Solve the System of Equations 9x^2+4y^2=36 y=9/2x-3

$9 x^{2} + 4 y^{2} = 36$ $y = \frac{9}{2} x - 3$

Combine $\frac{9}{2}$ and $x$ .

$y = \frac{9 x}{2} - 3$

$9 x^{2} + 4 y^{2} = 36$

Replace all occurrences of $y$ with $\frac{9 x}{2} - 3$ in each equation.

Replace all occurrences of $y$ in $9 x^{2} + 4 y^{2} = 36$ with $\frac{9 x}{2} - 3$ .

$9 x^{2} + 4 {(\frac{9 x}{2} - 3)}^{2} = 36$

$y = \frac{9 x}{2} - 3$

Simplify the left side.

Simplify $9 x^{2} + 4 {(\frac{9 x}{2} - 3)}^{2}$ .

Simplify each term.

Rewrite ${(\frac{9 x}{2} - 3)}^{2}$ as $(\frac{9 x}{2} - 3) (\frac{9 x}{2} - 3)$ .

$9 x^{2} + 4 ((\frac{9 x}{2} - 3) (\frac{9 x}{2} - 3)) = 36$

$y = \frac{9 x}{2} - 3$

Expand $(\frac{9 x}{2} - 3) (\frac{9 x}{2} - 3)$ using the FOIL Method.

Apply the distributive property.

$9 x^{2} + 4 (\frac{9 x}{2} \cdot (\frac{9 x}{2} - 3) - 3 (\frac{9 x}{2} - 3)) = 36$

$y = \frac{9 x}{2} - 3$

Apply the distributive property.

$9 x^{2} + 4 (\frac{9 x}{2} \cdot \frac{9 x}{2} + \frac{9 x}{2} \cdot - 3 - 3 (\frac{9 x}{2} - 3)) = 36$

$y = \frac{9 x}{2} - 3$

Apply the distributive property.

$9 x^{2} + 4 (\frac{9 x}{2} \cdot \frac{9 x}{2} + \frac{9 x}{2} \cdot - 3 - 3 \frac{9 x}{2} - 3 \cdot - 3) = 36$

$y = \frac{9 x}{2} - 3$

$9 x^{2} + 4 (\frac{9 x}{2} \cdot \frac{9 x}{2} + \frac{9 x}{2} \cdot - 3 - 3 \frac{9 x}{2} - 3 \cdot - 3) = 36$

$y = \frac{9 x}{2} - 3$

Simplify and combine like terms.

Simplify each term.

Multiply $\frac{9 x}{2} \cdot \frac{9 x}{2}$ .

Multiply $\frac{9 x}{2}$ by $\frac{9 x}{2}$ .

$9 x^{2} + 4 (\frac{9 x (9 x)}{2 \cdot 2} + \frac{9 x}{2} \cdot - 3 - 3 \frac{9 x}{2} - 3 \cdot - 3) = 36$

$y = \frac{9 x}{2} - 3$

Multiply $9$ by $9$ .

$9 x^{2} + 4 (\frac{81 x \cdot x}{2 \cdot 2} + \frac{9 x}{2} \cdot - 3 - 3 \frac{9 x}{2} - 3 \cdot - 3) = 36$

$y = \frac{9 x}{2} - 3$

Raise $x$ to the power of $1$ .

$9 x^{2} + 4 (\frac{81 (x \cdot x)}{2 \cdot 2} + \frac{9 x}{2} \cdot - 3 - 3 \frac{9 x}{2} - 3 \cdot - 3) = 36$

$y = \frac{9 x}{2} - 3$

Raise $x$ to the power of $1$ .

$9 x^{2} + 4 (\frac{81 (x \cdot x)}{2 \cdot 2} + \frac{9 x}{2} \cdot - 3 - 3 \frac{9 x}{2} - 3 \cdot - 3) = 36$

$y = \frac{9 x}{2} - 3$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$9 x^{2} + 4 (\frac{81 x^{1 + 1}}{2 \cdot 2} + \frac{9 x}{2} \cdot - 3 - 3 \frac{9 x}{2} - 3 \cdot - 3) = 36$

$y = \frac{9 x}{2} - 3$

Add $1$ and $1$ .

$9 x^{2} + 4 (\frac{81 x^{2}}{2 \cdot 2} + \frac{9 x}{2} \cdot - 3 - 3 \frac{9 x}{2} - 3 \cdot - 3) = 36$

$y = \frac{9 x}{2} - 3$

Multiply $2$ by $2$ .

