Solve {l}{-5y=-15x-110}{-5y=-12x-16}

$$\left. \begin{array} { l } { - 5 y = - 15 x - 110 } \\ { - 5 y = - 12 x - 16 } \end{array} \right.$$

Solve for y, x

$$x = -\frac{94}{3} = -31\frac{1}{3} \approx -31.333333333$$

$$y=-72$$

Steps Using Substitution

Consider the first equation. Add 15x to both sides.

$$-5y+15x=-110$$

Consider the second equation. Add 12x to both sides.

$$-5y+12x=-16$$

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.

$$-5y+15x=-110,-5y+12x=-16$$

Choose one of the equations and solve it for y by isolating y on the left hand side of the equal sign.

$$-5y+15x=-110$$

Subtract 15x from both sides of the equation.

$$-5y=-15x-110$$

Divide both sides by -5.

$$y=-\frac{1}{5}\left(-15x-110\right)$$

Multiply -\frac{1}{5} times -15x-110.

$$y=3x+22$$

Substitute 3x+22 for y in the other equation, -5y+12x=-16.

$$-5\left(3x+22\right)+12x=-16$$

Multiply -5 times 3x+22.

$$-15x-110+12x=-16$$

Add -15x to 12x.

$$-3x-110=-16$$

Add 110 to both sides of the equation.

$$-3x=94$$

Divide both sides by -3.

$$x=-\frac{94}{3}$$

Substitute -\frac{94}{3} for x in y=3x+22. Because the resulting equation contains only one variable, you can solve for y directly.

$$y=3\left(-\frac{94}{3}\right)+22$$

Multiply 3 times -\frac{94}{3}.

$$y=-94+22$$

Add 22 to -94.

$$y=-72$$

The system is now solved.

$$y=-72,x=-\frac{94}{3}$$

Steps Using Matrices

Consider the first equation. Add 15x to both sides.

$$-5y+15x=-110$$

Consider the second equation. Add 12x to both sides.

$$-5y+12x=-16$$

Put the equations in standard form and then use matrices to solve the system of equations.

$$-5y+15x=-110,-5y+12x=-16$$

Write the equations in matrix form.

$$\left(\begin{matrix}-5&15\\-5&12\end{matrix}\right)\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}-110\\-16\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}-5&15\\-5&12\end{matrix}\right).

$$inverse(\left(\begin{matrix}-5&15\\-5&12\end{matrix}\right))\left(\begin{matrix}-5&15\\-5&12\end{matrix}\right)\left(\begin{matrix}y\\x\end{matrix}\right)=inverse(\left(\begin{matrix}-5&15\\-5&12\end{matrix}\right))\left(\begin{matrix}-110\\-16\end{matrix}\right)$$

The product of a matrix and its inverse is the identity matrix.

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}y\\x\end{matrix}\right)=inverse(\left(\begin{matrix}-5&15\\-5&12\end{matrix}\right))\left(\begin{matrix}-110\\-16\end{matrix}\right)$$

Multiply the matrices on the left hand side of the equal sign.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=inverse(\left(\begin{matrix}-5&15\\-5&12\end{matrix}\right))\left(\begin{matrix}-110\\-16\end{matrix}\right)$$

For the 2\times 2 matrix \left(\begin{matrix}a&b\\c&d\end{matrix}\right), the inverse matrix is \left(\begin{matrix}\frac{d}{ad-bc}&\frac{-b}{ad-bc}\\\frac{-c}{ad-bc}&\frac{a}{ad-bc}\end{matrix}\right), so the matrix equation can be rewritten as a matrix multiplication problem.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}\frac{12}{-5\times 12-15\left(-5\right)}&-\frac{15}{-5\times 12-15\left(-5\right)}\\-\frac{-5}{-5\times 12-15\left(-5\right)}&-\frac{5}{-5\times 12-15\left(-5\right)}\end{matrix}\right)\left(\begin{matrix}-110\\-16\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}\frac{4}{5}&-1\\\frac{1}{3}&-\frac{1}{3}\end{matrix}\right)\left(\begin{matrix}-110\\-16\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}\frac{4}{5}\left(-110\right)-\left(-16\right)\\\frac{1}{3}\left(-110\right)-\frac{1}{3}\left(-16\right)\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}-72\\-\frac{94}{3}\end{matrix}\right)$$

Extract the matrix elements y and x.

$$y=-72,x=-\frac{94}{3}$$

Steps Using Elimination

Consider the first equation. Add 15x to both sides.

$$-5y+15x=-110$$

Consider the second equation. Add 12x to both sides.

$$-5y+12x=-16$$

In order to solve by elimination, coefficients of one of the variables must be the same in both equations so that the variable will cancel out when one equation is subtracted from the other.

$$-5y+15x=-110,-5y+12x=-16$$

Subtract -5y+12x=-16 from -5y+15x=-110 by subtracting like terms on each side of the equal sign.

$$-5y+5y+15x-12x=-110+16$$

Add -5y to 5y. Terms -5y and 5y cancel out, leaving an equation with only one variable that can be solved.

$$15x-12x=-110+16$$

Add 15x to -12x.

$$3x=-110+16$$

Add -110 to 16.

$$3x=-94$$

Divide both sides by 3.

$$x=-\frac{94}{3}$$

Substitute -\frac{94}{3} for x in -5y+12x=-16. Because the resulting equation contains only one variable, you can solve for y directly.

$$-5y+12\left(-\frac{94}{3}\right)=-16$$

Multiply 12 times -\frac{94}{3}.

$$-5y-376=-16$$

Add 376 to both sides of the equation.

$$-5y=360$$

Divide both sides by -5.

$$y=-72$$

The system is now solved.

$$y=-72,x=-\frac{94}{3}$$