Pre-Algebra Examples

$3 y (y + 4) = - 9$

Step 1

Divide each term in $3 y (y + 4) = - 9$ by $3$ and simplify.

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Step 1.1

Divide each term in $3 y (y + 4) = - 9$ by $3$ .

$\frac{3 y (y + 4)}{3} = \frac{- 9}{3}$

Step 1.2

Simplify the left side.

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Step 1.2.1

Cancel the common factor of $3$ .

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Step 1.2.1.1

Cancel the common factor.

$\frac{3 y (y + 4)}{3} = \frac{- 9}{3}$

Step 1.2.1.2

Divide $y (y + 4)$ by $1$ .

$y (y + 4) = \frac{- 9}{3}$

Step 1.2.2

Apply the distributive property.

$y \cdot y + y \cdot 4 = \frac{- 9}{3}$

Step 1.2.3

Simplify the expression.

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Step 1.2.3.1

Multiply $y$ by $y$ .

$y^{2} + y \cdot 4 = \frac{- 9}{3}$

Step 1.2.3.2

Move $4$ to the left of $y$ .

$y^{2} + 4 y = \frac{- 9}{3}$

Step 1.3

Simplify the right side.

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Step 1.3.1

Divide $- 9$ by $3$ .

$y^{2} + 4 y = - 3$

Step 2

Add $3$ to both sides of the equation.

$y^{2} + 4 y + 3 = 0$

Step 3

Factor $y^{2} + 4 y + 3$ using the AC method.

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Step 3.1

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 $3$ and whose sum is $4$ .

$1, 3$

Step 3.2

Write the factored form using these integers.

$(y + 1) (y + 3) = 0$

Step 4

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

$y + 1 = 0$

$y + 3 = 0$

Step 5

Set $y + 1$ equal to $0$ and solve for $y$ .

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Step 5.1

Set $y + 1$ equal to $0$ .

$y + 1 = 0$

Step 5.2

Subtract $1$ from both sides of the equation.

$y = - 1$

Step 6

Set $y + 3$ equal to $0$ and solve for $y$ .

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Step 6.1

Set $y + 3$ equal to $0$ .

$y + 3 = 0$

Step 6.2

Subtract $3$ from both sides of the equation.

$y = - 3$

Step 7

The final solution is all the values that make $(y + 1) (y + 3) = 0$ true.

$y = - 1, - 3$