Solve {l}{1.3x=1.3*(4/5sqrt{5})+3.3y}{1.3x^2=1.3*(4/5sqrt{5})^2+3.3y^2}

$$\left\{ \begin{array} { l } { 1.3 x = 1.3 \cdot ( \frac { 4 } { 5 } \sqrt { 5 } ) + 3.3 y } \\ { 1.3 x ^ { 2 } = 1.3 \cdot ( \frac { 4 } { 5 } \sqrt { 5 } ) ^ { 2 } + 3.3 y ^ { 2 } } \end{array} \right.$$

Solve for x, y

$$x=\frac{4\sqrt{5}}{5}\approx 1.788854382\text{, }y=0$$

$$x=-\frac{46\sqrt{5}}{25}\approx -4.114365079\text{, }y=-\frac{26\sqrt{5}}{25}\approx -2.325510697$$

Steps Using Substitution and Quadratic Formula

Consider the first equation. Multiply 1.3 and \frac{4}{5} to get \frac{26}{25}.

$$1.3x=\frac{26}{25}\sqrt{5}+3.3y$$

Subtract 3.3y from both sides.

$$1.3x-3.3y=\frac{26}{25}\sqrt{5}$$

Consider the second equation. Expand \left(\frac{4}{5}\sqrt{5}\right)^{2}.

$$1.3x^{2}=1.3\times \left(\frac{4}{5}\right)^{2}\left(\sqrt{5}\right)^{2}+3.3y^{2}$$

Calculate \frac{4}{5} to the power of 2 and get \frac{16}{25}.

$$1.3x^{2}=1.3\times \frac{16}{25}\left(\sqrt{5}\right)^{2}+3.3y^{2}$$

The square of \sqrt{5} is 5.

$$1.3x^{2}=1.3\times \frac{16}{25}\times 5+3.3y^{2}$$

Multiply \frac{16}{25} and 5 to get \frac{16}{5}.

$$1.3x^{2}=1.3\times \frac{16}{5}+3.3y^{2}$$

Multiply 1.3 and \frac{16}{5} to get \frac{104}{25}.

$$1.3x^{2}=\frac{104}{25}+3.3y^{2}$$

Subtract 3.3y^{2} from both sides.

$$1.3x^{2}-3.3y^{2}=\frac{104}{25}$$

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.

$$1.3x-3.3y=\frac{26\sqrt{5}}{25},-3.3y^{2}+1.3x^{2}=\frac{104}{25}$$

Solve 1.3x-3.3y=\frac{26\sqrt{5}}{25} for x by isolating x on the left hand side of the equal sign.

$$1.3x-3.3y=\frac{26\sqrt{5}}{25}$$

Subtract -3.3y from both sides of the equation.

$$1.3x=3.3y+\frac{26\sqrt{5}}{25}$$

Divide both sides of the equation by 1.3, which is the same as multiplying both sides by the reciprocal of the fraction.

$$x=\frac{33}{13}y+\frac{4\sqrt{5}}{5}$$

Substitute \frac{33}{13}y+\frac{4\sqrt{5}}{5} for x in the other equation, -3.3y^{2}+1.3x^{2}=\frac{104}{25}.

$$-3.3y^{2}+1.3\left(\frac{33}{13}y+\frac{4\sqrt{5}}{5}\right)^{2}=\frac{104}{25}$$

Square \frac{33}{13}y+\frac{4\sqrt{5}}{5}.

$$-3.3y^{2}+1.3\left(\frac{1089}{169}y^{2}+\frac{66}{13}\times \frac{4\sqrt{5}}{5}y+\left(\frac{4\sqrt{5}}{5}\right)^{2}\right)=\frac{104}{25}$$

Multiply 1.3 times \frac{1089}{169}y^{2}+\frac{66}{13}\times \frac{4\sqrt{5}}{5}y+\left(\frac{4\sqrt{5}}{5}\right)^{2}.

