Solve {l}{15x^2+10y^2=13.75}{3x+2y=-0.5}

$$\left\{ \begin{array} { l } { 15 x ^ { 2 } + 10 y ^ { 2 } = 13.75 } \\ { 3 x + 2 y = - 0.5 } \end{array} \right.$$

Solve for x, y

$$x=0.5\text{, }y=-1$$

$$x=-0.7\text{, }y=0.8$$

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.

$$3x+2y=-0.5,10y^{2}+15x^{2}=13.75$$

Solve 3x+2y=-0.5 for x by isolating x on the left hand side of the equal sign.

$$3x+2y=-0.5$$

Subtract 2y from both sides of the equation.

$$3x=-2y-0.5$$

Divide both sides by 3.

$$x=-\frac{2}{3}y-\frac{1}{6}$$

Substitute -\frac{2}{3}y-\frac{1}{6} for x in the other equation, 10y^{2}+15x^{2}=13.75.

$$10y^{2}+15\left(-\frac{2}{3}y-\frac{1}{6}\right)^{2}=13.75$$

Square -\frac{2}{3}y-\frac{1}{6}.

$$10y^{2}+15\left(\frac{4}{9}y^{2}+\frac{2}{9}y+\frac{1}{36}\right)=13.75$$

Multiply 15 times \frac{4}{9}y^{2}+\frac{2}{9}y+\frac{1}{36}.

$$10y^{2}+\frac{20}{3}y^{2}+\frac{10}{3}y+\frac{5}{12}=13.75$$

Add 10y^{2} to \frac{20}{3}y^{2}.

$$\frac{50}{3}y^{2}+\frac{10}{3}y+\frac{5}{12}=13.75$$

Subtract 13.75 from both sides of the equation.

$$\frac{50}{3}y^{2}+\frac{10}{3}y-\frac{40}{3}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 10+15\left(-\frac{2}{3}\right)^{2} for a, 15\left(-\frac{1}{6}\right)\left(-\frac{2}{3}\right)\times 2 for b, and -\frac{40}{3} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\frac{10}{3}±\sqrt{\left(\frac{10}{3}\right)^{2}-4\times \frac{50}{3}\left(-\frac{40}{3}\right)}}{2\times \frac{50}{3}}$$

Square 15\left(-\frac{1}{6}\right)\left(-\frac{2}{3}\right)\times 2.

$$y=\frac{-\frac{10}{3}±\sqrt{\frac{100}{9}-4\times \frac{50}{3}\left(-\frac{40}{3}\right)}}{2\times \frac{50}{3}}$$

Multiply -4 times 10+15\left(-\frac{2}{3}\right)^{2}.

$$y=\frac{-\frac{10}{3}±\sqrt{\frac{100}{9}-\frac{200}{3}\left(-\frac{40}{3}\right)}}{2\times \frac{50}{3}}$$

Multiply -\frac{200}{3} times -\frac{40}{3} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$y=\frac{-\frac{10}{3}±\sqrt{\frac{100+8000}{9}}}{2\times \frac{50}{3}}$$

Add \frac{100}{9} to \frac{8000}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$y=\frac{-\frac{10}{3}±\sqrt{900}}{2\times \frac{50}{3}}$$

Take the square root of 900.

$$y=\frac{-\frac{10}{3}±30}{2\times \frac{50}{3}}$$

Multiply 2 times 10+15\left(-\frac{2}{3}\right)^{2}.

$$y=\frac{-\frac{10}{3}±30}{\frac{100}{3}}$$

Now solve the equation y=\frac{-\frac{10}{3}±30}{\frac{100}{3}} when ± is plus. Add -\frac{10}{3} to 30.

$$y=\frac{\frac{80}{3}}{\frac{100}{3}}$$

Divide \frac{80}{3} by \frac{100}{3} by multiplying \frac{80}{3} by the reciprocal of \frac{100}{3}.

$$y=\frac{4}{5}$$

Now solve the equation y=\frac{-\frac{10}{3}±30}{\frac{100}{3}} when ± is minus. Subtract 30 from -\frac{10}{3}.

$$y=-\frac{\frac{100}{3}}{\frac{100}{3}}$$

Divide -\frac{100}{3} by \frac{100}{3} by multiplying -\frac{100}{3} by the reciprocal of \frac{100}{3}.

$$y=-1$$

There are two solutions for y: \frac{4}{5} and -1. Substitute \frac{4}{5} for y in the equation x=-\frac{2}{3}y-\frac{1}{6} to find the corresponding solution for x that satisfies both equations.

$$x=-\frac{2}{3}\times \frac{4}{5}-\frac{1}{6}$$

Multiply -\frac{2}{3} times \frac{4}{5} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$x=-\frac{8}{15}-\frac{1}{6}$$

Add -\frac{2}{3}\times \frac{4}{5} to -\frac{1}{6}.

$$x=-\frac{7}{10}$$

Now substitute -1 for y in the equation x=-\frac{2}{3}y-\frac{1}{6} and solve to find the corresponding solution for x that satisfies both equations.

$$x=-\frac{2}{3}\left(-1\right)-\frac{1}{6}$$

Multiply -\frac{2}{3} times -1.

$$x=\frac{2}{3}-\frac{1}{6}$$

Add -\left(-\frac{2}{3}\right) to -\frac{1}{6}.

$$x=\frac{1}{2}$$

The system is now solved.

$$x=-\frac{7}{10},y=\frac{4}{5}\text{ or }x=\frac{1}{2},y=-1$$