Solve g/9g+49=1/g+9

$$\frac { g } { 9 g + 49 } = \frac { 1 } { g + 9 }$$

Solve for g

$$g=-7$$

$$g=7$$

Steps by Finding Square Root

Variable g cannot be equal to any of the values -9,-\frac{49}{9} since division by zero is not defined. Multiply both sides of the equation by \left(g+9\right)\left(9g+49\right), the least common multiple of 9g+49,g+9.

$$\left(g+9\right)g=9g+49$$

Use the distributive property to multiply g+9 by g.

$$g^{2}+9g=9g+49$$

Subtract 9g from both sides.

$$g^{2}+9g-9g=49$$

Combine 9g and -9g to get 0.

$$g^{2}=49$$

Take the square root of both sides of the equation.

$$g=7 g=-7$$

Steps Using the Quadratic Formula

Variable g cannot be equal to any of the values -9,-\frac{49}{9} since division by zero is not defined. Multiply both sides of the equation by \left(g+9\right)\left(9g+49\right), the least common multiple of 9g+49,g+9.

$$\left(g+9\right)g=9g+49$$

Use the distributive property to multiply g+9 by g.

$$g^{2}+9g=9g+49$$

Subtract 9g from both sides.

$$g^{2}+9g-9g=49$$

Combine 9g and -9g to get 0.

$$g^{2}=49$$

Subtract 49 from both sides.

$$g^{2}-49=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 0 for b, and -49 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$g=\frac{0±\sqrt{0^{2}-4\left(-49\right)}}{2}$$

Square 0.

$$g=\frac{0±\sqrt{-4\left(-49\right)}}{2}$$

Multiply -4 times -49.

$$g=\frac{0±\sqrt{196}}{2}$$

Take the square root of 196.

$$g=\frac{0±14}{2}$$

Now solve the equation g=\frac{0±14}{2} when ± is plus. Divide 14 by 2.

$$g=7$$

Now solve the equation g=\frac{0±14}{2} when ± is minus. Divide -14 by 2.

$$g=-7$$

The equation is now solved.

$$g=7 g=-7$$