Solve f(x)=-9x/7x-2quadg(x)=-4/x

$$f ( x ) = - \frac { 9 x } { 7 x - 2 } \quad g ( x ) = - \frac { 4 } { x }$$

Solve for g

$$g=\frac{4\left(7x-2\right)}{9x^{3}}$$

$$x\neq \frac{2}{7}\text{ and }x\neq 0$$

Steps for Solving Linear Equation

Multiply both sides of the equation by x\left(7x-2\right), the least common multiple of 7x-2,x.

$$\left(-\frac{9x}{7x-2}\right)gxx\left(7x-2\right)=-\left(7x-2\right)\times 4$$

Multiply x and x to get x^{2}.

$$\left(-\frac{9x}{7x-2}\right)gx^{2}\left(7x-2\right)=-\left(7x-2\right)\times 4$$

Express \left(-\frac{9x}{7x-2}\right)g as a single fraction.

$$\frac{-9xg}{7x-2}x^{2}\left(7x-2\right)=-\left(7x-2\right)\times 4$$

Express \frac{-9xg}{7x-2}x^{2} as a single fraction.

$$\frac{-9xgx^{2}}{7x-2}\left(7x-2\right)=-\left(7x-2\right)\times 4$$

Express \frac{-9xgx^{2}}{7x-2}\left(7x-2\right) as a single fraction.

$$\frac{-9xgx^{2}\left(7x-2\right)}{7x-2}=-\left(7x-2\right)\times 4$$

Cancel out 7x-2 in both numerator and denominator.

$$-9gxx^{2}=-\left(7x-2\right)\times 4$$

To multiply powers of the same base, add their exponents. Add 1 and 2 to get 3.

$$-9gx^{3}=-\left(7x-2\right)\times 4$$

Use the distributive property to multiply 7x-2 by 4.

$$-9gx^{3}=-\left(28x-8\right)$$

To find the opposite of 28x-8, find the opposite of each term.

$$-9gx^{3}=-28x+8$$

The equation is in standard form.

$$\left(-9x^{3}\right)g=8-28x$$

Divide both sides by -9x^{3}.

$$\frac{\left(-9x^{3}\right)g}{-9x^{3}}=\frac{8-28x}{-9x^{3}}$$

Dividing by -9x^{3} undoes the multiplication by -9x^{3}.

$$g=\frac{8-28x}{-9x^{3}}$$

Divide -28x+8 by -9x^{3}.

$$g=-\frac{4\left(2-7x\right)}{9x^{3}}$$