Question

solve the system using the substitution method.

1. $\

$$\begin{cases}3x + y = -1\\\\2x + 3y = 18\\end{cases}$$

$

1. $\

$$\begin{cases}3x + 2y = 6\\\\x - 4y = -12\\end{cases}$$

$

Explanation:

For System 1: $3x + y = -1$; $2x + 3y = 18$

Step1: Isolate $y$ from first eq.

$y = -1 - 3x$

Step2: Substitute $y$ into second eq.

$2x + 3(-1 - 3x) = 18$

Step3: Simplify to solve for $x$

$2x - 3 - 9x = 18$
$-7x = 21$
$x = -3$

Step4: Substitute $x$ back to find $y$

$y = -1 - 3(-3) = -1 + 9 = 8$

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For System 2: $3x + 2y = 6$; $x - 4y = -12$

Step1: Isolate $x$ from second eq.

$x = 4y - 12$

Step2: Substitute $x$ into first eq.

$3(4y - 12) + 2y = 6$

Step3: Simplify to solve for $y$

$12y - 36 + 2y = 6$
$14y = 42$
$y = 3$

Step4: Substitute $y$ back to find $x$

$x = 4(3) - 12 = 12 - 12 = 0$

Answer:

1. $x = -3$, $y = 8$
2. $x = 0$, $y = 3$