Question

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

Answer 1

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

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

Explanation:

Part 1) we have

`\left|x+3\right|<12`

First solution case Positive

`+(x+3)<12`

`x<12-3`

`x<9`

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

Second solution case Negative

`-(x+3)<12`

`-x-3<12`

`-x<12+3`

`-x<15` ------> Multiply by -1 both sides

`x>-15`

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

The solution set is equal to

(-15,∞) ∩ (-∞,9)=(-15,9)

Part 2) we have

`y>-2` -------> inequality A

`x+y\leq 4` -----> 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

`5>-2` ------> is true

Inequality B

`1+5\leq 4`

`6\leq 4` -----> 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

`5>-2` ------> is true

Inequality B

`0+5\leq 4`

`5\leq 4` -----> 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

`-3>-2` ------> 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

`3>-2` ------> is true

Inequality B

`1+3\leq 4`

`4\leq 4` -----> is true

therefore

the ordered pair is a solution