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

$$\left. \begin{array} { l } { \frac { x ^ { 2 } } { 9 } + \frac { y ^ { 2 } } { 4 } = 1 } \\ { 3 x + 4 y = 1 } \end{array} \right.$$

Solve for x, y

$$x=\frac{63}{29}\approx 2.172413793\text{, }y=-\frac{40}{29}\approx -1.379310345$$

$$x=-\frac{9}{5}=-1.8\text{, }y=\frac{8}{5}=1.6$$

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.

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

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.

$$3x+4y=1,9y^{2}+4x^{2}=36$$

Solve 3x+4y=1 for x by isolating x on the left hand side of the equal sign.

$$3x+4y=1$$

Subtract 4y from both sides of the equation.

$$3x=-4y+1$$

Divide both sides by 3.

$$x=-\frac{4}{3}y+\frac{1}{3}$$

Substitute -\frac{4}{3}y+\frac{1}{3} for x in the other equation, 9y^{2}+4x^{2}=36.

$$9y^{2}+4\left(-\frac{4}{3}y+\frac{1}{3}\right)^{2}=36$$

Square -\frac{4}{3}y+\frac{1}{3}.

$$9y^{2}+4\left(\frac{16}{9}y^{2}-\frac{8}{9}y+\frac{1}{9}\right)=36$$

Multiply 4 times \frac{16}{9}y^{2}-\frac{8}{9}y+\frac{1}{9}.

$$9y^{2}+\frac{64}{9}y^{2}-\frac{32}{9}y+\frac{4}{9}=36$$

Add 9y^{2} to \frac{64}{9}y^{2}.

$$\frac{145}{9}y^{2}-\frac{32}{9}y+\frac{4}{9}=36$$

Subtract 36 from both sides of the equation.

$$\frac{145}{9}y^{2}-\frac{32}{9}y-\frac{320}{9}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 9+4\left(-\frac{4}{3}\right)^{2} for a, 4\times \frac{1}{3}\left(-\frac{4}{3}\right)\times 2 for b, and -\frac{320}{9} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-\frac{32}{9}\right)±\sqrt{\left(-\frac{32}{9}\right)^{2}-4\times \frac{145}{9}\left(-\frac{320}{9}\right)}}{2\times \frac{145}{9}}$$

Square 4\times \frac{1}{3}\left(-\frac{4}{3}\right)\times 2.

$$y=\frac{-\left(-\frac{32}{9}\right)±\sqrt{\frac{1024}{81}-4\times \frac{145}{9}\left(-\frac{320}{9}\right)}}{2\times \frac{145}{9}}$$

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

$$y=\frac{-\left(-\frac{32}{9}\right)±\sqrt{\frac{1024}{81}-\frac{580}{9}\left(-\frac{320}{9}\right)}}{2\times \frac{145}{9}}$$

Multiply -\frac{580}{9} times -\frac{320}{9} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$y=\frac{-\left(-\frac{32}{9}\right)±\sqrt{\frac{1024+185600}{81}}}{2\times \frac{145}{9}}$$

Add \frac{1024}{81} to \frac{185600}{81} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$y=\frac{-\left(-\frac{32}{9}\right)±\sqrt{2304}}{2\times \frac{145}{9}}$$

Take the square root of 2304.

$$y=\frac{-\left(-\frac{32}{9}\right)±48}{2\times \frac{145}{9}}$$

The opposite of 4\times \frac{1}{3}\left(-\frac{4}{3}\right)\times 2 is \frac{32}{9}.

$$y=\frac{\frac{32}{9}±48}{2\times \frac{145}{9}}$$

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

$$y=\frac{\frac{32}{9}±48}{\frac{290}{9}}$$

Now solve the equation y=\frac{\frac{32}{9}±48}{\frac{290}{9}} when ± is plus. Add \frac{32}{9} to 48.

$$y=\frac{\frac{464}{9}}{\frac{290}{9}}$$

Divide \frac{464}{9} by \frac{290}{9} by multiplying \frac{464}{9} by the reciprocal of \frac{290}{9}.

$$y=\frac{8}{5}$$

Now solve the equation y=\frac{\frac{32}{9}±48}{\frac{290}{9}} when ± is minus. Subtract 48 from \frac{32}{9}.

$$y=-\frac{\frac{400}{9}}{\frac{290}{9}}$$

Divide -\frac{400}{9} by \frac{290}{9} by multiplying -\frac{400}{9} by the reciprocal of \frac{290}{9}.

$$y=-\frac{40}{29}$$

There are two solutions for y: \frac{8}{5} and -\frac{40}{29}. Substitute \frac{8}{5} for y in the equation x=-\frac{4}{3}y+\frac{1}{3} to find the corresponding solution for x that satisfies both equations.

$$x=-\frac{4}{3}\times \frac{8}{5}+\frac{1}{3}$$

Multiply -\frac{4}{3} times \frac{8}{5} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$x=-\frac{32}{15}+\frac{1}{3}$$

Add -\frac{4}{3}\times \frac{8}{5} to \frac{1}{3}.

$$x=-\frac{9}{5}$$

Now substitute -\frac{40}{29} for y in the equation x=-\frac{4}{3}y+\frac{1}{3} and solve to find the corresponding solution for x that satisfies both equations.

$$x=-\frac{4}{3}\left(-\frac{40}{29}\right)+\frac{1}{3}$$

Multiply -\frac{4}{3} times -\frac{40}{29} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$x=\frac{160}{87}+\frac{1}{3}$$

Add -\frac{40}{29}\left(-\frac{4}{3}\right) to \frac{1}{3}.

$$x=\frac{63}{29}$$

The system is now solved.

$$x=-\frac{9}{5},y=\frac{8}{5}\text{ or }x=\frac{63}{29},y=-\frac{40}{29}$$