−15x+4≤109 OR −6x+70>−2minus, 15, x, plus, 4, is less than or equal to, 109, start color #ed5fa6, start text, space, O, R, space, end text, end color #ed5fa6, minus, 6, x, plus, 70, is greater than, minus, 2 Choose 1 answer: Choose 1 answer:
(Choice A) A x\geq-7x≥−7x, is greater than or equal to, minus, 7.
(Choice B) B -7\leq x<12−7≤x<12minus, 7, is less than or equal to, x, is less than, 12.
(Choice C) C x<12x<12x, is less than, 12.
(Choice D) D There are no solutions .
(Choice E) E All values of xxx are solutions.

Respuesta :

Answer:

Option E.

Step-by-step explanation:

The given inequalities are

[tex]-15x+4\leq 109[/tex]

[tex]-6x+70>-2[/tex]

We need to find the solution of given compound inequality.

Solve each inequality separately.

Solve first inequality:

[tex]-15x+4\leq 109[/tex]

Subtract 4 from both sides.

[tex]-15x\leq 109-4[/tex]

[tex]-15x\leq 105[/tex]

Divide both sides by -15. If we multiply or divide an inequality by a negative number, then the sign of inequality is changed.

[tex]x\geq -7[/tex]              .... (1)

Solve second inequality:

[tex]-6x+70>-2[/tex]

Subtract 70 from both sides.

[tex]-6x>-2-70[/tex]

[tex]-6x>-72[/tex]

Divide both sides by -6.

[tex]x<12[/tex]                   .... (2)

Form (1) and (2) we get

[tex]x\geq -7\text{ or }x<12=\text{All real numbers}[/tex]

Therefore, the correct option is E.

Hi There Answer Choice E

Step-by-step explanation:

−15x+4≤109 OR −6x+70>−2minus, 15, x, plus, 4, is less than or equal to, 109, start color #ed5fa6, start text, space, O, R, space, end text, end color #ed5fa6, minus, 6, x, plus, 70, is greater than, minus, 2 Choose 1 answer: Choose 1 answer:

(Choice A) A x\geq-7x≥−7x, is greater than or equal to, minus, 7.

(Choice B) B -7\leq x<12−7≤x<12minus, 7, is less than or equal to, x, is less than, 12.

(Choice C) C x<12x<12x, is less than, 12.

(Choice D) D There are no solutions .

(Choice E) E All values of xxx are solutions.

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