Basic Math Examples

$y = \frac{(b - 2) (b - 1)}{b - 2}$

Step 1

Multiply the equation by $b - 2$ .

$(b - 2) (y) = (\frac{(b - 2) (b - 1)}{b - 2}) (b - 2)$

Step 2

Simplify the left side.

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Step 2.1

Apply the distributive property.

$b y - 2 y = (\frac{(b - 2) (b - 1)}{b - 2}) (b - 2)$

Step 3

Simplify the right side.

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Step 3.1

Simplify $(\frac{(b - 2) (b - 1)}{b - 2}) (b - 2)$ .

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Step 3.1.1

Cancel the common factor of $b - 2$ .

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Step 3.1.1.1

Cancel the common factor.

$b y - 2 y = \frac{(b - 2) (b - 1)}{b - 2} (b - 2)$

Step 3.1.1.2

Rewrite the expression.

$b y - 2 y = (b - 2) (b - 1)$

Step 3.1.2

Expand $(b - 2) (b - 1)$ using the FOIL Method.

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Step 3.1.2.1

Apply the distributive property.

$b y - 2 y = b (b - 1) - 2 (b - 1)$

Step 3.1.2.2

Apply the distributive property.

$b y - 2 y = b \cdot b + b \cdot - 1 - 2 (b - 1)$

Step 3.1.2.3

Apply the distributive property.

$b y - 2 y = b \cdot b + b \cdot - 1 - 2 b - 2 \cdot - 1$

Step 3.1.3

Simplify and combine like terms.

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Step 3.1.3.1

Simplify each term.

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Step 3.1.3.1.1

Multiply $b$ by $b$ .

$b y - 2 y = b^{2} + b \cdot - 1 - 2 b - 2 \cdot - 1$

Step 3.1.3.1.2

Move $- 1$ to the left of $b$ .

$b y - 2 y = b^{2} - 1 \cdot b - 2 b - 2 \cdot - 1$

Step 3.1.3.1.3

Rewrite $- 1 b$ as $- b$ .

$b y - 2 y = b^{2} - b - 2 b - 2 \cdot - 1$

Step 3.1.3.1.4

Multiply $- 2$ by $- 1$ .

$b y - 2 y = b^{2} - b - 2 b + 2$

Step 3.1.3.2

Subtract $2 b$ from $- b$ .

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

Step 4

Solve for $b$ .

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Step 4.1

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

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

Step 4.2

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

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

Step 4.3

Add $2 y$ to both sides of the equation.

$b^{2} - 3 b + 2 - b y + 2 y = 0$

Step 4.4

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 4.5

Substitute the values $a = 1$ , $b = - 3 - y$ , and $c = 2 + 2 y$ into the quadratic formula and solve for $b$ .

$\frac{- (- 3 - y) \pm \sqrt{{(- 3 - y)}^{2} - 4 \cdot (1 \cdot (2 + 2 y))}}{2 \cdot 1}$

Step 4.6

Simplify.

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Step 4.6.1

Simplify the numerator.

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Step 4.6.1.1

Apply the distributive property.

$b = \frac{3 + y \pm \sqrt{{(- 3 - y)}^{2} - 4 \cdot 1 \cdot (2 + 2 y)}}{2 \cdot 1}$

Step 4.6.1.2

Multiply $- 1$ by $- 3$ .

$b = \frac{3 + y \pm \sqrt{{(- 3 - y)}^{2} - 4 \cdot 1 \cdot (2 + 2 y)}}{2 \cdot 1}$

Step 4.6.1.3

Multiply $- - y$ .

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Step 4.6.1.3.1

Multiply $- 1$ by $- 1$ .

$b = \frac{3 + 1 y \pm \sqrt{{(- 3 - y)}^{2} - 4 \cdot 1 \cdot (2 + 2 y)}}{2 \cdot 1}$

Step 4.6.1.3.2

Multiply $y$ by $1$ .

$b = \frac{3 + y \pm \sqrt{{(- 3 - y)}^{2} - 4 \cdot 1 \cdot (2 + 2 y)}}{2 \cdot 1}$

Step 4.6.1.4

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

$b = \frac{3 + y \pm \sqrt{(- 3 - y) (- 3 - y) - 4 \cdot 1 \cdot (2 + 2 y)}}{2 \cdot 1}$

Step 4.6.1.5

Expand $(- 3 - y) (- 3 - y)$ using the FOIL Method.

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Step 4.6.1.5.1

Apply the distributive property.

$b = \frac{3 + y \pm \sqrt{- 3 (- 3 - y) - y (- 3 - y) - 4 \cdot 1 \cdot (2 + 2 y)}}{2 \cdot 1}$

Step 4.6.1.5.2

Apply the distributive property.

$b = \frac{3 + y \pm \sqrt{- 3 \cdot - 3 - 3 (- y) - y (- 3 - y) - 4 \cdot 1 \cdot (2 + 2 y)}}{2 \cdot 1}$

Step 4.6.1.5.3

Apply the distributive property.

$b = \frac{3 + y \pm \sqrt{- 3 \cdot - 3 - 3 (- y) - y \cdot - 3 - y (- y) - 4 \cdot 1 \cdot (2 + 2 y)}}{2 \cdot 1}$

Step 4.6.1.6

Simplify and combine like terms.

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Step 4.6.1.6.1

Simplify each term.

