How to solve 7/x-6=5/5-x

1. Factor the equation (x + 2)(x + 3) = 0. 2. Set each factor equal to zero: x + 2 = 0 and x + 3 = 0. 3. Solve for x: x = -2 and x = -3.

Solve 7/x-6=5/5-x | Equatix

Question

$$7/x-6=5/5-x$$

Solve for x

$x = \frac{\sqrt{21} + 7}{2} \approx 5.791287847$
$x = \frac{7 - \sqrt{21}}{2} \approx 1.208712153$

Steps Using the Quadratic Formula

Steps for Completing the Square

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 $5x$, the least common multiple of $x,5$.

$$5\times 7+5x\left(-6\right)=x\times 5-x\times 5x$$

Multiply $5$ and $7$ to get $35$.

$$35+5x\left(-6\right)=x\times 5-x\times 5x$$

Multiply $5$ and $-6$ to get $-30$.

$$35-30x=x\times 5-x\times 5x$$

Multiply $x$ and $x$ to get $x^{2}$.

$$35-30x=x\times 5-x^{2}\times 5$$

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

$$35-30x=x\times 5-5x^{2}$$

Subtract $x\times 5$ from both sides.

$$35-30x-x\times 5=-5x^{2}$$

Combine $-30x$ and $-x\times 5$ to get $-35x$.

$$35-35x=-5x^{2}$$

Add $5x^{2}$ to both sides.

$$35-35x+5x^{2}=0$$

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.

$$5x^{2}-35x+35=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $5$ for $a$, $-35$ for $b$, and $35$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-\left(-35\right)±\sqrt{\left(-35\right)^{2}-4\times 5\times 35}}{2\times 5}$$

Square $-35$.

$$x=\frac{-\left(-35\right)±\sqrt{1225-4\times 5\times 35}}{2\times 5}$$

Multiply $-4$ times $5$.

$$x=\frac{-\left(-35\right)±\sqrt{1225-20\times 35}}{2\times 5}$$

Multiply $-20$ times $35$.

$$x=\frac{-\left(-35\right)±\sqrt{1225-700}}{2\times 5}$$

Add $1225$ to $-700$.

$$x=\frac{-\left(-35\right)±\sqrt{525}}{2\times 5}$$

Take the square root of $525$.

$$x=\frac{-\left(-35\right)±5\sqrt{21}}{2\times 5}$$

The opposite of $-35$ is $35$.

$$x=\frac{35±5\sqrt{21}}{2\times 5}$$

Multiply $2$ times $5$.

$$x=\frac{35±5\sqrt{21}}{10}$$

Now solve the equation $x=\frac{35±5\sqrt{21}}{10}$ when $±$ is plus. Add $35$ to $5\sqrt{21}$.

$$x=\frac{5\sqrt{21}+35}{10}$$

Divide $35+5\sqrt{21}$ by $10$.

$$x=\frac{\sqrt{21}+7}{2}$$

Now solve the equation $x=\frac{35±5\sqrt{21}}{10}$ when $±$ is minus. Subtract $5\sqrt{21}$ from $35$.

$$x=\frac{35-5\sqrt{21}}{10}$$

Divide $35-5\sqrt{21}$ by $10$.

$$x=\frac{7-\sqrt{21}}{2}$$

The equation is now solved.

$$x=\frac{\sqrt{21}+7}{2}$$ $$x=\frac{7-\sqrt{21}}{2}$$

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 $5x$, the least common multiple of $x,5$.

$$5\times 7+5x\left(-6\right)=x\times 5-x\times 5x$$

Multiply $5$ and $7$ to get $35$.

$$35+5x\left(-6\right)=x\times 5-x\times 5x$$

Multiply $5$ and $-6$ to get $-30$.

$$35-30x=x\times 5-x\times 5x$$

Multiply $x$ and $x$ to get $x^{2}$.

$$35-30x=x\times 5-x^{2}\times 5$$

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

$$35-30x=x\times 5-5x^{2}$$

Subtract $x\times 5$ from both sides.

$$35-30x-x\times 5=-5x^{2}$$

Combine $-30x$ and $-x\times 5$ to get $-35x$.

$$35-35x=-5x^{2}$$

Add $5x^{2}$ to both sides.

$$35-35x+5x^{2}=0$$

Subtract $35$ from both sides. Anything subtracted from zero gives its negation.

$$-35x+5x^{2}=-35$$

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

$$5x^{2}-35x=-35$$

Divide both sides by $5$.

$$\frac{5x^{2}-35x}{5}=-\frac{35}{5}$$

Dividing by $5$ undoes the multiplication by $5$.

$$x^{2}+\left(-\frac{35}{5}\right)x=-\frac{35}{5}$$

Divide $-35$ by $5$.

$$x^{2}-7x=-\frac{35}{5}$$

Divide $-35$ by $5$.

$$x^{2}-7x=-7$$

Divide $-7$, the coefficient of the $x$ term, by $2$ to get $-\frac{7}{2}$. Then add the square of $-\frac{7}{2}$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-7x+\left(-\frac{7}{2}\right)^{2}=-7+\left(-\frac{7}{2}\right)^{2}$$

Square $-\frac{7}{2}$ by squaring both the numerator and the denominator of the fraction.

$$x^{2}-7x+\frac{49}{4}=-7+\frac{49}{4}$$

Add $-7$ to $\frac{49}{4}$.

$$x^{2}-7x+\frac{49}{4}=\frac{21}{4}$$

Factor $x^{2}-7x+\frac{49}{4}$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(x-\frac{7}{2}\right)^{2}=\frac{21}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{7}{2}\right)^{2}}=\sqrt{\frac{21}{4}}$$

Simplify.

$$x-\frac{7}{2}=\frac{\sqrt{21}}{2}$$ $$x-\frac{7}{2}=-\frac{\sqrt{21}}{2}$$

Add $\frac{7}{2}$ to both sides of the equation.

$$x=\frac{\sqrt{21}+7}{2}$$ $$x=\frac{7-\sqrt{21}}{2}$$

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Steps Using the Quadratic Formula

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Steps Using the Quadratic Formula

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