Question

Which is the approximate solution for the system of equations 3x+4y=2 and 4x+y=-4

Answer 1

$3x+4y=2\qquad\text{subtract 3x from both sides}\\\\4y=-3x+2\qquad\text{divide both sides by 4}\\\\y=-\dfrac{3}{4}x+\dfrac{2}{4}\to y=-\dfrac{3}{4}x+\dfrac{1}{2}\\\\for\ x=0\to y=-\dfrac{3}{4}(0)+\dfrac{1}{2}=0+\dfrac{1}{2}=\dfrac{1}{2}\to\left(0,\ \dfrac{1}{2}\right)\\\\for\ x=2\to y=-\dfrac{3}{4}(2)+\dfrac{1}{2}=-\dfrac{3}{2}+\dfrac{1}{2}=-\dfrac{2}{2}=-1\to(2,\ -1)\\--------------------------------$

$4x+y=-4\qquad\text{subtract 4x from both sides}\\\\y=-4x-4\\\\for\ x=0\to y=-4(0)-4=0-4=-4\to(0,\ -4)\\\\for\ x=-1\to y=-4(-1)-4=4-4=0\to(-1,\ 0)\\\\Answer:\ \boxed{A.\ (-1.4,\ 1.5)}$

Answer 2

Answer:

A: -1.4, 1.5

Step-by-step explanation: