Respuesta :
Let [tex] t [/tex] be the inital number of Timmy's marbles, and [tex] b [/tex] be the inital number of Ben's marbles.
At the beginning, "Timmy had 3/4 of what Ben had", which means
[tex] t = \cfrac{3}{4}b \iff 4t=3b[/tex]
(I wrote the equation in an equivalent form to avoid denominators)
Then, Ben gives 36 marbles to Timmy. So, now Ben has [tex] b-36 [/tex] marbles, and Timmy has [tex] t+36 [/tex] marbles. The new ratio is 1 to 3 in Timmy's favour, so we have
[tex] b-36 = \cfrac{t+36}{3} \iff 3(b-36) = t+36 \iff 3b-108 = t+36 [/tex]
(same as before)
So, we have the following linear system:
[tex] \begin{cases}4t=3b\\3b-108=t+36\end{cases} [/tex]
From the first equation we know that [tex] 3b=4t [/tex], so the second equation becomes
[tex] 4t-108=t+36 \iff 3t = 144 \iff t = \cfrac{144}{3} = 48 [/tex]
Now that we know the value for [tex] t [/tex], we can simply plug it into the first equation to get
[tex] 3b = 4\times 48 = 192 \iff b=\cfrac{192}{3} = 64 [/tex]
