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I did this problem and i somehow got x=-36. help please? (Problem in comments)
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Solve for x in the proportion . x = 3 and x = −4 x = −3 and x = 4 x = −3 x = −12
cross multiply 3x^2 = -3x + 36
$$ \cfrac{3x}{3x-36}=\cfrac{-1}{x} \implies \cfrac{3x}{3(x-12)}=\cfrac{-1}{x}\\ \implies \cfrac{x}{x-12}=\cfrac{-1}{x}\\ \color{blue}{x(x-12)} \times \cfrac{x}{x-12}=\cfrac{-1}{x} \times \color{blue}{x(x-12)}\\ $$ what would that give you?
3x^2+3x-36= 0 now factor it 3(x^2+x-12 )=0 3(x+4)(x-3)= 0 x+4 = 0 or x-3=0 now solve for x
@Jim766 so x=-4 and x=3?
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right !
oh! okay! thanks!(:
yw
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