Write a system of equations to describe the situation below, solve using elimination, and fill in the blanks. $\newline$ Substitute teachers with Weston School District get paid by the day, although subs with teaching credentials earn a different amount than subs without credentials. Yesterday, $6$ non-credentialed subs and $3$ credentialed subs taught in the district. That cost the district $\$756$ . Today, $11$ non-credentialed subs and $3$ credentialed subs taught, receiving $\$1,151$ from the district. How much do subs get paid? $\newline$ Subs without credentials get paid $\$$ _ per day, and subs with credentials get paid $\$$ _ per day.

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Solve a system of equations using elimination: word problems

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Q. Write a system of equations to describe the situation below, solve using elimination, and fill in the blanks.

\newline

Substitute teachers with Weston School District get paid by the day, although subs with teaching credentials earn a different amount than subs without credentials. Yesterday,

6

non-credentialed subs and

3

credentialed subs taught in the district. That cost the district

\$756

. Today,

11

non-credentialed subs and

3

credentialed subs taught, receiving

\$1,151

from the district. How much do subs get paid?

\newline

Subs without credentials get paid

\$

_____ per day, and subs with credentials get paid

\$

_____ per day.

Write Equations: Let's denote the daily pay for non-credentialed subs as $x$ dollars and for credentialed subs as $y$ dollars. We can write two equations based on the given information: $\newline$ For yesterday: $\newline$ $6x + 3y = 756$ (Equation $1$ ) $\newline$ For today: $\newline$ $11x + 3y = 1151$ (Equation $2$ ) $\newline$ We will use the elimination method to solve this system of equations.
Eliminate y: First, we will multiply Equation $1$ by $-1$ to help eliminate the $y$ variable when we add it to Equation $2$ . $\newline$ $-1 \times (6x + 3y) = -1 \times 756$ $\newline$ $-6x - 3y = -756$ (Equation $3$ ) $\newline$ Now we have: $\newline$ $-6x - 3y = -756$ $\newline$ $11x + 3y = 1151$
Add Equations: Next, we add Equation $3$ to Equation $2$ to eliminate $y$ . $\newline$ $(-6x - 3y) + (11x + 3y) = (-756) + 1151$ $\newline$ $-6x + 11x = 1151 - 756$ $\newline$ $5x = 395$
Solve for x: Now we solve for x by dividing both sides of the equation by $5$ . $\newline$ $\frac{5x}{5} = \frac{395}{5}$ $\newline$ $x = 79$ $\newline$ So, non-credentialed subs get paid $\$79$ per day.
Substitute $x$ : We will now substitute the value of $x$ back into Equation $1$ to solve for $y$ . $\newline$ $6(79) + 3y = 756$ $\newline$ $474 + 3y = 756$
Solve for $y$ : Subtract $474$ from both sides of the equation to solve for $y$ . $\newline$ $3y = 756 - 474$ $\newline$ $3y = 282$
Solve for y: Subtract $474$ from both sides of the equation to solve for y. $\newline$ $3y = 756 - 474$ $\newline$ $3y = 282$ Finally, divide both sides by $3$ to find the value of $y$ . $\newline$ $\frac{3y}{3} = \frac{282}{3}$ $\newline$ $y = 94$ $\newline$ So, credentialed subs get paid $\$94$ per day.