Solve {l}{x^2+y^2=18}{x-2y=-3}

$$\left. \begin{array} { l } { x ^ { 2 } + y ^ { 2 } = 18 } \\ { x - 2 y = - 3 } \end{array} \right.$$

Solve for x, y

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

$$x=-\frac{21}{5}=-4.2\text{, }y=-\frac{3}{5}=-0.6$$

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.

$$x-2y=-3,y^{2}+x^{2}=18$$

Solve x-2y=-3 for x by isolating x on the left hand side of the equal sign.

$$x-2y=-3$$

Subtract -2y from both sides of the equation.

$$x=2y-3$$

Substitute 2y-3 for x in the other equation, y^{2}+x^{2}=18.

$$y^{2}+\left(2y-3\right)^{2}=18$$

Square 2y-3.

$$y^{2}+4y^{2}-12y+9=18$$

Add y^{2} to 4y^{2}.

$$5y^{2}-12y+9=18$$

Subtract 18 from both sides of the equation.

$$5y^{2}-12y-9=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1+1\times 2^{2} for a, 1\left(-3\right)\times 2\times 2 for b, and -9 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\times 5\left(-9\right)}}{2\times 5}$$

Square 1\left(-3\right)\times 2\times 2.

$$y=\frac{-\left(-12\right)±\sqrt{144-4\times 5\left(-9\right)}}{2\times 5}$$

Multiply -4 times 1+1\times 2^{2}.

$$y=\frac{-\left(-12\right)±\sqrt{144-20\left(-9\right)}}{2\times 5}$$

Multiply -20 times -9.

$$y=\frac{-\left(-12\right)±\sqrt{144+180}}{2\times 5}$$

Add 144 to 180.

$$y=\frac{-\left(-12\right)±\sqrt{324}}{2\times 5}$$

Take the square root of 324.

$$y=\frac{-\left(-12\right)±18}{2\times 5}$$

The opposite of 1\left(-3\right)\times 2\times 2 is 12.

$$y=\frac{12±18}{2\times 5}$$

Multiply 2 times 1+1\times 2^{2}.

$$y=\frac{12±18}{10}$$

Now solve the equation y=\frac{12±18}{10} when ± is plus. Add 12 to 18.

$$y=\frac{30}{10}$$

Divide 30 by 10.

$$y=3$$

Now solve the equation y=\frac{12±18}{10} when ± is minus. Subtract 18 from 12.

$$y=-\frac{6}{10}$$

Reduce the fraction \frac{-6}{10} to lowest terms by extracting and canceling out 2.

$$y=-\frac{3}{5}$$

There are two solutions for y: 3 and -\frac{3}{5}. Substitute 3 for y in the equation x=2y-3 to find the corresponding solution for x that satisfies both equations.

$$x=2\times 3-3$$

Multiply 2 times 3.

$$x=6-3$$

Add 2\times 3 to -3.

$$x=3$$

Now substitute -\frac{3}{5} for y in the equation x=2y-3 and solve to find the corresponding solution for x that satisfies both equations.

$$x=2\left(-\frac{3}{5}\right)-3$$

Multiply 2 times -\frac{3}{5}.

$$x=-\frac{6}{5}-3$$

Add -\frac{3}{5}\times 2 to -3.

$$x=-\frac{21}{5}$$

The system is now solved.

$$x=3,y=3\text{ or }x=-\frac{21}{5},y=-\frac{3}{5}$$