Solve Solve the following 1 . 4/x - 5/(x + 3) = 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

$$\left. \begin{array} { l } { \frac { 4 } { x } - \frac { 5 } { x + 3 } = 1 } \end{array} \right.$$

Solve for x

$x=-6$
$x=2$

Steps Using Factoring By Grouping

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using Factoring By Grouping

Variable $x$ cannot be equal to any of the values $-3,0$ since division by zero is not defined. Multiply both sides of the equation by $x\left(x+3\right)$, the least common multiple of $x,x+3$.

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

Use the distributive property to multiply $x+3$ by $4$.

$$4x+12-x\times 5=x\left(x+3\right)$$

Use the distributive property to multiply $x$ by $x+3$.

$$4x+12-x\times 5=x^{2}+3x$$

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

$$4x+12-x\times 5-x^{2}=3x$$

Subtract $3x$ from both sides.

$$4x+12-x\times 5-x^{2}-3x=0$$

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

$$x+12-x\times 5-x^{2}=0$$

Multiply $-1$ and $5$ to get $-5$.

$$x+12-5x-x^{2}=0$$

Combine $x$ and $-5x$ to get $-4x$.

$$-4x+12-x^{2}=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$-x^{2}-4x+12=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as $-x^{2}+ax+bx+12$. To find $a$ and $b$, set up a system to be solved.

$$a+b=-4$$ $$ab=-12=-12$$

Since $ab$ is negative, $a$ and $b$ have the opposite signs. Since $a+b$ is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product $-12$.

$$1,-12$$ $$2,-6$$ $$3,-4$$

Calculate the sum for each pair.

$$1-12=-11$$ $$2-6=-4$$ $$3-4=-1$$

The solution is the pair that gives sum $-4$.

$$a=2$$ $$b=-6$$

Rewrite $-x^{2}-4x+12$ as $\left(-x^{2}+2x\right)+\left(-6x+12\right)$.

$$\left(-x^{2}+2x\right)+\left(-6x+12\right)$$

Factor out $x$ in the first and $6$ in the second group.

$$x\left(-x+2\right)+6\left(-x+2\right)$$

Factor out common term $-x+2$ by using distributive property.

$$\left(-x+2\right)\left(x+6\right)$$

To find equation solutions, solve $-x+2=0$ and $x+6=0$.

$$x=2$$ $$x=-6$$

Steps Using the Quadratic Formula

Variable $x$ cannot be equal to any of the values $-3,0$ since division by zero is not defined. Multiply both sides of the equation by $x\left(x+3\right)$, the least common multiple of $x,x+3$.

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

Use the distributive property to multiply $x+3$ by $4$.

$$4x+12-x\times 5=x\left(x+3\right)$$

Use the distributive property to multiply $x$ by $x+3$.

$$4x+12-x\times 5=x^{2}+3x$$

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

$$4x+12-x\times 5-x^{2}=3x$$

Subtract $3x$ from both sides.

$$4x+12-x\times 5-x^{2}-3x=0$$

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

$$x+12-x\times 5-x^{2}=0$$

Multiply $-1$ and $5$ to get $-5$.

$$x+12-5x-x^{2}=0$$

Combine $x$ and $-5x$ to get $-4x$.

$$-4x+12-x^{2}=0$$

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.

$$-x^{2}-4x+12=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $-1$ for $a$, $-4$ for $b$, and $12$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\left(-1\right)\times 12}}{2\left(-1\right)}$$

Square $-4$.

$$x=\frac{-\left(-4\right)±\sqrt{16-4\left(-1\right)\times 12}}{2\left(-1\right)}$$

Multiply $-4$ times $-1$.

$$x=\frac{-\left(-4\right)±\sqrt{16+4\times 12}}{2\left(-1\right)}$$

Multiply $4$ times $12$.

$$x=\frac{-\left(-4\right)±\sqrt{16+48}}{2\left(-1\right)}$$

Add $16$ to $48$.

$$x=\frac{-\left(-4\right)±\sqrt{64}}{2\left(-1\right)}$$

Take the square root of $64$.

$$x=\frac{-\left(-4\right)±8}{2\left(-1\right)}$$

The opposite of $-4$ is $4$.

$$x=\frac{4±8}{2\left(-1\right)}$$

Multiply $2$ times $-1$.

$$x=\frac{4±8}{-2}$$

Now solve the equation $x=\frac{4±8}{-2}$ when $±$ is plus. Add $4$ to $8$.

$$x=\frac{12}{-2}$$

Divide $12$ by $-2$.

$$x=-6$$

Now solve the equation $x=\frac{4±8}{-2}$ when $±$ is minus. Subtract $8$ from $4$.

$$x=-\frac{4}{-2}$$

Divide $-4$ by $-2$.

$$x=2$$

The equation is now solved.

$$x=-6$$ $$x=2$$

Steps for Completing the Square

Variable $x$ cannot be equal to any of the values $-3,0$ since division by zero is not defined. Multiply both sides of the equation by $x\left(x+3\right)$, the least common multiple of $x,x+3$.

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

Use the distributive property to multiply $x+3$ by $4$.

$$4x+12-x\times 5=x\left(x+3\right)$$

Use the distributive property to multiply $x$ by $x+3$.

$$4x+12-x\times 5=x^{2}+3x$$

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

$$4x+12-x\times 5-x^{2}=3x$$

Subtract $3x$ from both sides.

$$4x+12-x\times 5-x^{2}-3x=0$$

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

$$x+12-x\times 5-x^{2}=0$$

Subtract $12$ from both sides. Anything subtracted from zero gives its negation.

$$x-x\times 5-x^{2}=-12$$

Multiply $-1$ and $5$ to get $-5$.

$$x-5x-x^{2}=-12$$

Combine $x$ and $-5x$ to get $-4x$.

$$-4x-x^{2}=-12$$

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 $x^{2}+bx=c$.

$$-x^{2}-4x=-12$$

Divide both sides by $-1$.

$$\frac{-x^{2}-4x}{-1}=-\frac{12}{-1}$$

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

$$x^{2}+\left(-\frac{4}{-1}\right)x=-\frac{12}{-1}$$

Divide $-4$ by $-1$.

$$x^{2}+4x=-\frac{12}{-1}$$

Divide $-12$ by $-1$.

$$x^{2}+4x=12$$

Divide $4$, the coefficient of the $x$ term, by $2$ to get $2$. Then add the square of $2$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+4x+2^{2}=12+2^{2}$$

Square $2$.

$$x^{2}+4x+4=12+4$$

Add $12$ to $4$.

$$x^{2}+4x+4=16$$

Factor $x^{2}+4x+4$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(x+2\right)^{2}=16$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+2\right)^{2}}=\sqrt{16}$$

Simplify.

$$x+2=4$$ $$x+2=-4$$

Subtract $2$ from both sides of the equation.

$$x=2$$ $$x=-6$$