Mrs. Griffith, the P.E. teacher, is pairing off students to race against each other. Dominic can run 4 meters per second, and Seth can run 6 meters per second. Mrs. Griffith decides to give Dominic a head start of 30 meters since he runs more slowly. Once the students start running, Seth will quickly catch up to Dominic. How long will that take? Write a system of equations, graph them, and type the solution. ____ seconds

Set Up Equations: Let's set up the equations for Dominic and Seth's running distances over time. Dominic's equation accounts for his head start: Dominic's distance = 4t + 30 (since he starts 30 meters ahead and runs at 4 meters per second), Seth's distance = 6t (since he starts from 0 meters and runs at 6 meters per second), We need to find the time $ t $ when their distances are equal, meaning they meet. Solve for t: Set the equations equal to solve for $ t $: 4t + 30 = 6t, Subtract 4t from both sides: 30 = 2t, Divide both sides by 2: t = 15. Check Solution: Check the solution by plugging $ t = 15 $ back into the original distance equations: Dominic's distance = 4(15) + 30 = 90 meters, Seth's distance = 6(15) = 90 meters, Both distances are equal, so the solution is correct.

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Mrs. Griffith, the P.E. teacher, is pairing off students to race against each other. Dominic can run $4 \, \text{meters per second}$ , and Seth can run $6 \, \text{meters per second}$ . Mrs. Griffith decides to give Dominic a head start of $30 \, \text{meters}$ since he runs more slowly. Once the students start running, Seth will quickly catch up to Dominic. How long will that take? $\newline$ Write a system of equations, graph them, and type the solution. $\newline$ $\_\_\_\_$ seconds $\newline$

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Algebra 1

Solve a system of equations using elimination: word problems

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Q. Mrs. Griffith, the P.E. teacher, is pairing off students to race against each other. Dominic can run

4 \, \text{meters per second}

, and Seth can run

6 \, \text{meters per second}

. Mrs. Griffith decides to give Dominic a head start of

30 \, \text{meters}

since he runs more slowly. Once the students start running, Seth will quickly catch up to Dominic. How long will that take?

\newline

Write a system of equations, graph them, and type the solution.

\newline

\_\_\_\_

seconds

\newline

Set Up Equations: Let's set up the equations for Dominic and Seth's running distances over time. Dominic's equation accounts for his head start: $\newline$ Dominic's distance = $4$ t + $30$ (since he starts $30$ meters ahead and runs at $4$ meters per second), $\newline$ Seth's distance = $6$ t (since he starts from $0$ meters and runs at $6$ meters per second), $\newline$ We need to find the time $t$ when their distances are equal, meaning they meet.
Solve for t: Set the equations equal to solve for $t$ : $\newline$ $4$ t + $30$ = $6$ t, $\newline$ Subtract $4$ t from both sides: $\newline$ $30$ = $2$ t, $\newline$ Divide both sides by $2$ : $\newline$ t = $15$ .
Check Solution: Check the solution by plugging $t = 15$ back into the original distance equations: $\newline$ Dominic's distance = $4$ ( $15$ ) + $30$ = $90$ meters, $\newline$ Seth's distance = $6$ ( $15$ ) = $90$ meters, $\newline$ Both distances are equal, so the solution is correct.