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:
![6R+3W=24](https://img.qammunity.org/2023/formulas/mathematics/college/lpoj5uw233grthk4r1owqvy5bb5sc9yt0e.png)
Now, when she requires 20 minutes to do 6 laps running and 2 laps walking, we have:
![6R+2W=20](https://img.qammunity.org/2023/formulas/mathematics/college/wygi4s7hfghcgqiyeth6t57hozs45h5lxn.png)
So, we have the following system of equations:
![\mleft\{\begin{aligned}6R+3W=24 \\ 6R+2W=20\end{aligned}\mright.](https://img.qammunity.org/2023/formulas/mathematics/college/uc0cizv1la5iyztcnpy9badyga7lgdms6t.png)
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:
![\begin{gathered} (6R+2W=20)(-1) \\ \Rightarrow-6R-2W=-20 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/covbki7wg9uksby9nxrll32d445yb8ccc7.png)
Then:
![\begin{gathered} 6R+3W=24\rbrack \\ -6R-2W=20 \\ \Rightarrow(6R-6R)+(3W-2W)=24-20 \\ \Rightarrow W=4 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/43251axjgcj39d6zn5lrvibvaxndnccn2i.png)
Now we use the value W=4 to find R:
![\begin{gathered} W=4 \\ 6R+3W=24 \\ \Rightarrow6R+3\cdot4=24 \\ \Rightarrow6R=24-12=12 \\ \Rightarrow6R=12 \\ \Rightarrow R=(12)/(6)=2 \\ R=2 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/p5i5l1i9kokvuoyadqzcxpo3idg0i0dpr1.png)
Therefore, it takes Tisha to complete a lap 2 minutes if she's running and 4 minutes if she's walking.