Solve for x (x^2)/(x^2-4)=(5x-6)/(x^2-4)

$\frac{x^{2}}{x^{2} - 4} = \frac{5 x - 6}{x^{2} - 4}$

Step 1

Since the expression on each side of the equation has the same denominator, the numerators must be equal.

$x^{2} = 5 x - 6$

Step 2

Subtract $5 x$ from both sides of the equation.

$x^{2} - 5 x = - 6$

Step 3

Add $6$ to both sides of the equation.

$x^{2} - 5 x + 6 = 0$

Step 4

Factor $x^{2} - 5 x + 6$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $6$ and whose sum is $- 5$ .

$- 3, - 2$

Write the factored form using these integers.

$(x - 3) (x - 2) = 0$

Step 5

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x - 3 = 0$

$x - 2 = 0$

Step 6

Set $x - 3$ equal to $0$ and solve for $x$ .

Set $x - 3$ equal to $0$ .

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

Step 7

Set $x - 2$ equal to $0$ and solve for $x$ .

Set $x - 2$ equal to $0$ .

$x - 2 = 0$

Add $2$ to both sides of the equation.

$x = 2$

Step 8

The final solution is all the values that make $(x - 3) (x - 2) = 0$ true.

$x = 3, 2$

Step 9

Exclude the solutions that do not make $\frac{x^{2}}{x^{2} - 4} = \frac{5 x - 6}{x^{2} - 4}$ true.

$x = 3$

Solve for x (x^2)/(x^2-4)=(5x-6)/(x^2-4)

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