$$-3.3y^{2}+\frac{1089}{130}y^{2}+\frac{33\times \frac{4\sqrt{5}}{5}}{5}y+\frac{13\times \left(\frac{4\sqrt{5}}{5}\right)^{2}}{10}=\frac{104}{25}$$

Add -3.3y^{2} to \frac{1089}{130}y^{2}.

$$\frac{66}{13}y^{2}+\frac{33\times \frac{4\sqrt{5}}{5}}{5}y+\frac{13\times \left(\frac{4\sqrt{5}}{5}\right)^{2}}{10}=\frac{104}{25}$$

Subtract \frac{104}{25} from both sides of the equation.

$$\frac{66}{13}y^{2}+\frac{33\times \frac{4\sqrt{5}}{5}}{5}y+\frac{13\times \left(\frac{4\sqrt{5}}{5}\right)^{2}}{10}-\frac{104}{25}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -3.3+1.3\times \frac{33}{13}^{2} for a, 1.3\times 2\times \frac{33}{13}\times \frac{4\sqrt{5}}{5} for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\frac{33\times \frac{4\sqrt{5}}{5}}{5}±\sqrt{\left(\frac{33\times \frac{4\sqrt{5}}{5}}{5}\right)^{2}}}{2\times \frac{66}{13}}$$

The square root of b^{2} is |b|. Substitute 1.3\times 2\times \frac{33}{13}\times \frac{4\sqrt{5}}{5} for b.

$$y=\frac{-\frac{33\times \frac{4\sqrt{5}}{5}}{5}±\frac{132\sqrt{5}}{25}}{2\times \frac{66}{13}}$$

Multiply 2 times -3.3+1.3\times \frac{33}{13}^{2}.

$$y=\frac{-\frac{132\sqrt{5}}{25}±\frac{132\sqrt{5}}{25}}{\frac{132}{13}}$$

Now solve the equation y=\frac{-\frac{132\sqrt{5}}{25}±\frac{132\sqrt{5}}{25}}{\frac{132}{13}} when ± is plus. Add -\frac{132\sqrt{5}}{25} to \frac{132\sqrt{5}}{25}.

$$y=\frac{0}{\frac{132}{13}}$$

Divide 0 by \frac{132}{13} by multiplying 0 by the reciprocal of \frac{132}{13}.

$$y=0$$

Now solve the equation y=\frac{-\frac{132\sqrt{5}}{25}±\frac{132\sqrt{5}}{25}}{\frac{132}{13}} when ± is minus. Subtract \frac{132\sqrt{5}}{25} from -\frac{132\sqrt{5}}{25}.

$$y=-\frac{\frac{264\sqrt{5}}{25}}{\frac{132}{13}}$$

Divide -\frac{264\sqrt{5}}{25} by \frac{132}{13} by multiplying -\frac{264\sqrt{5}}{25} by the reciprocal of \frac{132}{13}.

$$y=-\frac{26\sqrt{5}}{25}$$

There are two solutions for y: 0 and -\frac{26\sqrt{5}}{25}. Substitute 0 for y in the equation x=\frac{33}{13}y+\frac{4\sqrt{5}}{5} to find the corresponding solution for x that satisfies both equations.

$$x=\frac{4\sqrt{5}}{5}$$

Now substitute -\frac{26\sqrt{5}}{25} for y in the equation x=\frac{33}{13}y+\frac{4\sqrt{5}}{5} and solve to find the corresponding solution for x that satisfies both equations.

$$x=\frac{33}{13}\left(-\frac{26\sqrt{5}}{25}\right)+\frac{4\sqrt{5}}{5}$$

Multiply \frac{33}{13} times -\frac{26\sqrt{5}}{25}.

$$x=\frac{33\left(-\frac{26\sqrt{5}}{25}\right)}{13}+\frac{4\sqrt{5}}{5}$$

The system is now solved.

$$x=\frac{4\sqrt{5}}{5},y=0\text{ or }x=\frac{33\left(-\frac{26\sqrt{5}}{25}\right)}{13}+\frac{4\sqrt{5}}{5},y=-\frac{26\sqrt{5}}{25}$$