Write a system of equations to describe the situation below, solve using elimination, and fill in the blanks. $\newline$ The receivers for the Kingwood University football team are practicing running different routes on the field. They have to run a specific distance so that the quarterback knows exactly where to throw the ball. Craig ran $10$ post routes and $28$ slant routes, which meant he ran a total of $374$ yards. Pedro ran $12$ post routes and $17$ slant routes, which equaled a total of $316$ yards. How long is each route? $\newline$ A post route is $\_$ yards long and a slant route is $\_$ yards long.

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Algebra 1

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

The receivers for the Kingwood University football team are practicing running different routes on the field. They have to run a specific distance so that the quarterback knows exactly where to throw the ball. Craig ran

10

post routes and

28

slant routes, which meant he ran a total of

374

yards. Pedro ran

12

post routes and

17

slant routes, which equaled a total of

316

yards. How long is each route?

\newline

A post route is

\_

yards long and a slant route is

\_

yards long.

Define variables: Let's define the variables for the lengths of the routes. Let $x$ be the length of a post route in yards, and $y$ be the length of a slant route in yards. $\newline$ Craig's running distance can be represented by the equation: $10x + 28y = 374$ . $\newline$ Pedro's running distance can be represented by the equation: $12x + 17y = 316$ .
Write equations: Write the system of equations based on the information given. $\newline$ For Craig: $10x + 28y = 374$ $\newline$ For Pedro: $12x + 17y = 316$
Use elimination method: To use elimination, we need to make the coefficients of one of the variables the same in both equations. We can multiply the entire first equation by $12$ and the second equation by $10$ to make the coefficients of $x$ the same. $\newline$ Multiplying the first equation by $12$ : $12(10x + 28y) = 12(374)$ $\newline$ Multiplying the second equation by $10$ : $10(12x + 17y) = 10(316)$
Perform multiplication: Perform the multiplication from Step $3$ . $\newline$ First equation after multiplication: $120x + 336y = 4488$ $\newline$ Second equation after multiplication: $120x + 170y = 3160$
Subtract equations: Subtract the second equation from the first equation to eliminate $x$ . $\$120x + 336y$ - $120x + 170y$ = $4488$ - $3160$ \) $120x + 336y - 120x - 170y = 4488 - 3160$ $336y - 170y = 4488 - 3160$ $166y = 1328$
Solve for y: Solve for y by dividing both sides of the equation by $166$ . $\newline$ $\frac{166y}{166} = \frac{1328}{166}$ $\newline$ $y = 8$
Substitute and solve for $x$ : Now that we have the value for $y$ , we can substitute it back into one of the original equations to solve for $x$ . Let's use Craig's equation: $10x + 28y = 374$ . $\newline$ Substitute $y = 8$ into the equation: $10x + 28(8) = 374$
Substitute and solve for x: Now that we have the value for y, we can substitute it back into one of the original equations to solve for x. Let's use Craig's equation: $10x + 28y = 374$ . Substitute $y = 8$ into the equation: $10x + 28(8) = 374$ Solve for x. $10x + 224 = 374$ $10x = 374 - 224$ $10x = 150$ $x = \frac{150}{10}$ $x = 15$