Solve by Factoring 9y^2=16

$9 y^{2} = 16$

Step 1

Subtract $16$ from both sides of the equation.

$9 y^{2} - 16 = 0$

Step 2

Rewrite $9 y^{2}$ as ${(3 y)}^{2}$ .

${(3 y)}^{2} - 16 = 0$

Step 3

Rewrite $16$ as $4^{2}$ .

${(3 y)}^{2} - 4^{2} = 0$

Step 4

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 3 y$ and $b = 4$ .

$(3 y + 4) (3 y - 4) = 0$

Step 5

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

$3 y + 4 = 0$

$3 y - 4 = 0$

Step 6

Set $3 y + 4$ equal to $0$ and solve for $y$ .

Set $3 y + 4$ equal to $0$ .

$3 y + 4 = 0$

Solve $3 y + 4 = 0$ for $y$ .

Subtract $4$ from both sides of the equation.

$3 y = - 4$

Divide each term in $3 y = - 4$ by $3$ and simplify.

Divide each term in $3 y = - 4$ by $3$ .

$\frac{3 y}{3} = \frac{- 4}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 y}{3} = \frac{- 4}{3}$

Divide $y$ by $1$ .

$y = \frac{- 4}{3}$

Simplify the right side.

Move the negative in front of the fraction.

$y = - \frac{4}{3}$

Step 7

Set $3 y - 4$ equal to $0$ and solve for $y$ .

Set $3 y - 4$ equal to $0$ .

$3 y - 4 = 0$

Solve $3 y - 4 = 0$ for $y$ .

Add $4$ to both sides of the equation.

$3 y = 4$

Divide each term in $3 y = 4$ by $3$ and simplify.

Divide each term in $3 y = 4$ by $3$ .

$\frac{3 y}{3} = \frac{4}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 y}{3} = \frac{4}{3}$

Divide $y$ by $1$ .

$y = \frac{4}{3}$

Step 8

The final solution is all the values that make $(3 y + 4) (3 y - 4) = 0$ true.

$y = - \frac{4}{3}, \frac{4}{3}$

Solve by Factoring 9y^2=16

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