Algebra Examples

6 x^{2} + 3 y^{2} = 12

$6 x^{2} + 3 y^{2} = 12$ ,

x + y = 2

$x + y = 2$

Step 1

Subtract

y

$y$ from both sides of the equation.

x = 2 - y

$x = 2 - y$

6 x^{2} + 3 y^{2} = 12

$6 x^{2} + 3 y^{2} = 12$

Step 2

Replace all occurrences of

x

$x$ with

2 - y

$2 - y$ in each equation.

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Step 2.1

Replace all occurrences of

x

$x$ in

6 x^{2} + 3 y^{2} = 12

$6 x^{2} + 3 y^{2} = 12$ with

2 - y

$2 - y$ .

6 {(2 - y)}^{2} + 3 y^{2} = 12

$6 {(2 - y)}^{2} + 3 y^{2} = 12$

x = 2 - y

$x = 2 - y$

Step 2.2

Simplify the left side.

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Step 2.2.1

Simplify

6 {(2 - y)}^{2} + 3 y^{2}

$6 {(2 - y)}^{2} + 3 y^{2}$ .

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Step 2.2.1.1

Simplify each term.

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Step 2.2.1.1.1

Rewrite

{(2 - y)}^{2}

${(2 - y)}^{2}$ as

(2 - y) (2 - y)

$(2 - y) (2 - y)$ .

6 ((2 - y) (2 - y)) + 3 y^{2} = 12

$6 ((2 - y) (2 - y)) + 3 y^{2} = 12$

x = 2 - y

$x = 2 - y$

Step 2.2.1.1.2

Expand

(2 - y) (2 - y)

$(2 - y) (2 - y)$ using the FOIL Method.

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Step 2.2.1.1.2.1

Apply the distributive property.

6 (2 (2 - y) - y (2 - y)) + 3 y^{2} = 12

$6 (2 (2 - y) - y (2 - y)) + 3 y^{2} = 12$

x = 2 - y

$x = 2 - y$

Step 2.2.1.1.2.2

Apply the distributive property.

6 (2 \cdot 2 + 2 (- y) - y (2 - y)) + 3 y^{2} = 12

$6 (2 \cdot 2 + 2 (- y) - y (2 - y)) + 3 y^{2} = 12$

x = 2 - y

$x = 2 - y$

Step 2.2.1.1.2.3

Apply the distributive property.

6 (2 \cdot 2 + 2 (- y) - y \cdot 2 - y (- y)) + 3 y^{2} = 12

$6 (2 \cdot 2 + 2 (- y) - y \cdot 2 - y (- y)) + 3 y^{2} = 12$

x = 2 - y

$x = 2 - y$

6 (2 \cdot 2 + 2 (- y) - y \cdot 2 - y (- y)) + 3 y^{2} = 12

$6 (2 \cdot 2 + 2 (- y) - y \cdot 2 - y (- y)) + 3 y^{2} = 12$

x = 2 - y

$x = 2 - y$

Step 2.2.1.1.3

Simplify and combine like terms.

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Step 2.2.1.1.3.1

Simplify each term.

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Step 2.2.1.1.3.1.1

Multiply

2

$2$ by

2

$2$ .

6 (4 + 2 (- y) - y \cdot 2 - y (- y)) + 3 y^{2} = 12

$6 (4 + 2 (- y) - y \cdot 2 - y (- y)) + 3 y^{2} = 12$

x = 2 - y

$x = 2 - y$

Step 2.2.1.1.3.1.2

Multiply

- 1

$- 1$ by

2

$2$ .

6 (4 - 2 y - y \cdot 2 - y (- y)) + 3 y^{2} = 12

$6 (4 - 2 y - y \cdot 2 - y (- y)) + 3 y^{2} = 12$

x = 2 - y

$x = 2 - y$

Step 2.2.1.1.3.1.3

Multiply

2

$2$ by

- 1

$- 1$ .

6 (4 - 2 y - 2 y - y (- y)) + 3 y^{2} = 12

$6 (4 - 2 y - 2 y - y (- y)) + 3 y^{2} = 12$

x = 2 - y

$x = 2 - y$

Step 2.2.1.1.3.1.4

Rewrite using the commutative property of multiplication.

