Solve {l}{x+y=a}{x^2+y^2=a^2}

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

Solve for x, y

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

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

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.

$$x+y=a,y^{2}+x^{2}=a^{2}$$

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

$$x+y=a$$

Subtract y from both sides of the equation.

$$x=-y+a$$

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

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

Square -y+a.

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

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

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

Subtract a^{2} from both sides of the equation.

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

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

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

The square root of b^{2} is |b|. Substitute 1\left(-1\right)\times 2a for b.

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

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

$$y=\frac{2a±2|a|}{4}$$

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

$$y=\frac{2|a|+2a}{4}$$

Divide 2a+2|a| by 4.

$$y=\frac{|a|+a}{2}$$

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

$$y=\frac{-2|a|+2a}{4}$$

Divide 2a-2|a| by 4.

$$y=\frac{a-|a|}{2}$$

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

$$x=-\frac{|a|+a}{2}+a$$

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

$$x=-\frac{a-|a|}{2}+a$$

The system is now solved.

$$x=-\frac{|a|+a}{2}+a,y=\frac{|a|+a}{2}\text{ or }x=-\frac{a-|a|}{2}+a,y=\frac{a-|a|}{2}$$