Question

solve the system using the substitution method.

1. $\

$$\begin{cases}2x + 5y = 7 \\\\ x + 4y = 2\\end{cases}$$

$ \quad \quad \quad \quad \quad 2. $\

$$\begin{cases}3x + y = 16 \\\\ 2x - 3y = -4\\end{cases}$$

$

solve the system using the elimination method.

1. $\

$$\begin{cases}4x - 2y = -16 \\\\ -3x + 4y = 12\\end{cases}$$

$ \quad \quad \quad \quad \quad 4. $\

$$\begin{cases}2x + 5y = 14 \\\\ 3x - 2y = -36\\end{cases}$$

$

solve the system using any algebraic method.

1. $\

$$\begin{cases}3x + 7y = -1 \\\\ 2x + 3y = 6\\end{cases}$$

$ \quad \quad \quad \quad \quad 6. $\

$$\begin{cases}6x - 2y = 5 \\\\ -3x + y = 7\\end{cases}$$

$

Explanation:

System 1 (Substitution Method)

Step1: Isolate $x$ from second equation

$x = 2 - 4y$

Step2: Substitute $x$ into first equation

$2(2 - 4y) + 5y = 7$

Step3: Simplify to solve for $y$

$4 - 8y + 5y = 7 \implies -3y = 3 \implies y = -1$

Step4: Substitute $y=-1$ into $x=2-4y$

$x = 2 - 4(-1) = 6$

Step1: Isolate $y$ from first equation

$y = 16 - 3x$

Step2: Substitute $y$ into second equation

$2x - 3(16 - 3x) = -4$

Step3: Simplify to solve for $x$

$2x - 48 + 9x = -4 \implies 11x = 44 \implies x=4$

Step4: Substitute $x=4$ into $y=16-3x$

$y = 16 - 3(4) = 4$

Step1: Multiply first equation by 2

$8x - 4y = -32$

Step2: Add to second equation to eliminate $y$

$(8x - 4y) + (-3x + 4y) = -32 + 12$

Step3: Solve for $x$

$5x = -20 \implies x=-4$

Step4: Substitute $x=-4$ into first equation

$4(-4) - 2y = -16 \implies -16 -2y = -16 \implies y=0$

Answer:

$x=6$, $y=-1$

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System 2 (Substitution Method)