Respuesta :
Let's say first that duration of plan A is a, and duration for plan B is b.
We can now establish 2 equations. One for Wednesday and one for Thursday.
On Wednesday, frank trained clients for a total of 7 hours, being 3 clients with plan A, and 8 with plan B. Then:
[tex]3\cdot a+8\cdot b=7[/tex]On Thursday, frank trained clients for 6 hours, being 5 clients with plan A, and 2 clients with plan B. Then:
[tex]5\cdot a+2\cdot b=6[/tex]Now we have a system of 2 equations and 2 unknowns:
[tex]\begin{gathered} 3a+8b=7 \\ 5a+2b=6 \end{gathered}[/tex]It will be easy to solve the system by reduction method. We can eliminate terms 8b and 2b mmultiplying the second equation by -4:
[tex]\begin{gathered} -4\cdot(5a+2b)=-4\cdot6 \\ -20a-8b=-24 \end{gathered}[/tex]Now, the system is:
[tex]\begin{gathered} 3a+8b=7 \\ -20a-8b=-24 \end{gathered}[/tex]Combining two equations (adding them), we have:
[tex]\begin{gathered} (3-20)a+(8-8)b=7-24 \\ -17a+0b=-17 \\ -17a=-17 \\ a=\frac{-17}{-17} \\ a=1 \end{gathered}[/tex]Now, we know that plan A lasts 1 hour.
We can use any equation to solve for b, knowing that a = 1.
[tex]\begin{gathered} 5a+2b=6 \\ 5\cdot(1)+2b=6 \\ 5+2b=6 \\ 2b=6-5 \\ 2b=1 \\ b=\frac{1}{2} \end{gathered}[/tex]Now, we know plan B lasts half an hour (1/2 hour).
