Solve 15/(2n + 7) = (15 - 7n)/(2n) | 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

$$\frac{15}{2n+7}=\frac{15-7n}{2n}$$

Solve for n

$n=-5$
$n = \frac{3}{2} = 1\frac{1}{2} = 1.5$

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using the Quadratic Formula

Variable $n$ cannot be equal to any of the values $-\frac{7}{2},0$ since division by zero is not defined. Multiply both sides of the equation by $2n\left(2n+7\right)$, the least common multiple of $2n+7,2n$.

$$2n\times 15=\left(2n+7\right)\left(15-7n\right)$$

Multiply $2$ and $15$ to get $30$.

$$30n=\left(2n+7\right)\left(15-7n\right)$$

Use the distributive property to multiply $2n+7$ by $15-7n$ and combine like terms.

$$30n=-19n-14n^{2}+105$$

Add $19n$ to both sides.

$$30n+19n=-14n^{2}+105$$

Combine $30n$ and $19n$ to get $49n$.

$$49n=-14n^{2}+105$$

Add $14n^{2}$ to both sides.

$$49n+14n^{2}=105$$

Subtract $105$ from both sides.

$$49n+14n^{2}-105=0$$

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.

$$14n^{2}+49n-105=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $14$ for $a$, $49$ for $b$, and $-105$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$n=\frac{-49±\sqrt{49^{2}-4\times 14\left(-105\right)}}{2\times 14}$$

Square $49$.

$$n=\frac{-49±\sqrt{2401-4\times 14\left(-105\right)}}{2\times 14}$$

Multiply $-4$ times $14$.

$$n=\frac{-49±\sqrt{2401-56\left(-105\right)}}{2\times 14}$$

Multiply $-56$ times $-105$.

$$n=\frac{-49±\sqrt{2401+5880}}{2\times 14}$$

Add $2401$ to $5880$.

$$n=\frac{-49±\sqrt{8281}}{2\times 14}$$

Take the square root of $8281$.

$$n=\frac{-49±91}{2\times 14}$$

Multiply $2$ times $14$.

$$n=\frac{-49±91}{28}$$

Now solve the equation $n=\frac{-49±91}{28}$ when $±$ is plus. Add $-49$ to $91$.

$$n=\frac{42}{28}$$

Reduce the fraction $\frac{42}{28}$ to lowest terms by extracting and canceling out $14$.

$$n=\frac{3}{2}$$

Now solve the equation $n=\frac{-49±91}{28}$ when $±$ is minus. Subtract $91$ from $-49$.

$$n=-\frac{140}{28}$$

Divide $-140$ by $28$.

$$n=-5$$

The equation is now solved.

$$n=\frac{3}{2}$$ $$n=-5$$

Steps for Completing the Square

Variable $n$ cannot be equal to any of the values $-\frac{7}{2},0$ since division by zero is not defined. Multiply both sides of the equation by $2n\left(2n+7\right)$, the least common multiple of $2n+7,2n$.

$$2n\times 15=\left(2n+7\right)\left(15-7n\right)$$

Multiply $2$ and $15$ to get $30$.

$$30n=\left(2n+7\right)\left(15-7n\right)$$

Use the distributive property to multiply $2n+7$ by $15-7n$ and combine like terms.

$$30n=-19n-14n^{2}+105$$

Add $19n$ to both sides.

$$30n+19n=-14n^{2}+105$$

Combine $30n$ and $19n$ to get $49n$.

$$49n=-14n^{2}+105$$

Add $14n^{2}$ to both sides.

$$49n+14n^{2}=105$$

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

$$14n^{2}+49n=105$$

Divide both sides by $14$.

$$\frac{14n^{2}+49n}{14}=\frac{105}{14}$$

Dividing by $14$ undoes the multiplication by $14$.

$$n^{2}+\frac{49}{14}n=\frac{105}{14}$$

Reduce the fraction $\frac{49}{14}$ to lowest terms by extracting and canceling out $7$.

$$n^{2}+\frac{7}{2}n=\frac{105}{14}$$

Reduce the fraction $\frac{105}{14}$ to lowest terms by extracting and canceling out $7$.

$$n^{2}+\frac{7}{2}n=\frac{15}{2}$$

Divide $\frac{7}{2}$, the coefficient of the $x$ term, by $2$ to get $\frac{7}{4}$. Then add the square of $\frac{7}{4}$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$n^{2}+\frac{7}{2}n+\left(\frac{7}{4}\right)^{2}=\frac{15}{2}+\left(\frac{7}{4}\right)^{2}$$

Square $\frac{7}{4}$ by squaring both the numerator and the denominator of the fraction.

$$n^{2}+\frac{7}{2}n+\frac{49}{16}=\frac{15}{2}+\frac{49}{16}$$

Add $\frac{15}{2}$ to $\frac{49}{16}$ by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$n^{2}+\frac{7}{2}n+\frac{49}{16}=\frac{169}{16}$$

Factor $n^{2}+\frac{7}{2}n+\frac{49}{16}$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(n+\frac{7}{4}\right)^{2}=\frac{169}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(n+\frac{7}{4}\right)^{2}}=\sqrt{\frac{169}{16}}$$

Simplify.

$$n+\frac{7}{4}=\frac{13}{4}$$ $$n+\frac{7}{4}=-\frac{13}{4}$$

Subtract $\frac{7}{4}$ from both sides of the equation.

$$n=\frac{3}{2}$$ $$n=-5$$