Solve {l}{y=3/4x}{x^2+y^2=25}

$$\left\{ \begin{array} { l } { y = \frac { 3 } { 4 } x } \\ { x ^ { 2 } + y ^ { 2 } = 25 } \end{array} \right.$$

Solve for y, x

$$x=-4\text{, }y=-3$$

$$x=4\text{, }y=3$$

Steps Using Substitution and Quadratic Formula

Consider the first equation. Subtract \frac{3}{4}x from both sides.

$$y-\frac{3}{4}x=0$$

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.

$$y-\frac{3}{4}x=0,x^{2}+y^{2}=25$$

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

$$y-\frac{3}{4}x=0$$

Subtract -\frac{3}{4}x from both sides of the equation.

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

Substitute \frac{3}{4}x for y in the other equation, x^{2}+y^{2}=25.

$$x^{2}+\left(\frac{3}{4}x\right)^{2}=25$$

Square \frac{3}{4}x.

$$x^{2}+\frac{9}{16}x^{2}=25$$

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

$$\frac{25}{16}x^{2}=25$$

Subtract 25 from both sides of the equation.

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

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 -25 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

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

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

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

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

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

$$x=\frac{0±\sqrt{\frac{625}{4}}}{2\times \frac{25}{16}}$$

Take the square root of \frac{625}{4}.

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

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

$$x=\frac{0±\frac{25}{2}}{\frac{25}{8}}$$

Now solve the equation x=\frac{0±\frac{25}{2}}{\frac{25}{8}} when ± is plus.

$$x=4$$

Now solve the equation x=\frac{0±\frac{25}{2}}{\frac{25}{8}} when ± is minus.

$$x=-4$$

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

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

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

$$y=3$$

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

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

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

$$y=-3$$

The system is now solved.

$$y=3,x=4\text{ or }y=-3,x=-4$$