Solve (4x ^ 2)/(x - 1) - (1 - 3x)/4 = (20x)/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

$$\frac{4x^{2}}{x-1}-\frac{1-3x}{4}=\frac{20x}{3}$$

Solve for x

$x=3$
$x=-\frac{1}{23}\approx -0.043478261$

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 $1$ since division by zero is not defined. Multiply both sides of the equation by $12\left(x-1\right)$, the least common multiple of $x-1,4,3$.

$$12\times 4x^{2}-\left(3x-3\right)\left(1-3x\right)=\left(4x-4\right)\times 20x$$

Multiply $12$ and $4$ to get $48$.

$$48x^{2}-\left(3x-3\right)\left(1-3x\right)=\left(4x-4\right)\times 20x$$

Use the distributive property to multiply $3x-3$ by $1-3x$ and combine like terms.

$$48x^{2}-\left(12x-9x^{2}-3\right)=\left(4x-4\right)\times 20x$$

To find the opposite of $12x-9x^{2}-3$, find the opposite of each term.

$$48x^{2}-12x+9x^{2}+3=\left(4x-4\right)\times 20x$$

Combine $48x^{2}$ and $9x^{2}$ to get $57x^{2}$.

$$57x^{2}-12x+3=\left(4x-4\right)\times 20x$$

Use the distributive property to multiply $4x-4$ by $20$.

$$57x^{2}-12x+3=\left(80x-80\right)x$$

Use the distributive property to multiply $80x-80$ by $x$.

$$57x^{2}-12x+3=80x^{2}-80x$$

Subtract $80x^{2}$ from both sides.

$$57x^{2}-12x+3-80x^{2}=-80x$$

Combine $57x^{2}$ and $-80x^{2}$ to get $-23x^{2}$.

$$-23x^{2}-12x+3=-80x$$

Add $80x$ to both sides.

$$-23x^{2}-12x+3+80x=0$$

Combine $-12x$ and $80x$ to get $68x$.

$$-23x^{2}+68x+3=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as $-23x^{2}+ax+bx+3$. To find $a$ and $b$, set up a system to be solved.

$$a+b=68$$ $$ab=-23\times 3=-69$$

Since $ab$ is negative, $a$ and $b$ have the opposite signs. Since $a+b$ is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product $-69$.

$$-1,69$$ $$-3,23$$

Calculate the sum for each pair.

$$-1+69=68$$ $$-3+23=20$$

The solution is the pair that gives sum $68$.

$$a=69$$ $$b=-1$$

Rewrite $-23x^{2}+68x+3$ as $\left(-23x^{2}+69x\right)+\left(-x+3\right)$.

$$\left(-23x^{2}+69x\right)+\left(-x+3\right)$$

Factor out $23x$ in $-23x^{2}+69x$.

$$23x\left(-x+3\right)-x+3$$

Factor out common term $-x+3$ by using distributive property.

$$\left(-x+3\right)\left(23x+1\right)$$

To find equation solutions, solve $-x+3=0$ and $23x+1=0$.

$$x=3$$ $$x=-\frac{1}{23}$$

Steps Using the Quadratic Formula

Variable $x$ cannot be equal to $1$ since division by zero is not defined. Multiply both sides of the equation by $12\left(x-1\right)$, the least common multiple of $x-1,4,3$.

$$12\times 4x^{2}-\left(3x-3\right)\left(1-3x\right)=\left(4x-4\right)\times 20x$$

Multiply $12$ and $4$ to get $48$.

$$48x^{2}-\left(3x-3\right)\left(1-3x\right)=\left(4x-4\right)\times 20x$$

Use the distributive property to multiply $3x-3$ by $1-3x$ and combine like terms.

$$48x^{2}-\left(12x-9x^{2}-3\right)=\left(4x-4\right)\times 20x$$

To find the opposite of $12x-9x^{2}-3$, find the opposite of each term.

$$48x^{2}-12x+9x^{2}+3=\left(4x-4\right)\times 20x$$

Combine $48x^{2}$ and $9x^{2}$ to get $57x^{2}$.

$$57x^{2}-12x+3=\left(4x-4\right)\times 20x$$

Use the distributive property to multiply $4x-4$ by $20$.

$$57x^{2}-12x+3=\left(80x-80\right)x$$

Use the distributive property to multiply $80x-80$ by $x$.

$$57x^{2}-12x+3=80x^{2}-80x$$

Subtract $80x^{2}$ from both sides.

$$57x^{2}-12x+3-80x^{2}=-80x$$

Combine $57x^{2}$ and $-80x^{2}$ to get $-23x^{2}$.

$$-23x^{2}-12x+3=-80x$$

Add $80x$ to both sides.

$$-23x^{2}-12x+3+80x=0$$

Combine $-12x$ and $80x$ to get $68x$.

$$-23x^{2}+68x+3=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $-23$ for $a$, $68$ for $b$, and $3$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-68±\sqrt{68^{2}-4\left(-23\right)\times 3}}{2\left(-23\right)}$$

