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:Part 1) The solution set is (-15,∞) ∩ (-∞,9)=(-15,9)Part 2) The ordered pair (1,3) is a solution of the systemStep-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 Bwe know thatIf a ordered pair is a solution of the system of inequalities, then the ordered pair must satisfy both inequalitiesVerify each casecase 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 trueInequality B[tex]1+5\leq 4[/tex][tex]6\leq 4[/tex] -----> is not truethereforethe 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 trueInequality B[tex]0+5\leq 4[/tex][tex]5\leq 4[/tex] -----> is not truethereforethe 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 truethereforethe 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 trueInequality B[tex]1+3\leq 4[/tex][tex]4\leq 4[/tex] -----> is truethereforethe ordered pair is  a solution