Solve (2x + 1) = 5/(x - 3) | 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

$$(2x+1)=\frac{5}{x-3}$$

Solve for x

$x = \frac{\sqrt{89} + 5}{4} \approx 3.608495283$
$x=\frac{5-\sqrt{89}}{4}\approx -1.108495283$

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using the Quadratic Formula

Variable $x$ cannot be equal to $3$ since division by zero is not defined. Multiply both sides of the equation by $x-3$.

$$2x\left(x-3\right)+x-3=5$$

Use the distributive property to multiply $2x$ by $x-3$.

$$2x^{2}-6x+x-3=5$$

Combine $-6x$ and $x$ to get $-5x$.

$$2x^{2}-5x-3=5$$

Subtract $5$ from both sides.

$$2x^{2}-5x-3-5=0$$

Subtract $5$ from $-3$ to get $-8$.

$$2x^{2}-5x-8=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $2$ for $a$, $-5$ for $b$, and $-8$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 2\left(-8\right)}}{2\times 2}$$

Square $-5$.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\times 2\left(-8\right)}}{2\times 2}$$

Multiply $-4$ times $2$.

$$x=\frac{-\left(-5\right)±\sqrt{25-8\left(-8\right)}}{2\times 2}$$

Multiply $-8$ times $-8$.

$$x=\frac{-\left(-5\right)±\sqrt{25+64}}{2\times 2}$$

Add $25$ to $64$.

$$x=\frac{-\left(-5\right)±\sqrt{89}}{2\times 2}$$

The opposite of $-5$ is $5$.

$$x=\frac{5±\sqrt{89}}{2\times 2}$$

Multiply $2$ times $2$.

$$x=\frac{5±\sqrt{89}}{4}$$

Now solve the equation $x=\frac{5±\sqrt{89}}{4}$ when $±$ is plus. Add $5$ to $\sqrt{89}$.

$$x=\frac{\sqrt{89}+5}{4}$$

Now solve the equation $x=\frac{5±\sqrt{89}}{4}$ when $±$ is minus. Subtract $\sqrt{89}$ from $5$.

$$x=\frac{5-\sqrt{89}}{4}$$

The equation is now solved.

$$x=\frac{\sqrt{89}+5}{4}$$ $$x=\frac{5-\sqrt{89}}{4}$$

Steps for Completing the Square

Variable $x$ cannot be equal to $3$ since division by zero is not defined. Multiply both sides of the equation by $x-3$.

$$2x\left(x-3\right)+x-3=5$$

Use the distributive property to multiply $2x$ by $x-3$.

$$2x^{2}-6x+x-3=5$$

Combine $-6x$ and $x$ to get $-5x$.

$$2x^{2}-5x-3=5$$

Add $3$ to both sides.

$$2x^{2}-5x=5+3$$

Add $5$ and $3$ to get $8$.

$$2x^{2}-5x=8$$

Divide both sides by $2$.

$$\frac{2x^{2}-5x}{2}=\frac{8}{2}$$

Dividing by $2$ undoes the multiplication by $2$.

$$x^{2}-\frac{5}{2}x=\frac{8}{2}$$

Divide $8$ by $2$.

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

Divide $-\frac{5}{2}$, the coefficient of the $x$ term, by $2$ to get $-\frac{5}{4}$. Then add the square of $-\frac{5}{4}$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square $-\frac{5}{4}$ by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{5}{2}x+\frac{25}{16}=4+\frac{25}{16}$$

Add $4$ to $\frac{25}{16}$.

$$x^{2}-\frac{5}{2}x+\frac{25}{16}=\frac{89}{16}$$

Factor $x^{2}-\frac{5}{2}x+\frac{25}{16}$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(x-\frac{5}{4}\right)^{2}=\frac{89}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{4}\right)^{2}}=\sqrt{\frac{89}{16}}$$

Simplify.

$$x-\frac{5}{4}=\frac{\sqrt{89}}{4}$$ $$x-\frac{5}{4}=-\frac{\sqrt{89}}{4}$$

Add $\frac{5}{4}$ to both sides of the equation.

$$x=\frac{\sqrt{89}+5}{4}$$ $$x=\frac{5-\sqrt{89}}{4}$$