Solve {l}{x+y-1=0}{x^2+y^2-36=0}

$$\left. \begin{array} { l } { x + y - 1 = 0 } \\ { x ^ { 2 } + y ^ { 2 } - 36 = 0 } \end{array} \right.$$

Solve for x, y

$$x=\frac{\sqrt{71}+1}{2}\approx 4.713074887\text{, }y=\frac{1-\sqrt{71}}{2}\approx -3.713074887$$

$$x=\frac{1-\sqrt{71}}{2}\approx -3.713074887\text{, }y=\frac{\sqrt{71}+1}{2}\approx 4.713074887$$

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.

$$x+y-1=0,y^{2}+x^{2}-36=0$$

Solve x+y-1=0 for x by isolating x on the left hand side of the equal sign.

$$x+y-1=0$$

Add 1 to both sides of the equation.

$$x+y=1$$

Subtract y from both sides of the equation.

$$x=-y+1$$

Substitute -y+1 for x in the other equation, y^{2}+x^{2}-36=0.

$$y^{2}+\left(-y+1\right)^{2}-36=0$$

Square -y+1.

$$y^{2}+y^{2}-2y+1-36=0$$

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

$$2y^{2}-2y+1-36=0$$

Add 1\times 1^{2} to -36.

$$2y^{2}-2y-35=0$$

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 -35 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\times 2\left(-35\right)}}{2\times 2}$$

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

$$y=\frac{-\left(-2\right)±\sqrt{4-4\times 2\left(-35\right)}}{2\times 2}$$

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

$$y=\frac{-\left(-2\right)±\sqrt{4-8\left(-35\right)}}{2\times 2}$$

Multiply -8 times -35.

$$y=\frac{-\left(-2\right)±\sqrt{4+280}}{2\times 2}$$

Add 4 to 280.

$$y=\frac{-\left(-2\right)±\sqrt{284}}{2\times 2}$$

Take the square root of 284.

$$y=\frac{-\left(-2\right)±2\sqrt{71}}{2\times 2}$$

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

$$y=\frac{2±2\sqrt{71}}{2\times 2}$$

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

$$y=\frac{2±2\sqrt{71}}{4}$$

Now solve the equation y=\frac{2±2\sqrt{71}}{4} when ± is plus. Add 2 to 2\sqrt{71}.

$$y=\frac{2\sqrt{71}+2}{4}$$

Divide 2+2\sqrt{71} by 4.

$$y=\frac{\sqrt{71}+1}{2}$$

Now solve the equation y=\frac{2±2\sqrt{71}}{4} when ± is minus. Subtract 2\sqrt{71} from 2.

$$y=\frac{2-2\sqrt{71}}{4}$$

Divide 2-2\sqrt{71} by 4.

$$y=\frac{1-\sqrt{71}}{2}$$

There are two solutions for y: \frac{1+\sqrt{71}}{2} and \frac{1-\sqrt{71}}{2}. Substitute \frac{1+\sqrt{71}}{2} for y in the equation x=-y+1 to find the corresponding solution for x that satisfies both equations.

$$x=-\frac{\sqrt{71}+1}{2}+1$$

Now substitute \frac{1-\sqrt{71}}{2} for y in the equation x=-y+1 and solve to find the corresponding solution for x that satisfies both equations.

$$x=-\frac{1-\sqrt{71}}{2}+1$$

The system is now solved.

$$x=-\frac{\sqrt{71}+1}{2}+1,y=\frac{\sqrt{71}+1}{2}\text{ or }x=-\frac{1-\sqrt{71}}{2}+1,y=\frac{1-\sqrt{71}}{2}$$