Algebra Examples

$\frac{y}{3} = \frac{y + 8}{y + 5}$

Step 1

Multiply the numerator of the first fraction by the denominator of the second fraction. Set this equal to the product of the denominator of the first fraction and the numerator of the second fraction.

$y (y + 5) = 3 (y + 8)$

Step 2

Solve the equation for $y$ .

Tap for more steps...

Step 2.1

Simplify $y (y + 5)$ .

Tap for more steps...

Step 2.1.1

Rewrite.

$0 + 0 + y (y + 5) = 3 (y + 8)$

Step 2.1.2

Simplify by adding zeros.

$y (y + 5) = 3 (y + 8)$

Step 2.1.3

Apply the distributive property.

$y \cdot y + y \cdot 5 = 3 (y + 8)$

Step 2.1.4

Simplify the expression.

Tap for more steps...

Step 2.1.4.1

Multiply $y$ by $y$ .

$y^{2} + y \cdot 5 = 3 (y + 8)$

Step 2.1.4.2

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

$y^{2} + 5 y = 3 (y + 8)$

Step 2.2

Simplify $3 (y + 8)$ .

Tap for more steps...

Step 2.2.1

Apply the distributive property.

$y^{2} + 5 y = 3 y + 3 \cdot 8$

Step 2.2.2

Multiply $3$ by $8$ .

$y^{2} + 5 y = 3 y + 24$

Step 2.3

Move all terms containing $y$ to the left side of the equation.

Tap for more steps...

Step 2.3.1

Subtract $3 y$ from both sides of the equation.

$y^{2} + 5 y - 3 y = 24$

Step 2.3.2

Subtract $3 y$ from $5 y$ .

$y^{2} + 2 y = 24$

Step 2.4

Subtract $24$ from both sides of the equation.

$y^{2} + 2 y - 24 = 0$

Step 2.5

Factor $y^{2} + 2 y - 24$ using the AC method.

Tap for more steps...

Step 2.5.1

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 24$ and whose sum is $2$ .

$- 4, 6$

Step 2.5.2

Write the factored form using these integers.

$(y - 4) (y + 6) = 0$

Step 2.6

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$y - 4 = 0$

$y + 6 = 0$

Step 2.7

Set $y - 4$ equal to $0$ and solve for $y$ .

Tap for more steps...

Step 2.7.1

Set $y - 4$ equal to $0$ .

$y - 4 = 0$

Step 2.7.2

Add $4$ to both sides of the equation.

$y = 4$

Step 2.8

Set $y + 6$ equal to $0$ and solve for $y$ .

Tap for more steps...

Step 2.8.1

Set $y + 6$ equal to $0$ .

$y + 6 = 0$

Step 2.8.2

Subtract $6$ from both sides of the equation.

$y = - 6$

Step 2.9

The final solution is all the values that make $(y - 4) (y + 6) = 0$ true.

$y = 4, - 6$