Consider the equations y = |x − 1| and y = 3x + 2. The approximate solution of this system of equations is

. + λrh() e 1 1 2 4 4 4 4 4 3 4 4 4 4 4 + th−1(− s(x)2 dx ω + λrh−1() e 2 1 2 4 4

step-by-step explanation:

it would open into a square depending on the type of envelope i’m not sure about the second part