How to solve (7)/(2x+9)-6=5x

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)/(2x+9)-6=5x | Equatix

Question

$$\frac{ 7 }{ 2x+9 } -6=5x$$

Solve for x

$x = -\frac{47}{10} = -4\frac{7}{10} = -4.7$
$x=-1$

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 $-\frac{9}{2}$ since division by zero is not defined. Multiply both sides of the equation by $2x+9$.

$$7+\left(2x+9\right)\left(-6\right)=5x\left(2x+9\right)$$

Use the distributive property to multiply $2x+9$ by $-6$.

$$7-12x-54=5x\left(2x+9\right)$$

Subtract $54$ from $7$ to get $-47$.

$$-47-12x=5x\left(2x+9\right)$$

Use the distributive property to multiply $5x$ by $2x+9$.

$$-47-12x=10x^{2}+45x$$

Subtract $10x^{2}$ from both sides.

$$-47-12x-10x^{2}=45x$$

Subtract $45x$ from both sides.

$$-47-12x-10x^{2}-45x=0$$

Combine $-12x$ and $-45x$ to get $-57x$.

$$-47-57x-10x^{2}=0$$

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

$$-10x^{2}-57x-47=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as $-10x^{2}+ax+bx-47$. To find $a$ and $b$, set up a system to be solved.

$$a+b=-57$$ $$ab=-10\left(-47\right)=470$$

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 $470$.

$$-1,-470$$ $$-2,-235$$ $$-5,-94$$ $$-10,-47$$

Calculate the sum for each pair.

$$-1-470=-471$$ $$-2-235=-237$$ $$-5-94=-99$$ $$-10-47=-57$$

The solution is the pair that gives sum $-57$.

$$a=-10$$ $$b=-47$$

Rewrite $-10x^{2}-57x-47$ as $\left(-10x^{2}-10x\right)+\left(-47x-47\right)$.

$$\left(-10x^{2}-10x\right)+\left(-47x-47\right)$$

Factor out $10x$ in the first and $47$ in the second group.

$$10x\left(-x-1\right)+47\left(-x-1\right)$$

Factor out common term $-x-1$ by using distributive property.

$$\left(-x-1\right)\left(10x+47\right)$$

To find equation solutions, solve $-x-1=0$ and $10x+47=0$.

$$x=-1$$ $$x=-\frac{47}{10}$$

Steps Using the Quadratic Formula

Variable $x$ cannot be equal to $-\frac{9}{2}$ since division by zero is not defined. Multiply both sides of the equation by $2x+9$.

$$7+\left(2x+9\right)\left(-6\right)=5x\left(2x+9\right)$$

Use the distributive property to multiply $2x+9$ by $-6$.

$$7-12x-54=5x\left(2x+9\right)$$

Subtract $54$ from $7$ to get $-47$.

$$-47-12x=5x\left(2x+9\right)$$

Use the distributive property to multiply $5x$ by $2x+9$.

$$-47-12x=10x^{2}+45x$$

Subtract $10x^{2}$ from both sides.

$$-47-12x-10x^{2}=45x$$

Subtract $45x$ from both sides.

$$-47-12x-10x^{2}-45x=0$$

Combine $-12x$ and $-45x$ to get $-57x$.

$$-47-57x-10x^{2}=0$$

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.

$$-10x^{2}-57x-47=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $-10$ for $a$, $-57$ for $b$, and $-47$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-\left(-57\right)±\sqrt{\left(-57\right)^{2}-4\left(-10\right)\left(-47\right)}}{2\left(-10\right)}$$

Square $-57$.

$$x=\frac{-\left(-57\right)±\sqrt{3249-4\left(-10\right)\left(-47\right)}}{2\left(-10\right)}$$

Multiply $-4$ times $-10$.

$$x=\frac{-\left(-57\right)±\sqrt{3249+40\left(-47\right)}}{2\left(-10\right)}$$

Multiply $40$ times $-47$.

$$x=\frac{-\left(-57\right)±\sqrt{3249-1880}}{2\left(-10\right)}$$

Add $3249$ to $-1880$.

