How do you solve $\frac{3 x}{x + 9} + 1 = \frac{x}{x + 9}$ $(3x)/(x+9)+1=x/(x+9)$ and check for extraneous solutions?

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Answer 1 · 2016-11-23T07:05:40

$x = - 3$ $x=-3$

$\frac{3 x}{x + 9} + 1 = \frac{x}{x + 9}$ $(3x)/(x+9)+1=x/(x+9)$

Subtract $\frac{x}{x + 9}$ $x/(x+9)$ from both sides.

$\frac{3 x}{x + 9} - \frac{x}{x + 9} + 1 = 0$ $(3x)/(x+9)-x/(x+9)+1=0$

Subtract $1$ $1$ from both sides.

$\frac{3 x}{x + 9} - \frac{x}{x + 9} = - 1$ $(3x)/(x+9)-x/(x+9)=-1$

Combine the two terms on the left.

$\frac{3 x - x}{x + 9} = - 1$ $(3x-x)/(x+9)=-1$

$\frac{2 x}{x + 9} = - 1$ $(2x)/(x+9)=-1$

Multiply both sides by $(x + 9)$ $(x+9)$ .

$2 x = - 1 (x + 9)$ $2x=-1(x+9)$

$2 x = - x - 9$ $2x=-x-9$

Add $x$ $x$ to both sides.

$3 x = - 9$ $3x=-9$

Divide both sides by $3$ $3$ .

$x = - 3$ $x=-3$

How do you solve (3x)/(x+9)+1=x/(x+9)3xx+9+1=xx+9(3x)/(x+9)+1=x/(x+9) and check for extraneous solutions?