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Step 4.6.1.6.1.1

Multiply $- 3$ by $- 3$ .

$b = \frac{3 + y \pm \sqrt{9 - 3 (- y) - y \cdot - 3 - y (- y) - 4 \cdot 1 \cdot (2 + 2 y)}}{2 \cdot 1}$

Step 4.6.1.6.1.2

Multiply $- 1$ by $- 3$ .

$b = \frac{3 + y \pm \sqrt{9 + 3 y - y \cdot - 3 - y (- y) - 4 \cdot 1 \cdot (2 + 2 y)}}{2 \cdot 1}$

Step 4.6.1.6.1.3

Multiply $- 3$ by $- 1$ .

$b = \frac{3 + y \pm \sqrt{9 + 3 y + 3 y - y (- y) - 4 \cdot 1 \cdot (2 + 2 y)}}{2 \cdot 1}$

Step 4.6.1.6.1.4

Rewrite using the commutative property of multiplication.

$b = \frac{3 + y \pm \sqrt{9 + 3 y + 3 y - 1 \cdot (- 1 y \cdot y) - 4 \cdot 1 \cdot (2 + 2 y)}}{2 \cdot 1}$

Step 4.6.1.6.1.5

Multiply $y$ by $y$ by adding the exponents.

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Step 4.6.1.6.1.5.1

Move $y$ .

$b = \frac{3 + y \pm \sqrt{9 + 3 y + 3 y - 1 \cdot (- 1 (y \cdot y)) - 4 \cdot 1 \cdot (2 + 2 y)}}{2 \cdot 1}$

Step 4.6.1.6.1.5.2

Multiply $y$ by $y$ .

$b = \frac{3 + y \pm \sqrt{9 + 3 y + 3 y - 1 \cdot (- 1 y^{2}) - 4 \cdot 1 \cdot (2 + 2 y)}}{2 \cdot 1}$

Step 4.6.1.6.1.6

Multiply $- 1$ by $- 1$ .

$b = \frac{3 + y \pm \sqrt{9 + 3 y + 3 y + 1 y^{2} - 4 \cdot 1 \cdot (2 + 2 y)}}{2 \cdot 1}$

Step 4.6.1.6.1.7

Multiply $y^{2}$ by $1$ .

$b = \frac{3 + y \pm \sqrt{9 + 3 y + 3 y + y^{2} - 4 \cdot 1 \cdot (2 + 2 y)}}{2 \cdot 1}$

Step 4.6.1.6.2

Add $3 y$ and $3 y$ .

$b = \frac{3 + y \pm \sqrt{9 + 6 y + y^{2} - 4 \cdot 1 \cdot (2 + 2 y)}}{2 \cdot 1}$

Step 4.6.1.7

Multiply $- 4$ by $1$ .

$b = \frac{3 + y \pm \sqrt{9 + 6 y + y^{2} - 4 \cdot (2 + 2 y)}}{2 \cdot 1}$

Step 4.6.1.8

Apply the distributive property.

$b = \frac{3 + y \pm \sqrt{9 + 6 y + y^{2} - 4 \cdot 2 - 4 (2 y)}}{2 \cdot 1}$

Step 4.6.1.9

Multiply $- 4$ by $2$ .

$b = \frac{3 + y \pm \sqrt{9 + 6 y + y^{2} - 8 - 4 (2 y)}}{2 \cdot 1}$

Step 4.6.1.10

Multiply $2$ by $- 4$ .

$b = \frac{3 + y \pm \sqrt{9 + 6 y + y^{2} - 8 - 8 y}}{2 \cdot 1}$

Step 4.6.1.11

Subtract $8$ from $9$ .

$b = \frac{3 + y \pm \sqrt{6 y + y^{2} + 1 - 8 y}}{2 \cdot 1}$

Step 4.6.1.12

Subtract $8 y$ from $6 y$ .

$b = \frac{3 + y \pm \sqrt{- 2 y + y^{2} + 1}}{2 \cdot 1}$

Step 4.6.1.13

Reorder terms.

$b = \frac{3 + y \pm \sqrt{y^{2} - 2 y + 1}}{2 \cdot 1}$

Step 4.6.1.14

Factor using the perfect square rule.

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Step 4.6.1.14.1

Rewrite $1$ as $1^{2}$ .

$b = \frac{3 + y \pm \sqrt{y^{2} - 2 y + 1^{2}}}{2 \cdot 1}$

Step 4.6.1.14.2

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$2 y = 2 \cdot y \cdot 1$

Step 4.6.1.14.3

Rewrite the polynomial.

$b = \frac{3 + y \pm \sqrt{y^{2} - 2 \cdot y \cdot 1 + 1^{2}}}{2 \cdot 1}$

Step 4.6.1.14.4

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = y$ and $b = 1$ .

$b = \frac{3 + y \pm \sqrt{{(y - 1)}^{2}}}{2 \cdot 1}$

Step 4.6.1.15

Pull terms out from under the radical, assuming positive real numbers.

$b = \frac{3 + y \pm (y - 1)}{2 \cdot 1}$

Step 4.6.2

Multiply $2$ by $1$ .

$b = \frac{3 + y \pm (y - 1)}{2}$

Step 4.7

The final answer is the combination of both solutions.

$b = y + 1$

$b = 2$

$b = y + 1$

$b = 2$