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

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

Solve for x, y

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

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

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.

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

Solve x+y=1 for x by isolating x on the left hand side of the equal sign.

$$x+y=1$$

Subtract y from both sides of the equation.

$$x=-y+1$$

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

$$y^{2}+\left(-y+1\right)^{2}=25$$

Square -y+1.

$$y^{2}+y^{2}-2y+1=25$$

Add y^{2} to y^{2}.

$$2y^{2}-2y+1=25$$

Subtract 25 from both sides of the equation.

$$2y^{2}-2y-24=0$$

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

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

Square 1\times 1\left(-1\right)\times 2.

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

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

$$y=\frac{-\left(-2\right)±\sqrt{4-8\left(-24\right)}}{2\times 2}$$

Multiply -8 times -24.

$$y=\frac{-\left(-2\right)±\sqrt{4+192}}{2\times 2}$$

Add 4 to 192.

$$y=\frac{-\left(-2\right)±\sqrt{196}}{2\times 2}$$

Take the square root of 196.

$$y=\frac{-\left(-2\right)±14}{2\times 2}$$

The opposite of 1\times 1\left(-1\right)\times 2 is 2.

$$y=\frac{2±14}{2\times 2}$$

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

$$y=\frac{2±14}{4}$$

Now solve the equation y=\frac{2±14}{4} when ± is plus. Add 2 to 14.

$$y=\frac{16}{4}$$

Divide 16 by 4.

$$y=4$$

Now solve the equation y=\frac{2±14}{4} when ± is minus. Subtract 14 from 2.

$$y=-\frac{12}{4}$$

Divide -12 by 4.

$$y=-3$$

There are two solutions for y: 4 and -3. Substitute 4 for y in the equation x=-y+1 to find the corresponding solution for x that satisfies both equations.

$$x=-4+1$$

Add -4 to 1.

$$x=-3$$

Now substitute -3 for y in the equation x=-y+1 and solve to find the corresponding solution for x that satisfies both equations.

$$x=-\left(-3\right)+1$$

Multiply -1 times -3.

$$x=3+1$$

Add -3\left(-1\right) to 1.

$$x=4$$

The system is now solved.

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