Solve {c}{-15x=30y-90}{15x=45y+60}

$$\left. \begin{array} { c } { - 15 x = 30 y - 90 } \\ { 15 x = 45 y + 60 } \end{array} \right.$$

Solve for x, y

$$x = \frac{26}{5} = 5\frac{1}{5} = 5.2$$

$$y=\frac{2}{5}=0.4$$

Steps Using Substitution

Consider the first equation. Subtract 30y from both sides.

$$-15x-30y=-90$$

Consider the second equation. Subtract 45y from both sides.

$$15x-45y=60$$

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.

$$-15x-30y=-90,15x-45y=60$$

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

$$-15x-30y=-90$$

Add 30y to both sides of the equation.

$$-15x=30y-90$$

Divide both sides by -15.

$$x=-\frac{1}{15}\left(30y-90\right)$$

Multiply -\frac{1}{15} times -90+30y.

$$x=-2y+6$$

Substitute -2y+6 for x in the other equation, 15x-45y=60.

$$15\left(-2y+6\right)-45y=60$$

Multiply 15 times -2y+6.

$$-30y+90-45y=60$$

Add -30y to -45y.

$$-75y+90=60$$

Subtract 90 from both sides of the equation.

$$-75y=-30$$

Divide both sides by -75.

$$y=\frac{2}{5}$$

Substitute \frac{2}{5} for y in x=-2y+6. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-2\times \frac{2}{5}+6$$

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

$$x=-\frac{4}{5}+6$$

Add 6 to -\frac{4}{5}.

$$x=\frac{26}{5}$$

The system is now solved.

$$x=\frac{26}{5},y=\frac{2}{5}$$

Steps Using Matrices

Consider the first equation. Subtract 30y from both sides.

$$-15x-30y=-90$$

Consider the second equation. Subtract 45y from both sides.

$$15x-45y=60$$

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

$$-15x-30y=-90,15x-45y=60$$

Write the equations in matrix form.

$$\left(\begin{matrix}-15&-30\\15&-45\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-90\\60\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}-15&-30\\15&-45\end{matrix}\right).

$$inverse(\left(\begin{matrix}-15&-30\\15&-45\end{matrix}\right))\left(\begin{matrix}-15&-30\\15&-45\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}-15&-30\\15&-45\end{matrix}\right))\left(\begin{matrix}-90\\60\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}-15&-30\\15&-45\end{matrix}\right))\left(\begin{matrix}-90\\60\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}-15&-30\\15&-45\end{matrix}\right))\left(\begin{matrix}-90\\60\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{45}{-15\left(-45\right)-\left(-30\times 15\right)}&-\frac{-30}{-15\left(-45\right)-\left(-30\times 15\right)}\\-\frac{15}{-15\left(-45\right)-\left(-30\times 15\right)}&-\frac{15}{-15\left(-45\right)-\left(-30\times 15\right)}\end{matrix}\right)\left(\begin{matrix}-90\\60\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{25}&\frac{2}{75}\\-\frac{1}{75}&-\frac{1}{75}\end{matrix}\right)\left(\begin{matrix}-90\\60\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{25}\left(-90\right)+\frac{2}{75}\times 60\\-\frac{1}{75}\left(-90\right)-\frac{1}{75}\times 60\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{26}{5}\\\frac{2}{5}\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=\frac{26}{5},y=\frac{2}{5}$$

Steps Using Elimination

Consider the first equation. Subtract 30y from both sides.

$$-15x-30y=-90$$

Consider the second equation. Subtract 45y from both sides.

$$15x-45y=60$$

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.

$$-15x-30y=-90,15x-45y=60$$

To make -15x and 15x equal, multiply all terms on each side of the first equation by 15 and all terms on each side of the second by -15.

$$15\left(-15\right)x+15\left(-30\right)y=15\left(-90\right),-15\times 15x-15\left(-45\right)y=-15\times 60$$

Simplify.

$$-225x-450y=-1350,-225x+675y=-900$$

Subtract -225x+675y=-900 from -225x-450y=-1350 by subtracting like terms on each side of the equal sign.

$$-225x+225x-450y-675y=-1350+900$$

Add -225x to 225x. Terms -225x and 225x cancel out, leaving an equation with only one variable that can be solved.

$$-450y-675y=-1350+900$$

Add -450y to -675y.

$$-1125y=-1350+900$$

Add -1350 to 900.

$$-1125y=-450$$

Divide both sides by -1125.

$$y=\frac{2}{5}$$

Substitute \frac{2}{5} for y in 15x-45y=60. Because the resulting equation contains only one variable, you can solve for x directly.

$$15x-45\times \frac{2}{5}=60$$

Multiply -45 times \frac{2}{5}.

$$15x-18=60$$

Add 18 to both sides of the equation.

$$15x=78$$

Divide both sides by 15.

$$x=\frac{26}{5}$$

The system is now solved.

$$x=\frac{26}{5},y=\frac{2}{5}$$