Solve {l}{x=3y+2}{x^2/16+y^2/9=1}

$$\left. \begin{array} { l } { x = 3 y + 2 } \\ { \frac { x ^ { 2 } } { 16 } + \frac { y ^ { 2 } } { 9 } = 1 } \end{array} \right.$$

Solve for x, y

$$x=\frac{32-36\sqrt{93}}{97}\approx -3.249189973\text{, }y=\frac{-12\sqrt{93}-54}{97}\approx -1.749729991$$

$$x=\frac{36\sqrt{93}+32}{97}\approx 3.908983788\text{, }y=\frac{12\sqrt{93}-54}{97}\approx 0.636327929$$

Steps Using Substitution and Quadratic Formula

Consider the first equation. Subtract 3y from both sides.

$$x-3y=2$$

Consider the second equation. Multiply both sides of the equation by 144, the least common multiple of 16,9.

$$9x^{2}+16y^{2}=144$$

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.

$$x-3y=2,16y^{2}+9x^{2}=144$$

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

$$x-3y=2$$

Subtract -3y from both sides of the equation.

$$x=3y+2$$

Substitute 3y+2 for x in the other equation, 16y^{2}+9x^{2}=144.

$$16y^{2}+9\left(3y+2\right)^{2}=144$$

Square 3y+2.

$$16y^{2}+9\left(9y^{2}+12y+4\right)=144$$

Multiply 9 times 9y^{2}+12y+4.

$$16y^{2}+81y^{2}+108y+36=144$$

Add 16y^{2} to 81y^{2}.

$$97y^{2}+108y+36=144$$

Subtract 144 from both sides of the equation.

$$97y^{2}+108y-108=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 16+9\times 3^{2} for a, 9\times 2\times 2\times 3 for b, and -108 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-108±\sqrt{108^{2}-4\times 97\left(-108\right)}}{2\times 97}$$

Square 9\times 2\times 2\times 3.

$$y=\frac{-108±\sqrt{11664-4\times 97\left(-108\right)}}{2\times 97}$$

Multiply -4 times 16+9\times 3^{2}.

$$y=\frac{-108±\sqrt{11664-388\left(-108\right)}}{2\times 97}$$

Multiply -388 times -108.

$$y=\frac{-108±\sqrt{11664+41904}}{2\times 97}$$

Add 11664 to 41904.

$$y=\frac{-108±\sqrt{53568}}{2\times 97}$$

Take the square root of 53568.

$$y=\frac{-108±24\sqrt{93}}{2\times 97}$$

Multiply 2 times 16+9\times 3^{2}.

$$y=\frac{-108±24\sqrt{93}}{194}$$

Now solve the equation y=\frac{-108±24\sqrt{93}}{194} when ± is plus. Add -108 to 24\sqrt{93}.

$$y=\frac{24\sqrt{93}-108}{194}$$

Divide -108+24\sqrt{93} by 194.

$$y=\frac{12\sqrt{93}-54}{97}$$

Now solve the equation y=\frac{-108±24\sqrt{93}}{194} when ± is minus. Subtract 24\sqrt{93} from -108.

$$y=\frac{-24\sqrt{93}-108}{194}$$

Divide -108-24\sqrt{93} by 194.

$$y=\frac{-12\sqrt{93}-54}{97}$$

There are two solutions for y: \frac{-54+12\sqrt{93}}{97} and \frac{-54-12\sqrt{93}}{97}. Substitute \frac{-54+12\sqrt{93}}{97} for y in the equation x=3y+2 to find the corresponding solution for x that satisfies both equations.

$$x=3\times \frac{12\sqrt{93}-54}{97}+2$$

Now substitute \frac{-54-12\sqrt{93}}{97} for y in the equation x=3y+2 and solve to find the corresponding solution for x that satisfies both equations.

$$x=3\times \frac{-12\sqrt{93}-54}{97}+2$$

The system is now solved.

$$x=3\times \frac{12\sqrt{93}-54}{97}+2,y=\frac{12\sqrt{93}-54}{97}\text{ or }x=3\times \frac{-12\sqrt{93}-54}{97}+2,y=\frac{-12\sqrt{93}-54}{97}$$