$$x=\frac{-\left(-57\right)±\sqrt{1369}}{2\left(-10\right)}$$

Take the square root of $1369$.

$$x=\frac{-\left(-57\right)±37}{2\left(-10\right)}$$

The opposite of $-57$ is $57$.

$$x=\frac{57±37}{2\left(-10\right)}$$

Multiply $2$ times $-10$.

$$x=\frac{57±37}{-20}$$

Now solve the equation $x=\frac{57±37}{-20}$ when $±$ is plus. Add $57$ to $37$.

$$x=\frac{94}{-20}$$

Reduce the fraction $\frac{94}{-20}$ to lowest terms by extracting and canceling out $2$.

$$x=-\frac{47}{10}$$

Now solve the equation $x=\frac{57±37}{-20}$ when $±$ is minus. Subtract $37$ from $57$.

$$x=\frac{20}{-20}$$

Divide $20$ by $-20$.

$$x=-1$$

The equation is now solved.

$$x=-\frac{47}{10}$$ $$x=-1$$

Steps for Completing the Square

Variable $x$ cannot be equal to $-\frac{9}{2}$ since division by zero is not defined. Multiply both sides of the equation by $2x+9$.

$$7+\left(2x+9\right)\left(-6\right)=5x\left(2x+9\right)$$

Use the distributive property to multiply $2x+9$ by $-6$.

$$7-12x-54=5x\left(2x+9\right)$$

Subtract $54$ from $7$ to get $-47$.

$$-47-12x=5x\left(2x+9\right)$$

Use the distributive property to multiply $5x$ by $2x+9$.

$$-47-12x=10x^{2}+45x$$

Subtract $10x^{2}$ from both sides.

$$-47-12x-10x^{2}=45x$$

Subtract $45x$ from both sides.

$$-47-12x-10x^{2}-45x=0$$

Combine $-12x$ and $-45x$ to get $-57x$.

$$-47-57x-10x^{2}=0$$

Add $47$ to both sides. Anything plus zero gives itself.

$$-57x-10x^{2}=47$$

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

$$-10x^{2}-57x=47$$

Divide both sides by $-10$.

$$\frac{-10x^{2}-57x}{-10}=\frac{47}{-10}$$

Dividing by $-10$ undoes the multiplication by $-10$.

$$x^{2}+\left(-\frac{57}{-10}\right)x=\frac{47}{-10}$$

Divide $-57$ by $-10$.

$$x^{2}+\frac{57}{10}x=\frac{47}{-10}$$

Divide $47$ by $-10$.

$$x^{2}+\frac{57}{10}x=-\frac{47}{10}$$

Divide $\frac{57}{10}$, the coefficient of the $x$ term, by $2$ to get $\frac{57}{20}$. Then add the square of $\frac{57}{20}$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{57}{10}x+\left(\frac{57}{20}\right)^{2}=-\frac{47}{10}+\left(\frac{57}{20}\right)^{2}$$

Square $\frac{57}{20}$ by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{57}{10}x+\frac{3249}{400}=-\frac{47}{10}+\frac{3249}{400}$$

Add $-\frac{47}{10}$ to $\frac{3249}{400}$ by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{57}{10}x+\frac{3249}{400}=\frac{1369}{400}$$

Factor $x^{2}+\frac{57}{10}x+\frac{3249}{400}$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(x+\frac{57}{20}\right)^{2}=\frac{1369}{400}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{57}{20}\right)^{2}}=\sqrt{\frac{1369}{400}}$$

Simplify.

$$x+\frac{57}{20}=\frac{37}{20}$$ $$x+\frac{57}{20}=-\frac{37}{20}$$

Subtract $\frac{57}{20}$ from both sides of the equation.

$$x=-1$$ $$x=-\frac{47}{10}$$

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Example

Question

Solve for x

Steps Using Factoring By Grouping

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using Factoring By Grouping

Steps Using the Quadratic Formula

Steps for Completing the Square