What is (x^2-9)/5-(x^2-4}{15}<= \frac{1-2x)/3 ?

The solution to (x^2-9)/5-(x^2-4}{15}<= \frac{1-2x)/3 is -7<= x<= 2

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(x^2-9)/5-(x^2-4}{15}<= \frac{1-2x)/3

\frac{x ^{2} - 9}{5} - \frac{x ^{2} - 4}{15} \leq \frac{1 - 2 x}{3}

- 7 \leq x \leq 2

Solution steps

\frac{x ^{2} - 9}{5} - \frac{x ^{2} - 4}{15} \leq \frac{1 - 2 x}{3}

Rewrite in standard form

2 x^{2} + 10 x - 28 \leq 0

Factor

2 x^{2} + 10 x - 28 : 2 (x - 2) (x + 7)

2 (x - 2) (x + 7) \leq 0

Divide both sides by

2

\frac{2 ( x - 2 ) ( x + 7 )}{2} \leq \frac{0}{2}

Simplify

(x - 2) (x + 7) \leq 0

Identify the intervals

- 7 \leq x \leq 2

5 x + 4 (3 x - 5) = x - 4

\frac{( 3 - ( ( 2.001 ) ^{2} - 1 ) )}{2 - 2.001}

1 < y \leq 3

\frac{4}{5} - 1

(- 4)^{\frac{5}{2}}

What is (x^2-9)/5-(x^2-4}{15}<= \frac{1-2x)/3 ?
The solution to (x^2-9)/5-(x^2-4}{15}<= \frac{1-2x)/3 is -7<= x<= 2