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

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

Solve for x, y

$$x=3$$

$$y=4$$

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.

$$3x+4y=25,y^{2}+x^{2}=25$$

Solve 3x+4y=25 for x by isolating x on the left hand side of the equal sign.

$$3x+4y=25$$

Subtract 4y from both sides of the equation.

$$3x=-4y+25$$

Divide both sides by 3.

$$x=-\frac{4}{3}y+\frac{25}{3}$$

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

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

Square -\frac{4}{3}y+\frac{25}{3}.

$$y^{2}+\frac{16}{9}y^{2}-\frac{200}{9}y+\frac{625}{9}=25$$

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

$$\frac{25}{9}y^{2}-\frac{200}{9}y+\frac{625}{9}=25$$

Subtract 25 from both sides of the equation.

$$\frac{25}{9}y^{2}-\frac{200}{9}y+\frac{400}{9}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1+1\left(-\frac{4}{3}\right)^{2} for a, 1\times \frac{25}{3}\left(-\frac{4}{3}\right)\times 2 for b, and \frac{400}{9} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-\frac{200}{9}\right)±\sqrt{\left(-\frac{200}{9}\right)^{2}-4\times \frac{25}{9}\times \frac{400}{9}}}{2\times \frac{25}{9}}$$

Square 1\times \frac{25}{3}\left(-\frac{4}{3}\right)\times 2.

$$y=\frac{-\left(-\frac{200}{9}\right)±\sqrt{\frac{40000}{81}-4\times \frac{25}{9}\times \frac{400}{9}}}{2\times \frac{25}{9}}$$

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

$$y=\frac{-\left(-\frac{200}{9}\right)±\sqrt{\frac{40000}{81}-\frac{100}{9}\times \frac{400}{9}}}{2\times \frac{25}{9}}$$

Multiply -\frac{100}{9} times \frac{400}{9} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$y=\frac{-\left(-\frac{200}{9}\right)±\sqrt{\frac{40000-40000}{81}}}{2\times \frac{25}{9}}$$

Add \frac{40000}{81} to -\frac{40000}{81} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$y=\frac{-\left(-\frac{200}{9}\right)±\sqrt{0}}{2\times \frac{25}{9}}$$

Take the square root of 0.

$$y=-\frac{-\frac{200}{9}}{2\times \frac{25}{9}}$$

The opposite of 1\times \frac{25}{3}\left(-\frac{4}{3}\right)\times 2 is \frac{200}{9}.

$$y=\frac{\frac{200}{9}}{2\times \frac{25}{9}}$$

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

$$y=\frac{\frac{200}{9}}{\frac{50}{9}}$$

Divide \frac{200}{9} by \frac{50}{9} by multiplying \frac{200}{9} by the reciprocal of \frac{50}{9}.

$$y=4$$

There are two solutions for y: 4 and 4. Substitute 4 for y in the equation x=-\frac{4}{3}y+\frac{25}{3} to find the corresponding solution for x that satisfies both equations.

$$x=-\frac{4}{3}\times 4+\frac{25}{3}$$

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

$$x=\frac{-16+25}{3}$$

Add -\frac{4}{3}\times 4 to \frac{25}{3}.

$$x=3$$

The system is now solved.

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