Basic Math Examples

$c = \frac{y - 2}{c + 5}$

Step 1

Find the LCD of the terms in the equation.

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Step 1.1

Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.

$1, c + 5$

Step 1.2

Remove parentheses.

$1, c + 5$

Step 1.3

The LCM of one and any expression is the expression.

$c + 5$

Step 2

Multiply each term in $c = \frac{y - 2}{c + 5}$ by $c + 5$ to eliminate the fractions.

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Step 2.1

Multiply each term in $c = \frac{y - 2}{c + 5}$ by $c + 5$ .

$c (c + 5) = \frac{y - 2}{c + 5} (c + 5)$

Step 2.2

Simplify the left side.

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Step 2.2.1

Apply the distributive property.

$c \cdot c + c \cdot 5 = \frac{y - 2}{c + 5} (c + 5)$

Step 2.2.2

Simplify the expression.

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Step 2.2.2.1

Multiply $c$ by $c$ .

$c^{2} + c \cdot 5 = \frac{y - 2}{c + 5} (c + 5)$

Step 2.2.2.2

Move $5$ to the left of $c$ .

$c^{2} + 5 c = \frac{y - 2}{c + 5} (c + 5)$

Step 2.3

Simplify the right side.

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Step 2.3.1

Cancel the common factor of $c + 5$ .

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Step 2.3.1.1

Cancel the common factor.

$c^{2} + 5 c = \frac{y - 2}{c + 5} (c + 5)$

Step 2.3.1.2

Rewrite the expression.

$c^{2} + 5 c = y - 2$

Step 3

Solve the equation.

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Step 3.1

Move all the expressions to the left side of the equation.

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Step 3.1.1

Subtract $y$ from both sides of the equation.

$c^{2} + 5 c - y = - 2$

Step 3.1.2

Add $2$ to both sides of the equation.

$c^{2} + 5 c - y + 2 = 0$

Step 3.2

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 3.3

Substitute the values $a = 1$ , $b = 5$ , and $c = - y + 2$ into the quadratic formula and solve for $c$ .

$\frac{- 5 \pm \sqrt{5^{2} - 4 \cdot (1 \cdot (- y + 2))}}{2 \cdot 1}$

Step 3.4

Simplify.

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Step 3.4.1

Simplify the numerator.

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Step 3.4.1.1

Raise $5$ to the power of $2$ .

$c = \frac{- 5 \pm \sqrt{25 - 4 \cdot 1 \cdot (- y + 2)}}{2 \cdot 1}$

Step 3.4.1.2

Multiply $- 4$ by $1$ .

$c = \frac{- 5 \pm \sqrt{25 - 4 \cdot (- y + 2)}}{2 \cdot 1}$

Step 3.4.1.3

Apply the distributive property.

$c = \frac{- 5 \pm \sqrt{25 - 4 (- y) - 4 \cdot 2}}{2 \cdot 1}$

Step 3.4.1.4

Multiply $- 1$ by $- 4$ .

$c = \frac{- 5 \pm \sqrt{25 + 4 y - 4 \cdot 2}}{2 \cdot 1}$

Step 3.4.1.5

Multiply $- 4$ by $2$ .

$c = \frac{- 5 \pm \sqrt{25 + 4 y - 8}}{2 \cdot 1}$

Step 3.4.1.6

Subtract $8$ from $25$ .

$c = \frac{- 5 \pm \sqrt{4 y + 17}}{2 \cdot 1}$

Step 3.4.2

Multiply $2$ by $1$ .

$c = \frac{- 5 \pm \sqrt{4 y + 17}}{2}$

Step 3.5

The final answer is the combination of both solutions.

$c = - \frac{5 - \sqrt{4 y + 17}}{2}$

$c = - \frac{5 + \sqrt{4 y + 17}}{2}$

$c = - \frac{5 - \sqrt{4 y + 17}}{2}$

$c = - \frac{5 + \sqrt{4 y + 17}}{2}$