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Mathematics 17 Online
OpenStudy (anonymous):

I did this problem and i somehow got x=-36. help please? (Problem in comments)

OpenStudy (anonymous):

Solve for x in the proportion . x = 3 and x = −4 x = −3 and x = 4 x = −3 x = −12

OpenStudy (jim766):

cross multiply 3x^2 = -3x + 36

OpenStudy (jdoe0001):

$$ \cfrac{3x}{3x-36}=\cfrac{-1}{x} \implies \cfrac{3x}{3(x-12)}=\cfrac{-1}{x}\\ \implies \cfrac{x}{x-12}=\cfrac{-1}{x}\\ \color{blue}{x(x-12)} \times \cfrac{x}{x-12}=\cfrac{-1}{x} \times \color{blue}{x(x-12)}\\ $$ what would that give you?

OpenStudy (jim766):

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

OpenStudy (anonymous):

@Jim766 so x=-4 and x=3?

OpenStudy (jim766):

right !

OpenStudy (anonymous):

oh! okay! thanks!(:

OpenStudy (jim766):

yw

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