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Sagot :
Let R be the minutes Tisha runs and W be the minutes Tisha walks. Since she requires 24 minutes to do 6 laps running and 3 laps walking, we have the following equation:
[tex]6R+3W=24[/tex]Now, when she requires 20 minutes to do 6 laps running and 2 laps walking, we have:
[tex]6R+2W=20[/tex]So, we have the following system of equations:
[tex]\mleft\{\begin{aligned}6R+3W=24 \\ 6R+2W=20\end{aligned}\mright.[/tex]We are going to solve it by elimination. Notice that the coefficients of R are the same, so we just have to multiply one equation by -1 and add it altogether to the other equation:
[tex]\begin{gathered} (6R+2W=20)(-1) \\ \Rightarrow-6R-2W=-20 \end{gathered}[/tex]Then:
[tex]\begin{gathered} 6R+3W=24\rbrack \\ -6R-2W=20 \\ \Rightarrow(6R-6R)+(3W-2W)=24-20 \\ \Rightarrow W=4 \end{gathered}[/tex]Now we use the value W=4 to find R:
[tex]\begin{gathered} W=4 \\ 6R+3W=24 \\ \Rightarrow6R+3\cdot4=24 \\ \Rightarrow6R=24-12=12 \\ \Rightarrow6R=12 \\ \Rightarrow R=\frac{12}{6}=2 \\ R=2 \end{gathered}[/tex]Therefore, it takes Tisha to complete a lap 2 minutes if she's running and 4 minutes if she's walking.
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