$9 x^{2} + 4 (\frac{81 x^{2}}{4} + \frac{9 x}{2} \cdot - 3 - 3 \frac{9 x}{2} - 3 \cdot - 3) = 36$

$y = \frac{9 x}{2} - 3$

$9 x^{2} + 4 (\frac{81 x^{2}}{4} + \frac{9 x}{2} \cdot - 3 - 3 \frac{9 x}{2} - 3 \cdot - 3) = 36$

$y = \frac{9 x}{2} - 3$

Multiply $\frac{9 x}{2} \cdot - 3$ .

Combine $\frac{9 x}{2}$ and $- 3$ .

$9 x^{2} + 4 (\frac{81 x^{2}}{4} + \frac{9 x \cdot - 3}{2} - 3 \frac{9 x}{2} - 3 \cdot - 3) = 36$

$y = \frac{9 x}{2} - 3$

Multiply $- 3$ by $9$ .

$9 x^{2} + 4 (\frac{81 x^{2}}{4} + \frac{- 27 x}{2} - 3 \frac{9 x}{2} - 3 \cdot - 3) = 36$

$y = \frac{9 x}{2} - 3$

$9 x^{2} + 4 (\frac{81 x^{2}}{4} + \frac{- 27 x}{2} - 3 \frac{9 x}{2} - 3 \cdot - 3) = 36$

$y = \frac{9 x}{2} - 3$

Move the negative in front of the fraction.

$9 x^{2} + 4 (\frac{81 x^{2}}{4} - \frac{(27) x}{2} - 3 \frac{9 x}{2} - 3 \cdot - 3) = 36$

$y = \frac{9 x}{2} - 3$

Multiply $- 3 \frac{9 x}{2}$ .

Combine $- 3$ and $\frac{9 x}{2}$ .

$9 x^{2} + 4 (\frac{81 x^{2}}{4} - \frac{27 x}{2} + \frac{- 3 (9 x)}{2} - 3 \cdot - 3) = 36$

$y = \frac{9 x}{2} - 3$

Multiply $9$ by $- 3$ .

$9 x^{2} + 4 (\frac{81 x^{2}}{4} - \frac{27 x}{2} + \frac{- 27 x}{2} - 3 \cdot - 3) = 36$

$y = \frac{9 x}{2} - 3$

$9 x^{2} + 4 (\frac{81 x^{2}}{4} - \frac{27 x}{2} + \frac{- 27 x}{2} - 3 \cdot - 3) = 36$

$y = \frac{9 x}{2} - 3$

Move the negative in front of the fraction.

$9 x^{2} + 4 (\frac{81 x^{2}}{4} - \frac{27 x}{2} - \frac{(27) x}{2} - 3 \cdot - 3) = 36$

$y = \frac{9 x}{2} - 3$

Multiply $- 3$ by $- 3$ .

$9 x^{2} + 4 (\frac{81 x^{2}}{4} - \frac{27 x}{2} - \frac{27 x}{2} + 9) = 36$

$y = \frac{9 x}{2} - 3$

$9 x^{2} + 4 (\frac{81 x^{2}}{4} - \frac{27 x}{2} - \frac{27 x}{2} + 9) = 36$

$y = \frac{9 x}{2} - 3$

Subtract $\frac{27 x}{2}$ from $- \frac{27 x}{2}$ .

$9 x^{2} + 4 (\frac{81 x^{2}}{4} - 2 \frac{27 x}{2} + 9) = 36$

$y = \frac{9 x}{2} - 3$

$9 x^{2} + 4 (\frac{81 x^{2}}{4} - 2 \frac{27 x}{2} + 9) = 36$

$y = \frac{9 x}{2} - 3$

Simplify each term.

Cancel the common factor of $2$ .

Factor $2$ out of $- 2$ .

$9 x^{2} + 4 (\frac{81 x^{2}}{4} + 2 (- 1) (\frac{27 x}{2}) + 9) = 36$

$y = \frac{9 x}{2} - 3$

Cancel the common factor.

$9 x^{2} + 4 (\frac{81 x^{2}}{4} + 2 \cdot (- 1 \frac{27 x}{2}) + 9) = 36$

$y = \frac{9 x}{2} - 3$

Rewrite the expression.

$9 x^{2} + 4 (\frac{81 x^{2}}{4} - 1 (27 x) + 9) = 36$

$y = \frac{9 x}{2} - 3$

$9 x^{2} + 4 (\frac{81 x^{2}}{4} - 1 (27 x) + 9) = 36$

$y = \frac{9 x}{2} - 3$

Multiply $27$ by $- 1$ .

