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

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

Solve for y, x

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

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

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.

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

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

$$y-2x=3$$

Subtract -2x from both sides of the equation.

$$y=2x+3$$

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

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

Square 2x+3.

$$x^{2}+4x^{2}+12x+9=18$$

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

$$5x^{2}+12x+9=18$$

Subtract 18 from both sides of the equation.

$$5x^{2}+12x-9=0$$

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

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

Square 1\times 3\times 2\times 2.

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

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

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

Multiply -20 times -9.

$$x=\frac{-12±\sqrt{144+180}}{2\times 5}$$

Add 144 to 180.

$$x=\frac{-12±\sqrt{324}}{2\times 5}$$

Take the square root of 324.

$$x=\frac{-12±18}{2\times 5}$$

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

$$x=\frac{-12±18}{10}$$

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

$$x=\frac{6}{10}$$

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

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

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

$$x=-\frac{30}{10}$$

Divide -30 by 10.

$$x=-3$$

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

$$y=2\times \frac{3}{5}+3$$

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

$$y=\frac{6}{5}+3$$

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

$$y=\frac{21}{5}$$

Now substitute -3 for x in the equation y=2x+3 and solve to find the corresponding solution for y that satisfies both equations.

$$y=2\left(-3\right)+3$$

Multiply 2 times -3.

$$y=-6+3$$

Add -3\times 2 to 3.

$$y=-3$$

The system is now solved.

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