical points are left out: there the expression is 0 or undefined, not negative.
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5 Solution-5 < x < -3\qquad \left( -5, -3 \right)
Check with test values
| x | Left side | Right side | Holds? |
|---|---|---|---|
| −4 | −1 | 0 | ✓ true |
| −6 | 1/3 | 0 | ✗ false |
Change a number or try your own problem: the solver works it out the same way, with a graph and hint mode.