Solve for x (x^2+2x-3)/(2x^2+x-3)=0

$\frac{x^{2} + 2 x - 3}{2 x^{2} + x - 3} = 0$

Set the numerator equal to zero.

$x^{2} + 2 x - 3 = 0$

Solve the equation for $x$ .

Factor $x^{2} + 2 x - 3$ 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 $- 3$ and whose sum is $2$ .

$- 1, 3$

Write the factored form using these integers.

$(x - 1) (x + 3) = 0$

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

$x - 1 = 0$

$x + 3 = 0$

Set $x - 1$ equal to $0$ and solve for $x$ .

Set $x - 1$ equal to $0$ .

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

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

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

$x + 3 = 0$

Subtract $3$ from both sides of the equation.

$x = - 3$

The final solution is all the values that make $(x - 1) (x + 3) = 0$ true.

$x = 1, - 3$

Exclude the solutions that do not make $\frac{x^{2} + 2 x - 3}{2 x^{2} + x - 3} = 0$ true.

$x = - 3$

Solve for x (x^2+2x-3)/(2x^2+x-3)=0

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