Solve by Substitution 6x^2+3y^2=12 , x+y=2

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

Subtract $y$ from both sides of the equation.

$x = 2 - y$

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

Replace all occurrences of $x$ with $2 - y$ in each equation.

Replace all occurrences of $x$ in $6 x^{2} + 3 y^{2} = 12$ with $2 - y$ .

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

$x = 2 - y$

Simplify the left side.

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

Simplify each term.

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

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

$x = 2 - y$

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

Apply the distributive property.

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

$x = 2 - y$

Apply the distributive property.

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

$x = 2 - y$

Apply the distributive property.

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

$x = 2 - y$

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

$x = 2 - y$

Simplify and combine like terms.

Simplify each term.

Multiply $2$ by $2$ .

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

$x = 2 - y$

Multiply $- 1$ by $2$ .

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

$x = 2 - y$

Multiply $2$ by $- 1$ .

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

$x = 2 - y$

Rewrite using the commutative property of multiplication.

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

$x = 2 - y$

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

Move $y$ .

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

$x = 2 - y$

Multiply $y$ by $y$ .

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

$x = 2 - y$

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

$x = 2 - y$

Multiply $- 1$ by $- 1$ .

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

$x = 2 - y$

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

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

$x = 2 - y$

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

$x = 2 - y$

Subtract $2 y$ from $- 2 y$ .

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

$x = 2 - y$

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

$x = 2 - y$

Apply the distributive property.

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

$x = 2 - y$

Simplify.

Multiply $6$ by $4$ .

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

$x = 2 - y$

Multiply $- 4$ by $6$ .

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

$x = 2 - y$

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

$x = 2 - y$

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

$x = 2 - y$

Add $6 y^{2}$ and $3 y^{2}$ .

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

$x = 2 - y$

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

$x = 2 - y$

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

$x = 2 - y$

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

$x = 2 - y$

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

Subtract $12$ from both sides of the equation.

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

$x = 2 - y$

Subtract $12$ from $24$ .

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

$x = 2 - y$

Factor the left side of the equation.

Factor $3$ out of $- 24 y + 9 y^{2} + 12$ .

Factor $3$ out of $- 24 y$ .

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

$x = 2 - y$

Factor $3$ out of $9 y^{2}$ .

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

$x = 2 - y$

Factor $3$ out of $12$ .

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

$x = 2 - y$

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

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

$x = 2 - y$

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

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

$x = 2 - y$

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

$x = 2 - y$

Let $u = y$ . Substitute $u$ for all occurrences of $y$ .

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

$x = 2 - y$

Factor by grouping.

Reorder terms.

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

$x = 2 - y$

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 3 \cdot 4 = 12$ and whose sum is $b = - 8$ .

Factor $- 8$ out of $- 8 u$ .

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

$x = 2 - y$

Rewrite $- 8$ as $- 2$ plus $- 6$

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

$x = 2 - y$

Apply the distributive property.

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

$x = 2 - y$

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

$x = 2 - y$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

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

$x = 2 - y$

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

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

$x = 2 - y$

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

$x = 2 - y$

Factor the polynomial by factoring out the greatest common factor, $3 u - 2$ .

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

$x = 2 - y$

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

$x = 2 - y$

Factor.

Replace all occurrences of $u$ with $y$ .

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

$x = 2 - y$

Remove unnecessary parentheses.

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

$x = 2 - y$

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

$x = 2 - y$

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

$x = 2 - y$

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

$3 y - 2 = 0$

$y - 2 = 0$

$x = 2 - y$

Set $3 y - 2$ equal to $0$ and solve for $y$ .

Set $3 y - 2$ equal to $0$ .

$3 y - 2 = 0$

$x = 2 - y$

Solve $3 y - 2 = 0$ for $y$ .

Add $2$ to both sides of the equation.

$3 y = 2$

$x = 2 - y$

Divide each term in $3 y = 2$ by $3$ and simplify.

Divide each term in $3 y = 2$ by $3$ .

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

$x = 2 - y$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

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

$x = 2 - y$

Divide $y$ by $1$ .

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

$x = 2 - y$

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

$x = 2 - y$

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

$x = 2 - y$

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

$x = 2 - y$

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

$x = 2 - y$

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

$x = 2 - y$

Set $y - 2$ equal to $0$ and solve for $y$ .

Set $y - 2$ equal to $0$ .

$y - 2 = 0$

$x = 2 - y$

Add $2$ to both sides of the equation.

$y = 2$

$x = 2 - y$

$y = 2$

$x = 2 - y$

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

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

$x = 2 - y$

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

$x = 2 - y$

Replace all occurrences of $y$ with $\frac{2}{3}$ in each equation.

Replace all occurrences of $y$ in $x = 2 - y$ with $\frac{2}{3}$ .

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

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

Simplify the right side.

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

To write $2$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

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

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

Combine $2$ and $\frac{3}{3}$ .

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

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

Combine the numerators over the common denominator.

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

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

Simplify the numerator.

Multiply $2$ by $3$ .

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

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

Subtract $2$ from $6$ .

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

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

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

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

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

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

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

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

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

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

Replace all occurrences of $y$ with $2$ in each equation.

Replace all occurrences of $y$ in $x = 2 - y$ with $2$ .

$x = 2 - (2)$

$y = 2$

Simplify the right side.

Simplify $2 - (2)$ .

Multiply $- 1$ by $2$ .

$x = 2 - 2$

$y = 2$

Subtract $2$ from $2$ .

$x = 0$

$y = 2$

$x = 0$

$y = 2$

$x = 0$

$y = 2$

$x = 0$

$y = 2$

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

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

$(0, 2)$

The result can be shown in multiple forms.

Point Form:

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

Equation Form:

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

$x = 0, y = 2$

Solve by Substitution 6x^2+3y^2=12 , x+y=2

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