6 (4 - 2 y - 2 y - 1 \cdot (- 1 y \cdot y)) + 3 y^{2} = 12

$6 (4 - 2 y - 2 y - 1 \cdot (- 1 y \cdot y)) + 3 y^{2} = 12$

x = 2 - y

$x = 2 - y$

Step 2.2.1.1.3.1.5

Multiply

y

$y$ by

y

$y$ by adding the exponents.

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Step 2.2.1.1.3.1.5.1

Move

y

$y$ .

6 (4 - 2 y - 2 y - 1 \cdot (- 1 (y \cdot y))) + 3 y^{2} = 12

$6 (4 - 2 y - 2 y - 1 \cdot (- 1 (y \cdot y))) + 3 y^{2} = 12$

x = 2 - y

$x = 2 - y$

Step 2.2.1.1.3.1.5.2

Multiply

y

$y$ by

y

$y$ .

6 (4 - 2 y - 2 y - 1 \cdot (- 1 y^{2})) + 3 y^{2} = 12

$6 (4 - 2 y - 2 y - 1 \cdot (- 1 y^{2})) + 3 y^{2} = 12$

x = 2 - y

$x = 2 - y$

6 (4 - 2 y - 2 y - 1 \cdot (- 1 y^{2})) + 3 y^{2} = 12

$6 (4 - 2 y - 2 y - 1 \cdot (- 1 y^{2})) + 3 y^{2} = 12$

x = 2 - y

$x = 2 - y$

Step 2.2.1.1.3.1.6

Multiply

- 1

$- 1$ by

- 1

$- 1$ .

6 (4 - 2 y - 2 y + 1 y^{2}) + 3 y^{2} = 12

$6 (4 - 2 y - 2 y + 1 y^{2}) + 3 y^{2} = 12$

x = 2 - y

$x = 2 - y$

Step 2.2.1.1.3.1.7

Multiply

y^{2}

$y^{2}$ by

1

$1$ .

6 (4 - 2 y - 2 y + y^{2}) + 3 y^{2} = 12

$6 (4 - 2 y - 2 y + y^{2}) + 3 y^{2} = 12$

x = 2 - y

$x = 2 - y$

6 (4 - 2 y - 2 y + y^{2}) + 3 y^{2} = 12

$6 (4 - 2 y - 2 y + y^{2}) + 3 y^{2} = 12$

x = 2 - y

$x = 2 - y$

Step 2.2.1.1.3.2

Subtract

2 y

$2 y$ from

- 2 y

$- 2 y$ .

6 (4 - 4 y + y^{2}) + 3 y^{2} = 12

$6 (4 - 4 y + y^{2}) + 3 y^{2} = 12$

x = 2 - y

$x = 2 - y$

6 (4 - 4 y + y^{2}) + 3 y^{2} = 12

$6 (4 - 4 y + y^{2}) + 3 y^{2} = 12$

x = 2 - y

$x = 2 - y$

Step 2.2.1.1.4

Apply the distributive property.

6 \cdot 4 + 6 (- 4 y) + 6 y^{2} + 3 y^{2} = 12

$6 \cdot 4 + 6 (- 4 y) + 6 y^{2} + 3 y^{2} = 12$

x = 2 - y

$x = 2 - y$

Step 2.2.1.1.5

Simplify.

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Step 2.2.1.1.5.1

Multiply

6

$6$ by

4

$4$ .

24 + 6 (- 4 y) + 6 y^{2} + 3 y^{2} = 12

$24 + 6 (- 4 y) + 6 y^{2} + 3 y^{2} = 12$

x = 2 - y

$x = 2 - y$

Step 2.2.1.1.5.2

Multiply

- 4

$- 4$ by

6

$6$ .

24 - 24 y + 6 y^{2} + 3 y^{2} = 12

$24 - 24 y + 6 y^{2} + 3 y^{2} = 12$

x = 2 - y

$x = 2 - y$

24 - 24 y + 6 y^{2} + 3 y^{2} = 12

$24 - 24 y + 6 y^{2} + 3 y^{2} = 12$

x = 2 - y

$x = 2 - y$

24 - 24 y + 6 y^{2} + 3 y^{2} = 12

$24 - 24 y + 6 y^{2} + 3 y^{2} = 12$

x = 2 - y

$x = 2 - y$

Step 2.2.1.2

Add

6 y^{2}

$6 y^{2}$ and

3 y^{2}

$3 y^{2}$ .

