Question
Solve for $x; \frac{16}{x}-1=\frac{15}{x+1} ; x \neq 0,-1$
✓
Answer
Consider the equation:
$\frac{16}{x}-1=\frac{15}{x+1}$
$\Rightarrow \frac{16}{x}-\frac{15}{x+1}=1$
$\frac{16(x+1)-15 x}{x(x+1)}=1$
$\Rightarrow 16 x+16-15 x=x(x+1)$
$\Rightarrow x+16=x^2+x$
$\Rightarrow x^2=16$
Taking square root,
$x= \pm 4$
Therefore the solutions are $x= \pm 4$.
$\frac{16}{x}-1=\frac{15}{x+1}$
$\Rightarrow \frac{16}{x}-\frac{15}{x+1}=1$
$\frac{16(x+1)-15 x}{x(x+1)}=1$
$\Rightarrow 16 x+16-15 x=x(x+1)$
$\Rightarrow x+16=x^2+x$
$\Rightarrow x^2=16$
Taking square root,
$x= \pm 4$
Therefore the solutions are $x= \pm 4$.
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