If \frac{x - 3}{4} = \frac{9}{x - 3} and x > 3, what is the value of x? — SAT

If $\frac{x - 3}{4} = \frac{9}{x - 3}$ and $x > 3$ , what is the value of $x$ ?

Answer: 9

Cross-multiplying the equation

\frac{x - 3}{4} = \frac{9}{x - 3}

gives

(x - 3)^2 = 36

. Taking the square root of both sides gives

x - 3 = 6

x - 3 = -6

. Solving for

x

yields

x = 9

x = -3

. Since the question specifies the constraint

x > 3

, the only valid solution is

9

Cross-multiply the equation to eliminate the denominators.

(x - 3)^2 = 36

Multiplying both sides of the equation by

4(x - 3)

simplifies the rational equation into a quadratic form.

Take the square root of both sides of the equation.

x - 3 = 6

x - 3 = -6

Taking the square root of a squared term yields both positive and negative root options.

Solve each linear equation for

x

and apply the constraint

x > 3

x = 9

Adding

3

to both sides gives

x = 9

x = -3

. The constraint

x > 3

excludes the negative solution, leaving

9

as the only valid value.

Solving rational equations by cross-multiplication and factoring the resulting quadratic equation while adhering to domain constraints.

Question