24 - 24 y + 9 y^{2} = 12

$24 - 24 y + 9 y^{2} = 12$

x = 2 - y

$x = 2 - y$

24 - 24 y + 9 y^{2} = 12

$24 - 24 y + 9 y^{2} = 12$

x = 2 - y

$x = 2 - y$

24 - 24 y + 9 y^{2} = 12

$24 - 24 y + 9 y^{2} = 12$

x = 2 - y

$x = 2 - y$

24 - 24 y + 9 y^{2} = 12

$24 - 24 y + 9 y^{2} = 12$

x = 2 - y

$x = 2 - y$

Step 3

Solve for

y

$y$ in

24 - 24 y + 9 y^{2} = 12

$24 - 24 y + 9 y^{2} = 12$ .

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Step 3.1

Subtract

12

$12$ from both sides of the equation.

24 - 24 y + 9 y^{2} - 12 = 0

$24 - 24 y + 9 y^{2} - 12 = 0$

x = 2 - y

$x = 2 - y$

Step 3.2

Subtract

12

$12$ from

24

$24$ .

- 24 y + 9 y^{2} + 12 = 0

$- 24 y + 9 y^{2} + 12 = 0$

x = 2 - y

$x = 2 - y$

Step 3.3

Factor the left side of the equation.

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Step 3.3.1

Factor

3

$3$ out of

- 24 y + 9 y^{2} + 12

$- 24 y + 9 y^{2} + 12$ .

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Step 3.3.1.1

Factor

3

$3$ out of

- 24 y

$- 24 y$ .

3 (- 8 y) + 9 y^{2} + 12 = 0

$3 (- 8 y) + 9 y^{2} + 12 = 0$

x = 2 - y

$x = 2 - y$

Step 3.3.1.2

Factor

3

$3$ out of

9 y^{2}

$9 y^{2}$ .

3 (- 8 y) + 3 (3 y^{2}) + 12 = 0

$3 (- 8 y) + 3 (3 y^{2}) + 12 = 0$

x = 2 - y

$x = 2 - y$

Step 3.3.1.3

Factor

3

$3$ out of

12

$12$ .

3 (- 8 y) + 3 (3 y^{2}) + 3 (4) = 0

$3 (- 8 y) + 3 (3 y^{2}) + 3 (4) = 0$

x = 2 - y

$x = 2 - y$

Step 3.3.1.4

Factor

3

$3$ out of

3 (- 8 y) + 3 (3 y^{2})

$3 (- 8 y) + 3 (3 y^{2})$ .

3 (- 8 y + 3 y^{2}) + 3 (4) = 0

$3 (- 8 y + 3 y^{2}) + 3 (4) = 0$

x = 2 - y

$x = 2 - y$

Step 3.3.1.5

Factor

3

$3$ out of

3 (- 8 y + 3 y^{2}) + 3 (4)

$3 (- 8 y + 3 y^{2}) + 3 (4)$ .

3 (- 8 y + 3 y^{2} + 4) = 0

$3 (- 8 y + 3 y^{2} + 4) = 0$

x = 2 - y

$x = 2 - y$

3 (- 8 y + 3 y^{2} + 4) = 0

$3 (- 8 y + 3 y^{2} + 4) = 0$

x = 2 - y

$x = 2 - y$

Step 3.3.2

Let

u = y

$u = y$ . Substitute

u

$u$ for all occurrences of

y

$y$ .

3 (- 8 u + 3 u^{2} + 4) = 0

$3 (- 8 u + 3 u^{2} + 4) = 0$

x = 2 - y

$x = 2 - y$

Step 3.3.3

Factor by grouping.

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Step 3.3.3.1

Reorder terms.

3 (3 u^{2} - 8 u + 4) = 0

$3 (3 u^{2} - 8 u + 4) = 0$

x = 2 - y

$x = 2 - y$

Step 3.3.3.2

For a polynomial of the form

a x^{2} + b x + c

$a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is

a \cdot c = 3 \cdot 4 = 12

$a \cdot c = 3 \cdot 4 = 12$ and whose sum is

b = - 8

$b = - 8$ .

