Solve {l}{3x+4y=12}{x^2+y^2=16}

$$\left. \begin{array} { l } { 3 x + 4 y = 12 } \\ { x ^ { 2 } + y ^ { 2 } = 16 } \end{array} \right.$$

Solve for x, y

$$x=4\text{, }y=0$$

$$x=-\frac{28}{25}=-1.12\text{, }y=\frac{96}{25}=3.84$$

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.

$$3x+4y=12,y^{2}+x^{2}=16$$

Solve 3x+4y=12 for x by isolating x on the left hand side of the equal sign.

$$3x+4y=12$$

Subtract 4y from both sides of the equation.

$$3x=-4y+12$$

Divide both sides by 3.

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

Substitute -\frac{4}{3}y+4 for x in the other equation, y^{2}+x^{2}=16.

$$y^{2}+\left(-\frac{4}{3}y+4\right)^{2}=16$$

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

$$y^{2}+\frac{16}{9}y^{2}-\frac{32}{3}y+16=16$$

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

$$\frac{25}{9}y^{2}-\frac{32}{3}y+16=16$$

Subtract 16 from both sides of the equation.

$$\frac{25}{9}y^{2}-\frac{32}{3}y=0$$

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

$$y=\frac{-\left(-\frac{32}{3}\right)±\sqrt{\left(-\frac{32}{3}\right)^{2}}}{2\times \frac{25}{9}}$$

Take the square root of \left(-\frac{32}{3}\right)^{2}.

$$y=\frac{-\left(-\frac{32}{3}\right)±\frac{32}{3}}{2\times \frac{25}{9}}$$

The opposite of 1\times 4\left(-\frac{4}{3}\right)\times 2 is \frac{32}{3}.

$$y=\frac{\frac{32}{3}±\frac{32}{3}}{2\times \frac{25}{9}}$$

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

$$y=\frac{\frac{32}{3}±\frac{32}{3}}{\frac{50}{9}}$$

Now solve the equation y=\frac{\frac{32}{3}±\frac{32}{3}}{\frac{50}{9}} when ± is plus. Add \frac{32}{3} to \frac{32}{3} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$y=\frac{\frac{64}{3}}{\frac{50}{9}}$$

Divide \frac{64}{3} by \frac{50}{9} by multiplying \frac{64}{3} by the reciprocal of \frac{50}{9}.

$$y=\frac{96}{25}$$

Now solve the equation y=\frac{\frac{32}{3}±\frac{32}{3}}{\frac{50}{9}} when ± is minus. Subtract \frac{32}{3} from \frac{32}{3} by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$y=\frac{0}{\frac{50}{9}}$$

Divide 0 by \frac{50}{9} by multiplying 0 by the reciprocal of \frac{50}{9}.

$$y=0$$

There are two solutions for y: \frac{96}{25} and 0. Substitute \frac{96}{25} for y in the equation x=-\frac{4}{3}y+4 to find the corresponding solution for x that satisfies both equations.

$$x=-\frac{4}{3}\times \frac{96}{25}+4$$

Multiply -\frac{4}{3} times \frac{96}{25} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$x=-\frac{128}{25}+4$$

Add -\frac{4}{3}\times \frac{96}{25} to 4.

$$x=-\frac{28}{25}$$

Now substitute 0 for y in the equation x=-\frac{4}{3}y+4 and solve to find the corresponding solution for x that satisfies both equations.

$$x=4$$

The system is now solved.

$$x=-\frac{28}{25},y=\frac{96}{25}\text{ or }x=4,y=0$$