The figure is a trapezoid
Let's say Diana is "x" years old today.
in 3 years, she's going to be " x + 3 " years old.
33 years ago, she was " x - 33 " old.
how much is 4 times that anyway? well 4(x-33), that's 4 times as 33 years ago her age.
now, we know in 3 years, x+3, she'll be "<span>4 times as old as she was 33 years ago", 4(x-33), therefore, those amounts are equal then,
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![\bf \stackrel{\textit{3 years from now}}{x+3}~~=~~\stackrel{\textit{4 times as old as she was 33 years ago}}{4(x-33)} \\\\\\ x+3=4x-132\implies 135=3x\implies \cfrac{135}{3}=x\implies 45=x](https://tex.z-dn.net/?f=%5Cbf%20%5Cstackrel%7B%5Ctextit%7B3%20years%20from%20now%7D%7D%7Bx%2B3%7D~~%3D~~%5Cstackrel%7B%5Ctextit%7B4%20times%20as%20old%20as%20she%20was%2033%20years%20ago%7D%7D%7B4%28x-33%29%7D%0A%5C%5C%5C%5C%5C%5C%0Ax%2B3%3D4x-132%5Cimplies%20135%3D3x%5Cimplies%20%5Ccfrac%7B135%7D%7B3%7D%3Dx%5Cimplies%2045%3Dx)
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why is it most logical to use "x" for her age today?
well, because 3 years from now is just x + 3
and
33 years ago is x - 33, 4 times that is 4( x - 33 ).
works very well as a point of reference.</span>
X is always dependant to Y. so that should answer the 1st one. the 2nd looks to be around 17-18. 17.5 maybe. and the 3rd is yes cause he does not go to the pool so it isnt any cost.