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Step 3.3.3.2.1

Factor

- 8

$- 8$ out of

- 8 u

$- 8 u$ .

3 (3 u^{2} - 8 u + 4) = 0

$3 (3 u^{2} - 8 u + 4) = 0$

x = 2 - y

$x = 2 - y$

Step 3.3.3.2.2

Rewrite

- 8

$- 8$ as

- 2

$- 2$ plus

- 6

$- 6$

3 (3 u^{2} + (- 2 - 6) u + 4) = 0

$3 (3 u^{2} + (- 2 - 6) u + 4) = 0$

x = 2 - y

$x = 2 - y$

Step 3.3.3.2.3

Apply the distributive property.

3 (3 u^{2} - 2 u - 6 u + 4) = 0

$3 (3 u^{2} - 2 u - 6 u + 4) = 0$

x = 2 - y

$x = 2 - y$

3 (3 u^{2} - 2 u - 6 u + 4) = 0

$3 (3 u^{2} - 2 u - 6 u + 4) = 0$

x = 2 - y

$x = 2 - y$

Step 3.3.3.3

Factor out the greatest common factor from each group.

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Step 3.3.3.3.1

Group the first two terms and the last two terms.

3 ((3 u^{2} - 2 u) - 6 u + 4) = 0

$3 ((3 u^{2} - 2 u) - 6 u + 4) = 0$

x = 2 - y

$x = 2 - y$

Step 3.3.3.3.2

Factor out the greatest common factor (GCF) from each group.

3 (u (3 u - 2) - 2 (3 u - 2)) = 0

$3 (u (3 u - 2) - 2 (3 u - 2)) = 0$

x = 2 - y

$x = 2 - y$

3 (u (3 u - 2) - 2 (3 u - 2)) = 0

$3 (u (3 u - 2) - 2 (3 u - 2)) = 0$

x = 2 - y

$x = 2 - y$

Step 3.3.3.4

Factor the polynomial by factoring out the greatest common factor,

3 u - 2

$3 u - 2$ .

3 ((3 u - 2) (u - 2)) = 0

$3 ((3 u - 2) (u - 2)) = 0$

x = 2 - y

$x = 2 - y$

3 ((3 u - 2) (u - 2)) = 0

$3 ((3 u - 2) (u - 2)) = 0$

x = 2 - y

$x = 2 - y$

Step 3.3.4

Factor.

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Step 3.3.4.1

Replace all occurrences of

u

$u$ with

y

$y$ .

3 ((3 y - 2) (y - 2)) = 0

$3 ((3 y - 2) (y - 2)) = 0$

x = 2 - y

$x = 2 - y$

Step 3.3.4.2

Remove unnecessary parentheses.

3 (3 y - 2) (y - 2) = 0

$3 (3 y - 2) (y - 2) = 0$

x = 2 - y

$x = 2 - y$

3 (3 y - 2) (y - 2) = 0

$3 (3 y - 2) (y - 2) = 0$

x = 2 - y

$x = 2 - y$

3 (3 y - 2) (y - 2) = 0

$3 (3 y - 2) (y - 2) = 0$

x = 2 - y

$x = 2 - y$

Step 3.4

If any individual factor on the left side of the equation is equal to

0

$0$ , the entire expression will be equal to

0

$0$ .

3 y - 2 = 0

$3 y - 2 = 0$

y - 2 = 0

$y - 2 = 0$

x = 2 - y

$x = 2 - y$

Step 3.5

Set

3 y - 2

$3 y - 2$ equal to

0

$0$ and solve for

y

$y$ .

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Step 3.5.1

Set

3 y - 2

$3 y - 2$ equal to

0

$0$ .

3 y - 2 = 0

$3 y - 2 = 0$

x = 2 - y

$x = 2 - y$

Step 3.5.2

Solve

3 y - 2 = 0

$3 y - 2 = 0$ for

y

$y$ .

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Step 3.5.2.1

Add

2

$2$ to both sides of the equation.

3 y = 2

$3 y = 2$

x = 2 - y

$x = 2 - y$

Step 3.5.2.2

Divide each term in

3 y = 2

$3 y = 2$ by

3

$3$ and simplify.

