Maths Olympiad Prep

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Problem 978

AMC 12 late, AIME early
Algebra Difficulty 4.8 Find the answer HMMT November

Compute xw\frac{x}{w} if w0w \neq 0 and x+6y3z3x+4w=2y+zxw=23\frac{x+6 y-3 z}{-3 x+4 w}=\frac{-2 y+z}{x-w}=\frac{2}{3}.

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

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Official solution

We have x+6y3z=23(3x+4w)x+6 y-3 z=\frac{2}{3}(-3 x+4 w) and 2y+z=23(xw)-2 y+z=\frac{2}{3}(x-w), so xw=(x+6y3z)+3(2y+z)(3x+4w)+3(xw)=23(3x+4w)+323(xw)(3x+4w)+3(xw)=23[(3x+4w)+3(xw)](3x+4w)+3(xw)=23\frac{x}{w}=\frac{(x+6 y-3 z)+3(-2 y+z)}{(-3 x+4 w)+3(x-w)}=\frac{\frac{2}{3}(-3 x+4 w)+3 \cdot \frac{2}{3}(x-w)}{(-3 x+4 w)+3(x-w)}=\frac{\frac{2}{3}[(-3 x+4 w)+3(x-w)]}{(-3 x+4 w)+3(x-w)}=\frac{2}{3}

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