Solve {l}{x^2+y^2=25}{y=1/2x+1}

$$\left\{ \begin{array} { l } { x ^ { 2 } + y ^ { 2 } = 25 } \\ { y = \frac { 1 } { 2 } x + 1 } \end{array} \right.$$

Solve for x, y

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

$$x=-\frac{24}{5}=-4.8\text{, }y=-\frac{7}{5}=-1.4$$

Steps Using Substitution and Quadratic Formula

Consider the second equation. Subtract \frac{1}{2}x from both sides.

$$y-\frac{1}{2}x=1$$

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.

$$y-\frac{1}{2}x=1,x^{2}+y^{2}=25$$

Solve y-\frac{1}{2}x=1 for y by isolating y on the left hand side of the equal sign.

$$y-\frac{1}{2}x=1$$

Subtract -\frac{1}{2}x from both sides of the equation.

$$y=\frac{1}{2}x+1$$

Substitute \frac{1}{2}x+1 for y in the other equation, x^{2}+y^{2}=25.

$$x^{2}+\left(\frac{1}{2}x+1\right)^{2}=25$$

Square \frac{1}{2}x+1.

$$x^{2}+\frac{1}{4}x^{2}+x+1=25$$

Add x^{2} to \frac{1}{4}x^{2}.

$$\frac{5}{4}x^{2}+x+1=25$$

Subtract 25 from both sides of the equation.

$$\frac{5}{4}x^{2}+x-24=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1+1\times \left(\frac{1}{2}\right)^{2} for a, 1\times 1\times \frac{1}{2}\times 2 for b, and -24 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-1±\sqrt{1^{2}-4\times \frac{5}{4}\left(-24\right)}}{2\times \frac{5}{4}}$$

Square 1\times 1\times \frac{1}{2}\times 2.

$$x=\frac{-1±\sqrt{1-4\times \frac{5}{4}\left(-24\right)}}{2\times \frac{5}{4}}$$

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

$$x=\frac{-1±\sqrt{1-5\left(-24\right)}}{2\times \frac{5}{4}}$$

Multiply -5 times -24.

$$x=\frac{-1±\sqrt{1+120}}{2\times \frac{5}{4}}$$

Add 1 to 120.

$$x=\frac{-1±\sqrt{121}}{2\times \frac{5}{4}}$$

Take the square root of 121.

$$x=\frac{-1±11}{2\times \frac{5}{4}}$$

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

$$x=\frac{-1±11}{\frac{5}{2}}$$

Now solve the equation x=\frac{-1±11}{\frac{5}{2}} when ± is plus. Add -1 to 11.

$$x=\frac{10}{\frac{5}{2}}$$

Divide 10 by \frac{5}{2} by multiplying 10 by the reciprocal of \frac{5}{2}.

$$x=4$$

Now solve the equation x=\frac{-1±11}{\frac{5}{2}} when ± is minus. Subtract 11 from -1.

$$x=-\frac{12}{\frac{5}{2}}$$

Divide -12 by \frac{5}{2} by multiplying -12 by the reciprocal of \frac{5}{2}.

$$x=-\frac{24}{5}$$

There are two solutions for x: 4 and -\frac{24}{5}. Substitute 4 for x in the equation y=\frac{1}{2}x+1 to find the corresponding solution for y that satisfies both equations.

$$y=\frac{1}{2}\times 4+1$$

Multiply \frac{1}{2} times 4.

$$y=2+1$$

Add \frac{1}{2}\times 4 to 1.

$$y=3$$

Now substitute -\frac{24}{5} for x in the equation y=\frac{1}{2}x+1 and solve to find the corresponding solution for y that satisfies both equations.

$$y=\frac{1}{2}\left(-\frac{24}{5}\right)+1$$

Multiply \frac{1}{2} times -\frac{24}{5} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$y=-\frac{12}{5}+1$$

Add -\frac{24}{5}\times \frac{1}{2} to 1.

$$y=-\frac{7}{5}$$

The system is now solved.

$$y=3,x=4\text{ or }y=-\frac{7}{5},x=-\frac{24}{5}$$