Solve {l}{text{edbytheellipse}x^2/9+y^2/4=1}{text{e}x/3+y/2=1}

$$\left. \begin{array} { l } { \text { ed by the ellipse } \frac { x ^ { 2 } } { 9 } + \frac { y ^ { 2 } } { 4 } = 1 } \\ { \text { e } \frac { x } { 3 } + \frac { y } { 2 } = 1 } \end{array} \right.$$

Solve for x, y

$$x=0\text{, }y=2$$

$$x=3\text{, }y=0$$

Steps Using Substitution and Quadratic Formula

Consider the first equation. Multiply both sides of the equation by 36, the least common multiple of 9,4.

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

Consider the second equation. Multiply both sides of the equation by 6, the least common multiple of 3,2.

$$2x+3y=6$$

To solve a pair of equations using substitution, first solve one of the equations for one of the variables. Then substitute the result for that variable in the other equation.

$$2x+3y=6,9y^{2}+4x^{2}=36$$

Solve 2x+3y=6 for x by isolating x on the left hand side of the equal sign.

$$2x+3y=6$$

Subtract 3y from both sides of the equation.

$$2x=-3y+6$$

Divide both sides by 2.

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

Substitute -\frac{3}{2}y+3 for x in the other equation, 9y^{2}+4x^{2}=36.

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

Square -\frac{3}{2}y+3.

$$9y^{2}+4\left(\frac{9}{4}y^{2}-9y+9\right)=36$$

Multiply 4 times \frac{9}{4}y^{2}-9y+9.

$$9y^{2}+9y^{2}-36y+36=36$$

Add 9y^{2} to 9y^{2}.

$$18y^{2}-36y+36=36$$

Subtract 36 from both sides of the equation.

$$18y^{2}-36y=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 9+4\left(-\frac{3}{2}\right)^{2} for a, 4\times 3\left(-\frac{3}{2}\right)\times 2 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-36\right)±\sqrt{\left(-36\right)^{2}}}{2\times 18}$$

Take the square root of \left(-36\right)^{2}.

$$y=\frac{-\left(-36\right)±36}{2\times 18}$$

The opposite of 4\times 3\left(-\frac{3}{2}\right)\times 2 is 36.

$$y=\frac{36±36}{2\times 18}$$

Multiply 2 times 9+4\left(-\frac{3}{2}\right)^{2}.

$$y=\frac{36±36}{36}$$

Now solve the equation y=\frac{36±36}{36} when ± is plus. Add 36 to 36.

$$y=\frac{72}{36}$$

Divide 72 by 36.

$$y=2$$

Now solve the equation y=\frac{36±36}{36} when ± is minus. Subtract 36 from 36.

$$y=\frac{0}{36}$$

Divide 0 by 36.

$$y=0$$

There are two solutions for y: 2 and 0. Substitute 2 for y in the equation x=-\frac{3}{2}y+3 to find the corresponding solution for x that satisfies both equations.

$$x=-\frac{3}{2}\times 2+3$$

Multiply -\frac{3}{2} times 2.

$$x=-3+3$$

Add -\frac{3}{2}\times 2 to 3.

$$x=0$$

Now substitute 0 for y in the equation x=-\frac{3}{2}y+3 and solve to find the corresponding solution for x that satisfies both equations.

$$x=3$$

The system is now solved.

$$x=0,y=2\text{ or }x=3,y=0$$