Solve (3x)/((x)^(2)-1)=2((2x-1)/(x+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

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

Solve for x

$x=2$
$x=\frac{1}{4}=0.25$

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 $-1,1$ since division by zero is not defined. Multiply both sides of the equation by $\left(x-1\right)\left(x+1\right)$, the least common multiple of $x^{2}-1,x+1$.

$$3x=2\left(x-1\right)\left(2x-1\right)$$

Use the distributive property to multiply $2$ by $x-1$.

$$3x=\left(2x-2\right)\left(2x-1\right)$$

Use the distributive property to multiply $2x-2$ by $2x-1$ and combine like terms.

$$3x=4x^{2}-6x+2$$

Subtract $4x^{2}$ from both sides.

$$3x-4x^{2}=-6x+2$$

Add $6x$ to both sides.

$$3x-4x^{2}+6x=2$$

Combine $3x$ and $6x$ to get $9x$.

$$9x-4x^{2}=2$$

Subtract $2$ from both sides.

$$9x-4x^{2}-2=0$$

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

$$-4x^{2}+9x-2=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as $-4x^{2}+ax+bx-2$. To find $a$ and $b$, set up a system to be solved.

$$a+b=9$$ $$ab=-4\left(-2\right)=8$$

Since $ab$ is positive, $a$ and $b$ have the same sign. Since $a+b$ is positive, $a$ and $b$ are both positive. List all such integer pairs that give product $8$.

$$1,8$$ $$2,4$$

Calculate the sum for each pair.

$$1+8=9$$ $$2+4=6$$

The solution is the pair that gives sum $9$.

$$a=8$$ $$b=1$$

Rewrite $-4x^{2}+9x-2$ as $\left(-4x^{2}+8x\right)+\left(x-2\right)$.

$$\left(-4x^{2}+8x\right)+\left(x-2\right)$$

Factor out $4x$ in the first and $-1$ in the second group.

$$4x\left(-x+2\right)-\left(-x+2\right)$$

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

$$\left(-x+2\right)\left(4x-1\right)$$

To find equation solutions, solve $-x+2=0$ and $4x-1=0$.

$$x=2$$ $$x=\frac{1}{4}$$

Steps Using the Quadratic Formula

Variable $x$ cannot be equal to any of the values $-1,1$ since division by zero is not defined. Multiply both sides of the equation by $\left(x-1\right)\left(x+1\right)$, the least common multiple of $x^{2}-1,x+1$.

$$3x=2\left(x-1\right)\left(2x-1\right)$$

Use the distributive property to multiply $2$ by $x-1$.

$$3x=\left(2x-2\right)\left(2x-1\right)$$

Use the distributive property to multiply $2x-2$ by $2x-1$ and combine like terms.

$$3x=4x^{2}-6x+2$$

Subtract $4x^{2}$ from both sides.

$$3x-4x^{2}=-6x+2$$

Add $6x$ to both sides.

$$3x-4x^{2}+6x=2$$

Combine $3x$ and $6x$ to get $9x$.

$$9x-4x^{2}=2$$

Subtract $2$ from both sides.

$$9x-4x^{2}-2=0$$

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.

$$-4x^{2}+9x-2=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $-4$ for $a$, $9$ for $b$, and $-2$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-9±\sqrt{9^{2}-4\left(-4\right)\left(-2\right)}}{2\left(-4\right)}$$

Square $9$.

$$x=\frac{-9±\sqrt{81-4\left(-4\right)\left(-2\right)}}{2\left(-4\right)}$$

Multiply $-4$ times $-4$.

$$x=\frac{-9±\sqrt{81+16\left(-2\right)}}{2\left(-4\right)}$$

Multiply $16$ times $-2$.

$$x=\frac{-9±\sqrt{81-32}}{2\left(-4\right)}$$

Add $81$ to $-32$.

$$x=\frac{-9±\sqrt{49}}{2\left(-4\right)}$$

Take the square root of $49$.

$$x=\frac{-9±7}{2\left(-4\right)}$$

Multiply $2$ times $-4$.

$$x=\frac{-9±7}{-8}$$

Now solve the equation $x=\frac{-9±7}{-8}$ when $±$ is plus. Add $-9$ to $7$.

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

Reduce the fraction $\frac{-2}{-8}$ to lowest terms by extracting and canceling out $2$.

$$x=\frac{1}{4}$$

Now solve the equation $x=\frac{-9±7}{-8}$ when $±$ is minus. Subtract $7$ from $-9$.

$$x=-\frac{16}{-8}$$

Divide $-16$ by $-8$.

$$x=2$$

The equation is now solved.

$$x=\frac{1}{4}$$ $$x=2$$

Steps for Completing the Square

Variable $x$ cannot be equal to any of the values $-1,1$ since division by zero is not defined. Multiply both sides of the equation by $\left(x-1\right)\left(x+1\right)$, the least common multiple of $x^{2}-1,x+1$.

$$3x=2\left(x-1\right)\left(2x-1\right)$$

Use the distributive property to multiply $2$ by $x-1$.

$$3x=\left(2x-2\right)\left(2x-1\right)$$

Use the distributive property to multiply $2x-2$ by $2x-1$ and combine like terms.

$$3x=4x^{2}-6x+2$$

Subtract $4x^{2}$ from both sides.

$$3x-4x^{2}=-6x+2$$

Add $6x$ to both sides.

$$3x-4x^{2}+6x=2$$

Combine $3x$ and $6x$ to get $9x$.

$$9x-4x^{2}=2$$

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

$$-4x^{2}+9x=2$$

Divide both sides by $-4$.

$$\frac{-4x^{2}+9x}{-4}=\frac{2}{-4}$$

Dividing by $-4$ undoes the multiplication by $-4$.

$$x^{2}+\frac{9}{-4}x=\frac{2}{-4}$$

Divide $9$ by $-4$.

$$x^{2}-\frac{9}{4}x=\frac{2}{-4}$$

Reduce the fraction $\frac{2}{-4}$ to lowest terms by extracting and canceling out $2$.

$$x^{2}-\frac{9}{4}x=-\frac{1}{2}$$

Divide $-\frac{9}{4}$, the coefficient of the $x$ term, by $2$ to get $-\frac{9}{8}$. Then add the square of $-\frac{9}{8}$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{9}{4}x+\left(-\frac{9}{8}\right)^{2}=-\frac{1}{2}+\left(-\frac{9}{8}\right)^{2}$$

Square $-\frac{9}{8}$ by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{9}{4}x+\frac{81}{64}=-\frac{1}{2}+\frac{81}{64}$$

Add $-\frac{1}{2}$ to $\frac{81}{64}$ by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{9}{4}x+\frac{81}{64}=\frac{49}{64}$$

Factor $x^{2}-\frac{9}{4}x+\frac{81}{64}$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(x-\frac{9}{8}\right)^{2}=\frac{49}{64}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{9}{8}\right)^{2}}=\sqrt{\frac{49}{64}}$$

Simplify.

$$x-\frac{9}{8}=\frac{7}{8}$$ $$x-\frac{9}{8}=-\frac{7}{8}$$

Add $\frac{9}{8}$ to both sides of the equation.

$$x=2$$ $$x=\frac{1}{4}$$