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Step 3.5.2.2.1

Divide each term in

3 y = 2

$3 y = 2$ by

3

$3$ .

\frac{3 y}{3} = \frac{2}{3}

$\frac{3 y}{3} = \frac{2}{3}$

x = 2 - y

$x = 2 - y$

Step 3.5.2.2.2

Simplify the left side.

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Step 3.5.2.2.2.1

Cancel the common factor of

3

$3$ .

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Step 3.5.2.2.2.1.1

Cancel the common factor.

$\frac{3 y}{3} = \frac{2}{3}$

$x = 2 - y$

Step 3.5.2.2.2.1.2

Divide

$y$ by

$1$ .

$y = \frac{2}{3}$

$x = 2 - y$

$y = \frac{2}{3}$

$x = 2 - y$

$y = \frac{2}{3}$

$x = 2 - y$

$y = \frac{2}{3}$

$x = 2 - y$

$y = \frac{2}{3}$

$x = 2 - y$

$y = \frac{2}{3}$

$x = 2 - y$

Step 3.6

Set

$y - 2$ equal to

$0$ and solve for

$y$ .

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Step 3.6.1

Set

$y - 2$ equal to

$0$ .

$y - 2 = 0$

$x = 2 - y$

Step 3.6.2

Add

$2$ to both sides of the equation.

$y = 2$

$x = 2 - y$

$y = 2$

$x = 2 - y$

Step 3.7

The final solution is all the values that make

$3 (3 y - 2) (y - 2) = 0$ true.

$y = \frac{2}{3}, 2$

$x = 2 - y$

$y = \frac{2}{3}, 2$

$x = 2 - y$

Step 4

Replace all occurrences of

$y$ with

$\frac{2}{3}$ in each equation.

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Step 4.1

Replace all occurrences of

$y$ in

$x = 2 - y$ with

$\frac{2}{3}$ .

$x = 2 - (\frac{2}{3})$

$y = \frac{2}{3}$

Step 4.2

Simplify the right side.

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Step 4.2.1

Simplify

$2 - (\frac{2}{3})$ .

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Step 4.2.1.1

To write

$2$ as a fraction with a common denominator, multiply by

$\frac{3}{3}$ .

$x = 2 \cdot \frac{3}{3} - \frac{2}{3}$

$y = \frac{2}{3}$

Step 4.2.1.2

Combine

$2$ and

$\frac{3}{3}$ .

$x = \frac{2 \cdot 3}{3} - \frac{2}{3}$

$y = \frac{2}{3}$

Step 4.2.1.3

Combine the numerators over the common denominator.

$x = \frac{2 \cdot 3 - 2}{3}$

$y = \frac{2}{3}$

Step 4.2.1.4

Simplify the numerator.

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Step 4.2.1.4.1

Multiply

$2$ by

$3$ .

$x = \frac{6 - 2}{3}$

$y = \frac{2}{3}$

Step 4.2.1.4.2

Subtract

$2$ from

$6$ .

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

$y = \frac{2}{3}$

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

$y = \frac{2}{3}$

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

$y = \frac{2}{3}$

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

$y = \frac{2}{3}$

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

$y = \frac{2}{3}$

Step 5

Replace all occurrences of

$y$ with

$2$ in each equation.

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Step 5.1

Replace all occurrences of

$y$ in

$x = 2 - y$ with

$2$ .

$x = 2 - (2)$

$y = 2$

Step 5.2

Simplify the right side.

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Step 5.2.1

Simplify

$2 - (2)$ .

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Step 5.2.1.1

Multiply

$- 1$ by

$2$ .

$x = 2 - 2$

$y = 2$

Step 5.2.1.2

Subtract

$2$ from

$2$ .

$x = 0$

$y = 2$

$x = 0$

$y = 2$

$x = 0$

$y = 2$

$x = 0$

$y = 2$

Step 6

The solution to the system is the complete set of ordered pairs that are valid solutions.

$(\frac{4}{3}, \frac{2}{3})$

$(0, 2)$

Step 7

The result can be shown in multiple forms.

Point Form:

$(\frac{4}{3}, \frac{2}{3}), (0, 2)$

Equation Form:

$x = \frac{4}{3}, y = \frac{2}{3}$

$x = 0, y = 2$

Step 8

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