Solve 6/x + (x - 3)/4 = 2 | 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

$$\frac{6}{x}+\frac{x-3}{4}=2$$

Solve for x

$x=3$
$x=8$

Steps Using Factoring

Steps Using Factoring By Grouping

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using Factoring

Variable $x$ cannot be equal to $0$ since division by zero is not defined. Multiply both sides of the equation by $4x$, the least common multiple of $x,4$.

$$4\times 6+x\left(x-3\right)=8x$$

Multiply $4$ and $6$ to get $24$.

$$24+x\left(x-3\right)=8x$$

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

$$24+x^{2}-3x=8x$$

Subtract $8x$ from both sides.

$$24+x^{2}-3x-8x=0$$

Combine $-3x$ and $-8x$ to get $-11x$.

$$24+x^{2}-11x=0$$

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

$$x^{2}-11x+24=0$$

To solve the equation, factor $x^{2}-11x+24$ using formula $x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right)$. To find $a$ and $b$, set up a system to be solved.

$$a+b=-11$$ $$ab=24$$

Since $ab$ is positive, $a$ and $b$ have the same sign. Since $a+b$ is negative, $a$ and $b$ are both negative. List all such integer pairs that give product $24$.

$$-1,-24$$ $$-2,-12$$ $$-3,-8$$ $$-4,-6$$

Calculate the sum for each pair.

$$-1-24=-25$$ $$-2-12=-14$$ $$-3-8=-11$$ $$-4-6=-10$$

The solution is the pair that gives sum $-11$.

$$a=-8$$ $$b=-3$$

Rewrite factored expression $\left(x+a\right)\left(x+b\right)$ using the obtained values.

$$\left(x-8\right)\left(x-3\right)$$

To find equation solutions, solve $x-8=0$ and $x-3=0$.

$$x=8$$ $$x=3$$

Steps Using Factoring By Grouping

Variable $x$ cannot be equal to $0$ since division by zero is not defined. Multiply both sides of the equation by $4x$, the least common multiple of $x,4$.

$$4\times 6+x\left(x-3\right)=8x$$

Multiply $4$ and $6$ to get $24$.

$$24+x\left(x-3\right)=8x$$

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

$$24+x^{2}-3x=8x$$

Subtract $8x$ from both sides.

$$24+x^{2}-3x-8x=0$$

Combine $-3x$ and $-8x$ to get $-11x$.

$$24+x^{2}-11x=0$$

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

$$x^{2}-11x+24=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as $x^{2}+ax+bx+24$. To find $a$ and $b$, set up a system to be solved.

$$a+b=-11$$ $$ab=1\times 24=24$$

Since $ab$ is positive, $a$ and $b$ have the same sign. Since $a+b$ is negative, $a$ and $b$ are both negative. List all such integer pairs that give product $24$.

$$-1,-24$$ $$-2,-12$$ $$-3,-8$$ $$-4,-6$$

Calculate the sum for each pair.

$$-1-24=-25$$ $$-2-12=-14$$ $$-3-8=-11$$ $$-4-6=-10$$

The solution is the pair that gives sum $-11$.

$$a=-8$$ $$b=-3$$

Rewrite $x^{2}-11x+24$ as $\left(x^{2}-8x\right)+\left(-3x+24\right)$.

$$\left(x^{2}-8x\right)+\left(-3x+24\right)$$

Factor out $x$ in the first and $-3$ in the second group.

$$x\left(x-8\right)-3\left(x-8\right)$$

Factor out common term $x-8$ by using distributive property.

$$\left(x-8\right)\left(x-3\right)$$

To find equation solutions, solve $x-8=0$ and $x-3=0$.

$$x=8$$ $$x=3$$

Steps Using the Quadratic Formula

Variable $x$ cannot be equal to $0$ since division by zero is not defined. Multiply both sides of the equation by $4x$, the least common multiple of $x,4$.

$$4\times 6+x\left(x-3\right)=8x$$

Multiply $4$ and $6$ to get $24$.

$$24+x\left(x-3\right)=8x$$

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

$$24+x^{2}-3x=8x$$

Subtract $8x$ from both sides.

$$24+x^{2}-3x-8x=0$$

Combine $-3x$ and $-8x$ to get $-11x$.

$$24+x^{2}-11x=0$$

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.

$$x^{2}-11x+24=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $1$ for $a$, $-11$ for $b$, and $24$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-\left(-11\right)±\sqrt{\left(-11\right)^{2}-4\times 24}}{2}$$

Square $-11$.

$$x=\frac{-\left(-11\right)±\sqrt{121-4\times 24}}{2}$$

Multiply $-4$ times $24$.

$$x=\frac{-\left(-11\right)±\sqrt{121-96}}{2}$$

Add $121$ to $-96$.

$$x=\frac{-\left(-11\right)±\sqrt{25}}{2}$$

Take the square root of $25$.

$$x=\frac{-\left(-11\right)±5}{2}$$

The opposite of $-11$ is $11$.

$$x=\frac{11±5}{2}$$

Now solve the equation $x=\frac{11±5}{2}$ when $±$ is plus. Add $11$ to $5$.

$$x=\frac{16}{2}$$

Divide $16$ by $2$.

$$x=8$$

Now solve the equation $x=\frac{11±5}{2}$ when $±$ is minus. Subtract $5$ from $11$.

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

Divide $6$ by $2$.

$$x=3$$

The equation is now solved.

$$x=8$$ $$x=3$$

Steps for Completing the Square

Variable $x$ cannot be equal to $0$ since division by zero is not defined. Multiply both sides of the equation by $4x$, the least common multiple of $x,4$.

$$4\times 6+x\left(x-3\right)=8x$$

Multiply $4$ and $6$ to get $24$.

$$24+x\left(x-3\right)=8x$$

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

$$24+x^{2}-3x=8x$$

Subtract $8x$ from both sides.

$$24+x^{2}-3x-8x=0$$

Combine $-3x$ and $-8x$ to get $-11x$.

$$24+x^{2}-11x=0$$

Subtract $24$ from both sides. Anything subtracted from zero gives its negation.

$$x^{2}-11x=-24$$

Divide $-11$, the coefficient of the $x$ term, by $2$ to get $-\frac{11}{2}$. Then add the square of $-\frac{11}{2}$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-11x+\left(-\frac{11}{2}\right)^{2}=-24+\left(-\frac{11}{2}\right)^{2}$$

Square $-\frac{11}{2}$ by squaring both the numerator and the denominator of the fraction.

$$x^{2}-11x+\frac{121}{4}=-24+\frac{121}{4}$$

Add $-24$ to $\frac{121}{4}$.

$$x^{2}-11x+\frac{121}{4}=\frac{25}{4}$$

Factor $x^{2}-11x+\frac{121}{4}$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(x-\frac{11}{2}\right)^{2}=\frac{25}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{11}{2}\right)^{2}}=\sqrt{\frac{25}{4}}$$

Simplify.

$$x-\frac{11}{2}=\frac{5}{2}$$ $$x-\frac{11}{2}=-\frac{5}{2}$$

Add $\frac{11}{2}$ to both sides of the equation.

$$x=8$$ $$x=3$$