Solve {l}{18x+15y=12}{30x-15y=-100}

$$\left. \begin{array} { l } { 18 x + 15 y = 12 } \\ { 30 x - 15 y = - 100 } \end{array} \right.$$

Solve for x, y

$$x = -\frac{11}{6} = -1\frac{5}{6} \approx -1.833333333$$

$$y=3$$

Steps Using Substitution

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.

$$18x+15y=12,30x-15y=-100$$

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

$$18x+15y=12$$

Subtract 15y from both sides of the equation.

$$18x=-15y+12$$

Divide both sides by 18.

$$x=\frac{1}{18}\left(-15y+12\right)$$

Multiply \frac{1}{18} times -15y+12.

$$x=-\frac{5}{6}y+\frac{2}{3}$$

Substitute -\frac{5y}{6}+\frac{2}{3} for x in the other equation, 30x-15y=-100.

$$30\left(-\frac{5}{6}y+\frac{2}{3}\right)-15y=-100$$

Multiply 30 times -\frac{5y}{6}+\frac{2}{3}.

$$-25y+20-15y=-100$$

Add -25y to -15y.

$$-40y+20=-100$$

Subtract 20 from both sides of the equation.

$$-40y=-120$$

Divide both sides by -40.

$$y=3$$

Substitute 3 for y in x=-\frac{5}{6}y+\frac{2}{3}. Because the resulting equation contains only one variable, you can solve for x directly.

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

Multiply -\frac{5}{6} times 3.

$$x=-\frac{5}{2}+\frac{2}{3}$$

Add \frac{2}{3} to -\frac{5}{2} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=-\frac{11}{6}$$

The system is now solved.

$$x=-\frac{11}{6},y=3$$

Steps Using Matrices

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

$$18x+15y=12,30x-15y=-100$$

Write the equations in matrix form.

$$\left(\begin{matrix}18&15\\30&-15\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}12\\-100\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}18&15\\30&-15\end{matrix}\right).

$$inverse(\left(\begin{matrix}18&15\\30&-15\end{matrix}\right))\left(\begin{matrix}18&15\\30&-15\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}18&15\\30&-15\end{matrix}\right))\left(\begin{matrix}12\\-100\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}18&15\\30&-15\end{matrix}\right))\left(\begin{matrix}12\\-100\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}18&15\\30&-15\end{matrix}\right))\left(\begin{matrix}12\\-100\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{15}{18\left(-15\right)-15\times 30}&-\frac{15}{18\left(-15\right)-15\times 30}\\-\frac{30}{18\left(-15\right)-15\times 30}&\frac{18}{18\left(-15\right)-15\times 30}\end{matrix}\right)\left(\begin{matrix}12\\-100\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{48}&\frac{1}{48}\\\frac{1}{24}&-\frac{1}{40}\end{matrix}\right)\left(\begin{matrix}12\\-100\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{48}\times 12+\frac{1}{48}\left(-100\right)\\\frac{1}{24}\times 12-\frac{1}{40}\left(-100\right)\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{11}{6}\\3\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=-\frac{11}{6},y=3$$

Steps Using Elimination

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.

$$18x+15y=12,30x-15y=-100$$

To make 18x and 30x equal, multiply all terms on each side of the first equation by 30 and all terms on each side of the second by 18.

$$30\times 18x+30\times 15y=30\times 12,18\times 30x+18\left(-15\right)y=18\left(-100\right)$$

Simplify.

$$540x+450y=360,540x-270y=-1800$$

Subtract 540x-270y=-1800 from 540x+450y=360 by subtracting like terms on each side of the equal sign.

$$540x-540x+450y+270y=360+1800$$

Add 540x to -540x. Terms 540x and -540x cancel out, leaving an equation with only one variable that can be solved.

$$450y+270y=360+1800$$

Add 450y to 270y.

$$720y=360+1800$$

Add 360 to 1800.

$$720y=2160$$

Divide both sides by 720.

$$y=3$$

Substitute 3 for y in 30x-15y=-100. Because the resulting equation contains only one variable, you can solve for x directly.

$$30x-15\times 3=-100$$

Multiply -15 times 3.

$$30x-45=-100$$

Add 45 to both sides of the equation.

$$30x=-55$$

Divide both sides by 30.

$$x=-\frac{11}{6}$$

The system is now solved.

$$x=-\frac{11}{6},y=3$$