Basic Math Examples

$6 (y + 5) = y (y + 13)$

Step 1

Since $y$ is on the right side of the equation, switch the sides so it is on the left side of the equation.

$y (y + 13) = 6 (y + 5)$

Step 2

Simplify $y (y + 13)$ .

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Step 2.1

Rewrite.

$0 + 0 + y (y + 13) = 6 (y + 5)$

Step 2.2

Simplify by adding zeros.

$y (y + 13) = 6 (y + 5)$

Step 2.3

Apply the distributive property.

$y \cdot y + y \cdot 13 = 6 (y + 5)$

Step 2.4

Simplify the expression.

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Step 2.4.1

Multiply $y$ by $y$ .

$y^{2} + y \cdot 13 = 6 (y + 5)$

Step 2.4.2

Move $13$ to the left of $y$ .

$y^{2} + 13 y = 6 (y + 5)$

Step 3

Simplify $6 (y + 5)$ .

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Step 3.1

Apply the distributive property.

$y^{2} + 13 y = 6 y + 6 \cdot 5$

Step 3.2

Multiply $6$ by $5$ .

$y^{2} + 13 y = 6 y + 30$

Step 4

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

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Step 4.1

Subtract $6 y$ from both sides of the equation.

$y^{2} + 13 y - 6 y = 30$

Step 4.2

Subtract $6 y$ from $13 y$ .

$y^{2} + 7 y = 30$

Step 5

Subtract $30$ from both sides of the equation.

$y^{2} + 7 y - 30 = 0$

Step 6

Factor $y^{2} + 7 y - 30$ using the AC method.

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Step 6.1

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 $- 30$ and whose sum is $7$ .

$- 3, 10$

Step 6.2

Write the factored form using these integers.

$(y - 3) (y + 10) = 0$

Step 7

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

$y - 3 = 0$

$y + 10 = 0$

Step 8

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

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Step 8.1

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

$y - 3 = 0$

Step 8.2

Add $3$ to both sides of the equation.

$y = 3$

Step 9

Set $y + 10$ equal to $0$ and solve for $y$ .

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Step 9.1

Set $y + 10$ equal to $0$ .

$y + 10 = 0$

Step 9.2

Subtract $10$ from both sides of the equation.

$y = - 10$

Step 10

The final solution is all the values that make $(y - 3) (y + 10) = 0$ true.

$y = 3, - 10$