Question

directions: solve each system by elimination.

1. $\

$$\begin{cases} y=3x+4 \\\\ y=x-2 \\end{cases}$$

$

1. $\

$$\begin{cases} x+4y=13 \\\\ x-y=3 \\end{cases}$$

$

1. $\

$$\begin{cases} 3x-10y=14 \\\\ 3x-9y=15 \\end{cases}$$

$

1. $\

$$\begin{cases} 4x+2y=6 \\\\ -2x+2y=18 \\end{cases}$$

$

Explanation:

Problem 1

Step1: Set equations equal

$3x + 4 = x - 2$

Step2: Isolate x terms

$3x - x = -2 - 4$

Step3: Solve for x

$2x = -6 \implies x = -3$

Step4: Substitute x to find y

$y = (-3) - 2 = -5$

Problem 2

Step1: Subtract equations

$(x + 4y) - (x - y) = 13 - 3$

Step2: Simplify to solve y

$5y = 10 \implies y = 2$

Step3: Substitute y to find x

$x - 2 = 3 \implies x = 5$

Problem 3

Step1: Subtract equations

$(3x - 9y) - (3x - 10y) = 15 - 14$

Step2: Simplify to solve y

$y = 1$

Step3: Substitute y to find x

$3x - 9(1) = 15 \implies 3x = 24 \implies x = 8$

Problem 4

Step1: Subtract equations

$(4x + 2y) - (-2x + 2y) = 6 - 18$

Step2: Simplify to solve x

$6x = -12 \implies x = -2$

Step3: Substitute x to find y

$4(-2) + 2y = 6 \implies -8 + 2y = 6 \implies 2y = 14 \implies y = 7$

Answer:

1. $x=-3$, $y=-5$
2. $x=5$, $y=2$
3. $x=8$, $y=1$
4. $x=-2$, $y=7$