Question

unit 3, lesson 6: elimination or substitution?
learning focus: identify when to use elimination or substitution to solve a system of equations.
solve each system using any method to solve the system. circle which method you decided to use. graph the to check your solution.

1. \\(\

$$\begin{cases} 3x + 4y = 23 \\\\ 5x + 3y = 31 \\end{cases}$$

\\)

Explanation:

Step1: Scale equations for elimination

Multiply first equation by 3: $3(3x + 4y) = 3(23) \implies 9x + 12y = 69$
Multiply second equation by -4: $-4(5x + 3y) = -4(31) \implies -20x - 12y = -124$

Step2: Add equations to eliminate y

Add the two new equations:
$(9x + 12y) + (-20x - 12y) = 69 + (-124)$
$-11x = -55$

Step3: Solve for x

Divide both sides by -11:
$x = \frac{-55}{-11} = 5$

Step4: Substitute x into first equation

Plug $x=5$ into $3x + 4y = 23$:
$3(5) + 4y = 23 \implies 15 + 4y = 23$

Step5: Solve for y

Subtract 15, then divide by 4:
$4y = 23 - 15 = 8 \implies y = \frac{8}{4} = 2$

Step6: Verify the solution

Check in second equation $5x + 3y = 31$:
$5(5) + 3(2) = 25 + 6 = 31$, which matches.

Answer:

$x=5$, $y=2$