Solve {l}{x+y=3}{4x^2-y^2=4}

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

Solve for x, y

$$x=\frac{4\sqrt{3}}{3}-1\approx 1.309401077\text{, }y=-\frac{4\sqrt{3}}{3}+4\approx 1.690598923$$

$$x=-\frac{4\sqrt{3}}{3}-1\approx -3.309401077\text{, }y=\frac{4\sqrt{3}}{3}+4\approx 6.309401077$$

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.

$$x+y=3,-y^{2}+4x^{2}=4$$

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

$$x+y=3$$

Subtract y from both sides of the equation.

$$x=-y+3$$

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

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

Square -y+3.

$$-y^{2}+4\left(y^{2}-6y+9\right)=4$$

Multiply 4 times y^{2}-6y+9.

$$-y^{2}+4y^{2}-24y+36=4$$

Add -y^{2} to 4y^{2}.

$$3y^{2}-24y+36=4$$

Subtract 4 from both sides of the equation.

$$3y^{2}-24y+32=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1+4\left(-1\right)^{2} for a, 4\times 3\left(-1\right)\times 2 for b, and 32 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-24\right)±\sqrt{\left(-24\right)^{2}-4\times 3\times 32}}{2\times 3}$$

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

$$y=\frac{-\left(-24\right)±\sqrt{576-4\times 3\times 32}}{2\times 3}$$

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

$$y=\frac{-\left(-24\right)±\sqrt{576-12\times 32}}{2\times 3}$$

Multiply -12 times 32.

$$y=\frac{-\left(-24\right)±\sqrt{576-384}}{2\times 3}$$

Add 576 to -384.

$$y=\frac{-\left(-24\right)±\sqrt{192}}{2\times 3}$$

Take the square root of 192.

$$y=\frac{-\left(-24\right)±8\sqrt{3}}{2\times 3}$$

The opposite of 4\times 3\left(-1\right)\times 2 is 24.

$$y=\frac{24±8\sqrt{3}}{2\times 3}$$

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

$$y=\frac{24±8\sqrt{3}}{6}$$

Now solve the equation y=\frac{24±8\sqrt{3}}{6} when ± is plus. Add 24 to 8\sqrt{3}.

$$y=\frac{8\sqrt{3}+24}{6}$$

Divide 24+8\sqrt{3} by 6.

$$y=\frac{4\sqrt{3}}{3}+4$$

Now solve the equation y=\frac{24±8\sqrt{3}}{6} when ± is minus. Subtract 8\sqrt{3} from 24.

$$y=\frac{24-8\sqrt{3}}{6}$$

Divide 24-8\sqrt{3} by 6.

$$y=-\frac{4\sqrt{3}}{3}+4$$

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

$$x=-\left(\frac{4\sqrt{3}}{3}+4\right)+3$$

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

$$x=-\left(-\frac{4\sqrt{3}}{3}+4\right)+3$$

The system is now solved.

$$x=-\left(\frac{4\sqrt{3}}{3}+4\right)+3,y=\frac{4\sqrt{3}}{3}+4\text{ or }x=-\left(-\frac{4\sqrt{3}}{3}+4\right)+3,y=-\frac{4\sqrt{3}}{3}+4$$