$9 x^{2} + 4 (\frac{81 x^{2}}{4} - 27 x + 9) = 36$

$y = \frac{9 x}{2} - 3$

$9 x^{2} + 4 (\frac{81 x^{2}}{4} - 27 x + 9) = 36$

$y = \frac{9 x}{2} - 3$

Apply the distributive property.

$9 x^{2} + 4 (\frac{81 x^{2}}{4}) + 4 (- 27 x) + 4 \cdot 9 = 36$

$y = \frac{9 x}{2} - 3$

Simplify.

Cancel the common factor of $4$ .

Cancel the common factor.

$9 x^{2} + 4 (\frac{81 x^{2}}{4}) + 4 (- 27 x) + 4 \cdot 9 = 36$

$y = \frac{9 x}{2} - 3$

Rewrite the expression.

$9 x^{2} + 81 x^{2} + 4 (- 27 x) + 4 \cdot 9 = 36$

$y = \frac{9 x}{2} - 3$

$9 x^{2} + 81 x^{2} + 4 (- 27 x) + 4 \cdot 9 = 36$

$y = \frac{9 x}{2} - 3$

Multiply $- 27$ by $4$ .

$9 x^{2} + 81 x^{2} - 108 x + 4 \cdot 9 = 36$

$y = \frac{9 x}{2} - 3$

Multiply $4$ by $9$ .

$9 x^{2} + 81 x^{2} - 108 x + 36 = 36$

$y = \frac{9 x}{2} - 3$

$9 x^{2} + 81 x^{2} - 108 x + 36 = 36$

$y = \frac{9 x}{2} - 3$

$9 x^{2} + 81 x^{2} - 108 x + 36 = 36$

$y = \frac{9 x}{2} - 3$

Add $9 x^{2}$ and $81 x^{2}$ .

$90 x^{2} - 108 x + 36 = 36$

$y = \frac{9 x}{2} - 3$

$90 x^{2} - 108 x + 36 = 36$

$y = \frac{9 x}{2} - 3$

$90 x^{2} - 108 x + 36 = 36$

$y = \frac{9 x}{2} - 3$

$90 x^{2} - 108 x + 36 = 36$

$y = \frac{9 x}{2} - 3$

Solve for $x$ in $90 x^{2} - 108 x + 36 = 36$ .

Subtract $36$ from both sides of the equation.

$90 x^{2} - 108 x + 36 - 36 = 0$

$y = \frac{9 x}{2} - 3$

Combine the opposite terms in $90 x^{2} - 108 x + 36 - 36$ .

Subtract $36$ from $36$ .

$90 x^{2} - 108 x + 0 = 0$

$y = \frac{9 x}{2} - 3$

Add $90 x^{2} - 108 x$ and $0$ .

$90 x^{2} - 108 x = 0$

$y = \frac{9 x}{2} - 3$

$90 x^{2} - 108 x = 0$

$y = \frac{9 x}{2} - 3$

Factor $18 x$ out of $90 x^{2} - 108 x$ .

Factor $18 x$ out of $90 x^{2}$ .

$18 x (5 x) - 108 x = 0$

$y = \frac{9 x}{2} - 3$

Factor $18 x$ out of $- 108 x$ .

$18 x (5 x) + 18 x (- 6) = 0$

$y = \frac{9 x}{2} - 3$

Factor $18 x$ out of $18 x (5 x) + 18 x (- 6)$ .

$18 x (5 x - 6) = 0$

$y = \frac{9 x}{2} - 3$

$18 x (5 x - 6) = 0$

$y = \frac{9 x}{2} - 3$

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

$x = 0$

$5 x - 6 = 0$

$y = \frac{9 x}{2} - 3$

Set $x$ equal to $0$ .

$x = 0$

$y = \frac{9 x}{2} - 3$

Set $5 x - 6$ equal to $0$ and solve for $x$ .

Set $5 x - 6$ equal to $0$ .

$5 x - 6 = 0$

$y = \frac{9 x}{2} - 3$

Solve $5 x - 6 = 0$ for $x$ .

Add $6$ to both sides of the equation.

$5 x = 6$

$y = \frac{9 x}{2} - 3$

Divide each term in $5 x = 6$ by $5$ and simplify.

Divide each term in $5 x = 6$ by $5$ .

$\frac{5 x}{5} = \frac{6}{5}$

$y = \frac{9 x}{2} - 3$

Simplify the left side.

Cancel the common factor of $5$ .

Cancel the common factor.

