Solve the Rational Equation for x (x^2-10)/(x-1)=(-14-5x)/(x-1)

$\frac{x^{2} - 10}{x - 1} = \frac{- 14 - 5 x}{x - 1}$

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

$x^{2} - 10 = - 14 - 5 x$

Add $5 x$ to both sides of the equation.

$x^{2} - 10 + 5 x = - 14$

Add $14$ to both sides of the equation.

$x^{2} - 10 + 5 x + 14 = 0$

Add $- 10$ and $14$ .

$x^{2} + 5 x + 4 = 0$

Factor $x^{2} + 5 x + 4$ 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 $4$ and whose sum is $5$ .

$1, 4$

Write the factored form using these integers.

$(x + 1) (x + 4) = 0$

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

$x + 1 = 0$

$x + 4 = 0$

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

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

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

Set $x + 4$ equal to $0$ and solve for $x$ .

Set $x + 4$ equal to $0$ .

$x + 4 = 0$

Subtract $4$ from both sides of the equation.

$x = - 4$

The final solution is all the values that make $(x + 1) (x + 4) = 0$ true.

$x = - 1, - 4$

Solve the Rational Equation for x (x^2-10)/(x-1)=(-14-5x)/(x-1)

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