Solve 3/(x - 6) - 4/(x - 5) = 1 | Equatix - Math Solver

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Example

3(r+2s)=2t-4

a⋅(b-2)=3b

3(r+2s)=2t-4

a⋅(b-2)=3b

Question

$$\frac{3}{x-6}-\frac{4}{x-5}=1$$

Solve for x

$x=3$
$x=7$

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using the Quadratic Formula

Variable $x$ cannot be equal to any of the values $5,6$ since division by zero is not defined. Multiply both sides of the equation by $\left(x-6\right)\left(x-5\right)$, the least common multiple of $x-6,x-5$.

$$\left(x-5\right)\times 3-\left(x-6\right)\times 4=\left(x-6\right)\left(x-5\right)$$

Use the distributive property to multiply $x-5$ by $3$.

$$3x-15-\left(x-6\right)\times 4=\left(x-6\right)\left(x-5\right)$$

Use the distributive property to multiply $x-6$ by $4$.

$$3x-15-\left(4x-24\right)=\left(x-6\right)\left(x-5\right)$$

To find the opposite of $4x-24$, find the opposite of each term.

$$3x-15-4x+24=\left(x-6\right)\left(x-5\right)$$

Combine $3x$ and $-4x$ to get $-x$.

$$-x-15+24=\left(x-6\right)\left(x-5\right)$$

Add $-15$ and $24$ to get $9$.

$$-x+9=\left(x-6\right)\left(x-5\right)$$

Use the distributive property to multiply $x-6$ by $x-5$ and combine like terms.

$$-x+9=x^{2}-11x+30$$

Subtract $x^{2}$ from both sides.

$$-x+9-x^{2}=-11x+30$$

Add $11x$ to both sides.

$$-x+9-x^{2}+11x=30$$

Combine $-x$ and $11x$ to get $10x$.

$$10x+9-x^{2}=30$$

Subtract $30$ from both sides.

$$10x+9-x^{2}-30=0$$

Subtract $30$ from $9$ to get $-21$.

$$10x-21-x^{2}=0$$

All equations of the form $ax^{2}+bx+c=0$ can be solved using the quadratic formula: $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$. The quadratic formula gives two solutions, one when $±$ is addition and one when it is subtraction.

$$-x^{2}+10x-21=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $-1$ for $a$, $10$ for $b$, and $-21$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-10±\sqrt{10^{2}-4\left(-1\right)\left(-21\right)}}{2\left(-1\right)}$$

Square $10$.

$$x=\frac{-10±\sqrt{100-4\left(-1\right)\left(-21\right)}}{2\left(-1\right)}$$

Multiply $-4$ times $-1$.

$$x=\frac{-10±\sqrt{100+4\left(-21\right)}}{2\left(-1\right)}$$

Multiply $4$ times $-21$.

$$x=\frac{-10±\sqrt{100-84}}{2\left(-1\right)}$$

Add $100$ to $-84$.

$$x=\frac{-10±\sqrt{16}}{2\left(-1\right)}$$

Take the square root of $16$.

$$x=\frac{-10±4}{2\left(-1\right)}$$

Multiply $2$ times $-1$.

$$x=\frac{-10±4}{-2}$$

Now solve the equation $x=\frac{-10±4}{-2}$ when $±$ is plus. Add $-10$ to $4$.

$$x=-\frac{6}{-2}$$

Divide $-6$ by $-2$.

$$x=3$$

Now solve the equation $x=\frac{-10±4}{-2}$ when $±$ is minus. Subtract $4$ from $-10$.

$$x=-\frac{14}{-2}$$

Divide $-14$ by $-2$.

$$x=7$$

The equation is now solved.

$$x=3$$ $$x=7$$

Steps for Completing the Square

Variable $x$ cannot be equal to any of the values $5,6$ since division by zero is not defined. Multiply both sides of the equation by $\left(x-6\right)\left(x-5\right)$, the least common multiple of $x-6,x-5$.

$$\left(x-5\right)\times 3-\left(x-6\right)\times 4=\left(x-6\right)\left(x-5\right)$$

Use the distributive property to multiply $x-5$ by $3$.

$$3x-15-\left(x-6\right)\times 4=\left(x-6\right)\left(x-5\right)$$

Use the distributive property to multiply $x-6$ by $4$.

$$3x-15-\left(4x-24\right)=\left(x-6\right)\left(x-5\right)$$

To find the opposite of $4x-24$, find the opposite of each term.

$$3x-15-4x+24=\left(x-6\right)\left(x-5\right)$$

Combine $3x$ and $-4x$ to get $-x$.

$$-x-15+24=\left(x-6\right)\left(x-5\right)$$

Add $-15$ and $24$ to get $9$.

$$-x+9=\left(x-6\right)\left(x-5\right)$$

Use the distributive property to multiply $x-6$ by $x-5$ and combine like terms.

$$-x+9=x^{2}-11x+30$$

Subtract $x^{2}$ from both sides.

$$-x+9-x^{2}=-11x+30$$

Add $11x$ to both sides.

$$-x+9-x^{2}+11x=30$$

Combine $-x$ and $11x$ to get $10x$.

$$10x+9-x^{2}=30$$

Subtract $9$ from both sides.

$$10x-x^{2}=30-9$$

Subtract $9$ from $30$ to get $21$.

$$10x-x^{2}=21$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form $x^{2}+bx=c$.

$$-x^{2}+10x=21$$

Divide both sides by $-1$.

$$\frac{-x^{2}+10x}{-1}=\frac{21}{-1}$$

Dividing by $-1$ undoes the multiplication by $-1$.

$$x^{2}+\frac{10}{-1}x=\frac{21}{-1}$$

Divide $10$ by $-1$.

$$x^{2}-10x=\frac{21}{-1}$$

Divide $21$ by $-1$.

$$x^{2}-10x=-21$$

Divide $-10$, the coefficient of the $x$ term, by $2$ to get $-5$. Then add the square of $-5$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-10x+\left(-5\right)^{2}=-21+\left(-5\right)^{2}$$

Square $-5$.

$$x^{2}-10x+25=-21+25$$

Add $-21$ to $25$.

$$x^{2}-10x+25=4$$

Factor $x^{2}-10x+25$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(x-5\right)^{2}=4$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-5\right)^{2}}=\sqrt{4}$$

Simplify.

$$x-5=2$$ $$x-5=-2$$

Add $5$ to both sides of the equation.

$$x=7$$ $$x=3$$