Question

name
solve each system by substitution.

1. $3x + 4y = 11$

$y = -4x + 6$
$3x + 4(-4x+6)=11$
$3x + -16x+24 = 11$
$-13x +24=11$
$-13x = -13$
$x = 1$
$y=-4(1)+6$

1. $y = -4x - 1$

$3x + 3y = 6$
$3x+3(-4x-1)=6$
$3x + -12x-3 = 6$
$-9x -3 = 6$
$-9x = 9$
$x=-1$
$y=-4(-1)-1$
$y=3$

1. $y = 2x - 8$

$2x + 4y = -12$
$2x+4(2x-8)=-12$
$2x + 8x-32 = -12$
$10x = 20$
$x=2$
$y=2(2)-8$
$y=-4$

1. $-3x + 4y = 8$

$y = -4x + 2$
$-3x+4(-4x+2)=8$
$-3x + -16x+8 = 8$
$-19x = 0$
$x=0$

Explanation:

Step1: Substitute $y$ into first equation

For 1): $3x + 4(-4x+6)=11$
For 2): $3x + 3(-4x-1)=6$
For 3): $2x + 4(2x-8)=-12$
For 4): $-3x + 4(-4x+2)=8$

Step2: Simplify and solve for $x$

For 1):
$3x-16x+24=11$
$-13x=11-24$
$-13x=-13$
$x=1$

For 2):
$3x-12x-3=6$
$-9x=6+3$
$-9x=9$
$x=-1$

For 3):
$2x+8x-32=-12$
$10x=-12+32$
$10x=20$
$x=2$

For 4):
$-3x-16x+8=8$
$-19x=8-8$
$-19x=0$
$x=0$

Step3: Substitute $x$ to find $y$

For 1): $y=-4(1)+6=2$
For 2): $y=-4(-1)-1=3$
For 3): $y=2(2)-8=-4$
For 4): $y=-4(0)+2=2$

Answer:

1. $x=1$, $y=2$
2. $x=-1$, $y=3$
3. $x=2$, $y=-4$
4. $x=0$, $y=2$