Solve {l}{x^2+y^2=100}{4x-3y=0}

$$\left.\begin{array} { l } { x ^ { 2 } + y ^ { 2 } = 100 } \\ { 4 x - 3 y = 0 } \end{array} \right\}$$

Solve for x, y

$$x=-6\text{, }y=-8$$

$$x=6\text{, }y=8$$

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.

$$4x-3y=0,y^{2}+x^{2}=100$$

Solve 4x-3y=0 for x by isolating x on the left hand side of the equal sign.

$$4x-3y=0$$

Subtract -3y from both sides of the equation.

$$4x=3y$$

Divide both sides by 4.

$$x=\frac{3}{4}y$$

Substitute \frac{3}{4}y for x in the other equation, y^{2}+x^{2}=100.

$$y^{2}+\left(\frac{3}{4}y\right)^{2}=100$$

Square \frac{3}{4}y.

$$y^{2}+\frac{9}{16}y^{2}=100$$

Add y^{2} to \frac{9}{16}y^{2}.

$$\frac{25}{16}y^{2}=100$$

Subtract 100 from both sides of the equation.

$$\frac{25}{16}y^{2}-100=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1+1\times \left(\frac{3}{4}\right)^{2} for a, 1\times 0\times \frac{3}{4}\times 2 for b, and -100 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{0±\sqrt{0^{2}-4\times \frac{25}{16}\left(-100\right)}}{2\times \frac{25}{16}}$$

Square 1\times 0\times \frac{3}{4}\times 2.

$$y=\frac{0±\sqrt{-4\times \frac{25}{16}\left(-100\right)}}{2\times \frac{25}{16}}$$

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

$$y=\frac{0±\sqrt{-\frac{25}{4}\left(-100\right)}}{2\times \frac{25}{16}}$$

Multiply -\frac{25}{4} times -100.

$$y=\frac{0±\sqrt{625}}{2\times \frac{25}{16}}$$

Take the square root of 625.

$$y=\frac{0±25}{2\times \frac{25}{16}}$$

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

$$y=\frac{0±25}{\frac{25}{8}}$$

Now solve the equation y=\frac{0±25}{\frac{25}{8}} when ± is plus. Divide 25 by \frac{25}{8} by multiplying 25 by the reciprocal of \frac{25}{8}.

$$y=8$$

Now solve the equation y=\frac{0±25}{\frac{25}{8}} when ± is minus. Divide -25 by \frac{25}{8} by multiplying -25 by the reciprocal of \frac{25}{8}.

$$y=-8$$

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

$$x=\frac{3}{4}\times 8$$

Multiply \frac{3}{4} times 8.

$$x=6$$

Now substitute -8 for y in the equation x=\frac{3}{4}y and solve to find the corresponding solution for x that satisfies both equations.

$$x=\frac{3}{4}\left(-8\right)$$

Multiply \frac{3}{4} times -8.

$$x=-6$$

The system is now solved.

$$x=6,y=8\text{ or }x=-6,y=-8$$