$\frac{5 x}{5} = \frac{6}{5}$

$y = \frac{9 x}{2} - 3$

Divide $x$ by $1$ .

$x = \frac{6}{5}$

$y = \frac{9 x}{2} - 3$

$x = \frac{6}{5}$

$y = \frac{9 x}{2} - 3$

$x = \frac{6}{5}$

$y = \frac{9 x}{2} - 3$

$x = \frac{6}{5}$

$y = \frac{9 x}{2} - 3$

$x = \frac{6}{5}$

$y = \frac{9 x}{2} - 3$

$x = \frac{6}{5}$

$y = \frac{9 x}{2} - 3$

The final solution is all the values that make $18 x (5 x - 6) = 0$ true.

$x = 0, \frac{6}{5}$

$y = \frac{9 x}{2} - 3$

$x = 0, \frac{6}{5}$

$y = \frac{9 x}{2} - 3$

Replace all occurrences of $x$ with $0$ in each equation.

Replace all occurrences of $x$ in $y = \frac{9 x}{2} - 3$ with $0$ .

$y = \frac{9 (0)}{2} - 3$

$x = 0$

Simplify the right side.

Simplify $\frac{9 (0)}{2} - 3$ .

Multiply $9$ by $0$ .

$y = \frac{0}{2} - 3$

$x = 0$

Divide $0$ by $2$ .

$y = 0 - 3$

$x = 0$

Subtract $3$ from $0$ .

$y = - 3$

$x = 0$

$y = - 3$

$x = 0$

$y = - 3$

$x = 0$

$y = - 3$

$x = 0$

Replace all occurrences of $x$ with $\frac{6}{5}$ in each equation.

Replace all occurrences of $x$ in $y = \frac{9 x}{2} - 3$ with $\frac{6}{5}$ .

$y = \frac{9 (\frac{6}{5})}{2} - 3$

$x = \frac{6}{5}$

Simplify the right side.

Simplify $\frac{9 (\frac{6}{5})}{2} - 3$ .

Simplify each term.

Combine $9$ and $\frac{6}{5}$ .

$y = \frac{\frac{9 \cdot 6}{5}}{2} - 3$

$x = \frac{6}{5}$

Multiply $9$ by $6$ .

$y = \frac{\frac{54}{5}}{2} - 3$

$x = \frac{6}{5}$

Multiply the numerator by the reciprocal of the denominator.

$y = \frac{54}{5} \cdot \frac{1}{2} - 3$

$x = \frac{6}{5}$

Cancel the common factor of $2$ .

Factor $2$ out of $54$ .

$y = \frac{2 (27)}{5} \cdot \frac{1}{2} - 3$

$x = \frac{6}{5}$

Cancel the common factor.

$y = \frac{2 \cdot 27}{5} \cdot \frac{1}{2} - 3$

$x = \frac{6}{5}$

Rewrite the expression.

$y = \frac{27}{5} - 3$

$x = \frac{6}{5}$

$y = \frac{27}{5} - 3$

$x = \frac{6}{5}$

$y = \frac{27}{5} - 3$

$x = \frac{6}{5}$

To write $- 3$ as a fraction with a common denominator, multiply by $\frac{5}{5}$ .

$y = \frac{27}{5} - 3 \cdot \frac{5}{5}$

$x = \frac{6}{5}$

Combine $- 3$ and $\frac{5}{5}$ .

$y = \frac{27}{5} + \frac{- 3 \cdot 5}{5}$

$x = \frac{6}{5}$

Combine the numerators over the common denominator.

$y = \frac{27 - 3 \cdot 5}{5}$

$x = \frac{6}{5}$

Simplify the numerator.

Multiply $- 3$ by $5$ .

$y = \frac{27 - 15}{5}$

$x = \frac{6}{5}$

Subtract $15$ from $27$ .

$y = \frac{12}{5}$

$x = \frac{6}{5}$

$y = \frac{12}{5}$

$x = \frac{6}{5}$

$y = \frac{12}{5}$

$x = \frac{6}{5}$

$y = \frac{12}{5}$

$x = \frac{6}{5}$

$y = \frac{12}{5}$

$x = \frac{6}{5}$

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

$(0, - 3)$

$(\frac{6}{5}, \frac{12}{5})$

The result can be shown in multiple forms.

Point Form:

$(0, - 3), (\frac{6}{5}, \frac{12}{5})$

Equation Form:

$x = 0, y = - 3$

$x = \frac{6}{5}, y = \frac{12}{5}$

Solve the System of Equations 9x^2+4y^2=36 y=9/2x-3

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