Square $68$.

$$x=\frac{-68±\sqrt{4624-4\left(-23\right)\times 3}}{2\left(-23\right)}$$

Multiply $-4$ times $-23$.

$$x=\frac{-68±\sqrt{4624+92\times 3}}{2\left(-23\right)}$$

Multiply $92$ times $3$.

$$x=\frac{-68±\sqrt{4624+276}}{2\left(-23\right)}$$

Add $4624$ to $276$.

$$x=\frac{-68±\sqrt{4900}}{2\left(-23\right)}$$

Take the square root of $4900$.

$$x=\frac{-68±70}{2\left(-23\right)}$$

Multiply $2$ times $-23$.

$$x=\frac{-68±70}{-46}$$

Now solve the equation $x=\frac{-68±70}{-46}$ when $±$ is plus. Add $-68$ to $70$.

$$x=\frac{2}{-46}$$

Reduce the fraction $\frac{2}{-46}$ to lowest terms by extracting and canceling out $2$.

$$x=-\frac{1}{23}$$

Now solve the equation $x=\frac{-68±70}{-46}$ when $±$ is minus. Subtract $70$ from $-68$.

$$x=-\frac{138}{-46}$$

Divide $-138$ by $-46$.

$$x=3$$

The equation is now solved.

$$x=-\frac{1}{23}$$ $$x=3$$

Steps for Completing the Square

Variable $x$ cannot be equal to $1$ since division by zero is not defined. Multiply both sides of the equation by $12\left(x-1\right)$, the least common multiple of $x-1,4,3$.

$$12\times 4x^{2}-\left(3x-3\right)\left(1-3x\right)=\left(4x-4\right)\times 20x$$

Multiply $12$ and $4$ to get $48$.

$$48x^{2}-\left(3x-3\right)\left(1-3x\right)=\left(4x-4\right)\times 20x$$

Use the distributive property to multiply $3x-3$ by $1-3x$ and combine like terms.

$$48x^{2}-\left(12x-9x^{2}-3\right)=\left(4x-4\right)\times 20x$$

To find the opposite of $12x-9x^{2}-3$, find the opposite of each term.

$$48x^{2}-12x+9x^{2}+3=\left(4x-4\right)\times 20x$$

Combine $48x^{2}$ and $9x^{2}$ to get $57x^{2}$.

$$57x^{2}-12x+3=\left(4x-4\right)\times 20x$$

Use the distributive property to multiply $4x-4$ by $20$.

$$57x^{2}-12x+3=\left(80x-80\right)x$$

Use the distributive property to multiply $80x-80$ by $x$.

$$57x^{2}-12x+3=80x^{2}-80x$$

Subtract $80x^{2}$ from both sides.

$$57x^{2}-12x+3-80x^{2}=-80x$$

Combine $57x^{2}$ and $-80x^{2}$ to get $-23x^{2}$.

$$-23x^{2}-12x+3=-80x$$

Add $80x$ to both sides.

$$-23x^{2}-12x+3+80x=0$$

Combine $-12x$ and $80x$ to get $68x$.

$$-23x^{2}+68x+3=0$$

Subtract $3$ from both sides. Anything subtracted from zero gives its negation.

$$-23x^{2}+68x=-3$$

Divide both sides by $-23$.

$$\frac{-23x^{2}+68x}{-23}=-\frac{3}{-23}$$

Dividing by $-23$ undoes the multiplication by $-23$.

$$x^{2}+\frac{68}{-23}x=-\frac{3}{-23}$$

Divide $68$ by $-23$.

$$x^{2}-\frac{68}{23}x=-\frac{3}{-23}$$

Divide $-3$ by $-23$.

$$x^{2}-\frac{68}{23}x=\frac{3}{23}$$

Divide $-\frac{68}{23}$, the coefficient of the $x$ term, by $2$ to get $-\frac{34}{23}$. Then add the square of $-\frac{34}{23}$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{68}{23}x+\left(-\frac{34}{23}\right)^{2}=\frac{3}{23}+\left(-\frac{34}{23}\right)^{2}$$

Square $-\frac{34}{23}$ by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{68}{23}x+\frac{1156}{529}=\frac{3}{23}+\frac{1156}{529}$$

Add $\frac{3}{23}$ to $\frac{1156}{529}$ by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{68}{23}x+\frac{1156}{529}=\frac{1225}{529}$$

Factor $x^{2}-\frac{68}{23}x+\frac{1156}{529}$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(x-\frac{34}{23}\right)^{2}=\frac{1225}{529}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{34}{23}\right)^{2}}=\sqrt{\frac{1225}{529}}$$

Simplify.

$$x-\frac{34}{23}=\frac{35}{23}$$ $$x-\frac{34}{23}=-\frac{35}{23}$$

Add $\frac{34}{23}$ to both sides of the equation.

$$x=3$$ $$x=-\frac{1}{23}$$