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Solve for x. 1/x + 1/x+16= 1/15
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\[\frac{1}{x}+\frac{1}{x+16}=\frac{1}{15}\] \[15(x+16)+15x=x(x+16)\] \[15x+240+15x=x^2+16x\] \[30x+240=x^2+16x\] \[0=x^2+16x-30x-240\] \[0=x^2-14x-240\] \[x^2-14x-240=0\] \[(x-24)(x+10)=0\] \[x-24=0 \ \textrm{or} \ x+10=0\] \[x=24 \ \textrm{or} \ x=-10\] I'll leave the check to you.
Thanks Jim
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