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Which is the solution set to the given inequality |x+3|<12? x infinity(-15,9), x infinity {-15,9], x infinity (00,-15)U(9,00), x infinity (-00,-15)n(9,00)

Which ordered pair is a solution to the system?
{y>-2
{x + y less than or equal to 4

(1,5)
(0,5)
(-2,-3)
(1,3)

I really need help I am failing algebra II and just can't seem to understand it or graphing

Respuesta :

calculista

Answer:

Part 1) The solution set is (-15,∞) ∩ (-∞,9)=(-15,9)

Part 2) The ordered pair (1,3) is a solution of the system

Step-by-step explanation:

Part 1) we have

[tex]\left|x+3\right|<12[/tex]

First solution case Positive

[tex]+(x+3)<12[/tex]

[tex]x<12-3[/tex]

[tex]x<9[/tex]

The solution first case is the interval -------> (-∞,9)

Second solution case Negative

[tex]-(x+3)<12[/tex]

[tex]-x-3<12[/tex]

[tex]-x<12+3[/tex]

[tex]-x<15[/tex] ------> Multiply by -1 both sides

[tex]x>-15[/tex]

The solution second case is the interval -------> (-15,∞)

The solution set is equal to

(-15,∞) ∩ (-∞,9)=(-15,9)  

Part 2) we have

[tex]y>-2[/tex] -------> inequality A

[tex]x+y\leq 4[/tex] -----> inequality B

we know that

If a ordered pair is a solution of the system of inequalities, then the ordered pair must satisfy both inequalities

Verify each case

case a) (1,5)

Substitute the value of x and the value of y in the inequality and then compare

Inequality A

[tex]5>-2[/tex] ------> is true

Inequality B

[tex]1+5\leq 4[/tex]

[tex]6\leq 4[/tex] -----> is not true

therefore

the ordered pair is not a solution

case b) (0,5)

Substitute the value of x and the value of y in the inequality and then compare

Inequality A

[tex]5>-2[/tex] ------> is true

Inequality B

[tex]0+5\leq 4[/tex]

[tex]5\leq 4[/tex] -----> is not true

therefore

the ordered pair is not a solution

case c) (-2,-3)

Substitute the value of x and the value of y in the inequality and then compare

Inequality A

[tex]-3>-2[/tex] ------> is not true

therefore

the ordered pair is not a solution

case d) (1,3)

Substitute the value of x and the value of y in the inequality and then compare

Inequality A

[tex]3>-2[/tex] ------> is true

Inequality B

[tex]1+3\leq 4[/tex]

[tex]4\leq 4[/tex] -----> is true

therefore

the ordered pair is  a solution

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