Basic Math Examples

$1 r (r + 1) = r + (r + 1) + 5$

Step 1

Simplify $1 r (r + 1)$ .

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Step 1.1

Rewrite.

$0 + 0 + 1 r (r + 1) = r + r + 1 + 5$

Step 1.2

Multiply $r$ by $1$ .

$r (r + 1) = r + r + 1 + 5$

Step 1.3

Apply the distributive property.

$r \cdot r + r \cdot 1 = r + r + 1 + 5$

Step 1.4

Simplify the expression.

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Step 1.4.1

Multiply $r$ by $r$ .

$r^{2} + r \cdot 1 = r + r + 1 + 5$

Step 1.4.2

Multiply $r$ by $1$ .

$r^{2} + r = r + r + 1 + 5$

Step 2

Simplify $r + r + 1 + 5$ .

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Step 2.1

Add $r$ and $r$ .

$r^{2} + r = 2 r + 1 + 5$

Step 2.2

Add $1$ and $5$ .

$r^{2} + r = 2 r + 6$

Step 3

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

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Step 3.1

Subtract $2 r$ from both sides of the equation.

$r^{2} + r - 2 r = 6$

Step 3.2

Subtract $2 r$ from $r$ .

$r^{2} - r = 6$

Step 4

Subtract $6$ from both sides of the equation.

$r^{2} - r - 6 = 0$

Step 5

Factor $r^{2} - r - 6$ using the AC method.

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Step 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 $- 6$ and whose sum is $- 1$ .

$- 3, 2$

Step 5.2

Write the factored form using these integers.

$(r - 3) (r + 2) = 0$

Step 6

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

$r - 3 = 0$

$r + 2 = 0$

Step 7

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

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Step 7.1

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

$r - 3 = 0$

Step 7.2

Add $3$ to both sides of the equation.

$r = 3$

Step 8

Set $r + 2$ equal to $0$ and solve for $r$ .

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Step 8.1

Set $r + 2$ equal to $0$ .

$r + 2 = 0$

Step 8.2

Subtract $2$ from both sides of the equation.

$r = - 2$

Step 9

The final solution is all the values that make $(r - 3) (r + 2) = 0$ true.

$r = 3, - 2$