Solve {l}{x^2+y^2=41}{2x+3y=2}

$$\left. \begin{array} { l } { x ^ { 2 } + y ^ { 2 } = 41 } \\ { 2 x + 3 y = 2 } \end{array} \right.$$

Solve for x, y

$$x=\frac{73}{13}\approx 5.615384615\text{, }y=-\frac{40}{13}\approx -3.076923077$$

$$x=-5\text{, }y=4$$

Steps Using Substitution and Quadratic Formula

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.

$$2x+3y=2,y^{2}+x^{2}=41$$

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

$$2x+3y=2$$

Subtract 3y from both sides of the equation.

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

Divide both sides by 2.

$$x=-\frac{3}{2}y+1$$

Substitute -\frac{3}{2}y+1 for x in the other equation, y^{2}+x^{2}=41.

$$y^{2}+\left(-\frac{3}{2}y+1\right)^{2}=41$$

Square -\frac{3}{2}y+1.

$$y^{2}+\frac{9}{4}y^{2}-3y+1=41$$

Add y^{2} to \frac{9}{4}y^{2}.

$$\frac{13}{4}y^{2}-3y+1=41$$

Subtract 41 from both sides of the equation.

$$\frac{13}{4}y^{2}-3y-40=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1+1\left(-\frac{3}{2}\right)^{2} for a, 1\times 1\left(-\frac{3}{2}\right)\times 2 for b, and -40 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\times \frac{13}{4}\left(-40\right)}}{2\times \frac{13}{4}}$$

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

$$y=\frac{-\left(-3\right)±\sqrt{9-4\times \frac{13}{4}\left(-40\right)}}{2\times \frac{13}{4}}$$

Multiply -4 times 1+1\left(-\frac{3}{2}\right)^{2}.

$$y=\frac{-\left(-3\right)±\sqrt{9-13\left(-40\right)}}{2\times \frac{13}{4}}$$

Multiply -13 times -40.

$$y=\frac{-\left(-3\right)±\sqrt{9+520}}{2\times \frac{13}{4}}$$

Add 9 to 520.

$$y=\frac{-\left(-3\right)±\sqrt{529}}{2\times \frac{13}{4}}$$

Take the square root of 529.

$$y=\frac{-\left(-3\right)±23}{2\times \frac{13}{4}}$$

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

$$y=\frac{3±23}{2\times \frac{13}{4}}$$

Multiply 2 times 1+1\left(-\frac{3}{2}\right)^{2}.

$$y=\frac{3±23}{\frac{13}{2}}$$

Now solve the equation y=\frac{3±23}{\frac{13}{2}} when ± is plus. Add 3 to 23.

$$y=\frac{26}{\frac{13}{2}}$$

Divide 26 by \frac{13}{2} by multiplying 26 by the reciprocal of \frac{13}{2}.

$$y=4$$

Now solve the equation y=\frac{3±23}{\frac{13}{2}} when ± is minus. Subtract 23 from 3.

$$y=-\frac{20}{\frac{13}{2}}$$

Divide -20 by \frac{13}{2} by multiplying -20 by the reciprocal of \frac{13}{2}.

$$y=-\frac{40}{13}$$

There are two solutions for y: 4 and -\frac{40}{13}. Substitute 4 for y in the equation x=-\frac{3}{2}y+1 to find the corresponding solution for x that satisfies both equations.

$$x=-\frac{3}{2}\times 4+1$$

Multiply -\frac{3}{2} times 4.

$$x=-6+1$$

Add -\frac{3}{2}\times 4 to 1.

$$x=-5$$

Now substitute -\frac{40}{13} for y in the equation x=-\frac{3}{2}y+1 and solve to find the corresponding solution for x that satisfies both equations.

$$x=-\frac{3}{2}\left(-\frac{40}{13}\right)+1$$

Multiply -\frac{3}{2} times -\frac{40}{13} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$x=\frac{60}{13}+1$$

Add -\frac{40}{13}\left(-\frac{3}{2}\right) to 1.

$$x=\frac{73}{13}$$

The system is now solved.

$$x=-5,y=4\text{ or }x=\frac{73}{13},y=-\frac{40}{13}$$