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

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

Solve for x, y

$$x=-5$$

$$y = -\frac{13}{2} = -6\frac{1}{2} = -6.5$$

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.

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

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

$$60x-60y=90$$

Add 60y to both sides of the equation.

$$60x=60y+90$$

Divide both sides by 60.

$$x=\frac{1}{60}\left(60y+90\right)$$

Multiply \frac{1}{60} times 60y+90.

$$x=y+\frac{3}{2}$$

Substitute y+\frac{3}{2} for x in the other equation, -45x+30y=30.

$$-45\left(y+\frac{3}{2}\right)+30y=30$$

Multiply -45 times y+\frac{3}{2}.

$$-45y-\frac{135}{2}+30y=30$$

Add -45y to 30y.

$$-15y-\frac{135}{2}=30$$

Add \frac{135}{2} to both sides of the equation.

$$-15y=\frac{195}{2}$$

Divide both sides by -15.

$$y=-\frac{13}{2}$$

Substitute -\frac{13}{2} for y in x=y+\frac{3}{2}. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=\frac{-13+3}{2}$$

Add \frac{3}{2} to -\frac{13}{2} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=-5$$

The system is now solved.

$$x=-5,y=-\frac{13}{2}$$

Steps Using Matrices

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

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

Write the equations in matrix form.

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

Left multiply the equation by the inverse matrix of \left(\begin{matrix}60&-60\\-45&30\end{matrix}\right).

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

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

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

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

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

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.

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

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{30}&-\frac{1}{15}\\-\frac{1}{20}&-\frac{1}{15}\end{matrix}\right)\left(\begin{matrix}90\\30\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{30}\times 90-\frac{1}{15}\times 30\\-\frac{1}{20}\times 90-\frac{1}{15}\times 30\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-5\\-\frac{13}{2}\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=-5,y=-\frac{13}{2}$$

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.

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

To make 60x and -45x equal, multiply all terms on each side of the first equation by -45 and all terms on each side of the second by 60.

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

Simplify.

$$-2700x+2700y=-4050,-2700x+1800y=1800$$

Subtract -2700x+1800y=1800 from -2700x+2700y=-4050 by subtracting like terms on each side of the equal sign.

$$-2700x+2700x+2700y-1800y=-4050-1800$$

Add -2700x to 2700x. Terms -2700x and 2700x cancel out, leaving an equation with only one variable that can be solved.

$$2700y-1800y=-4050-1800$$

Add 2700y to -1800y.

$$900y=-4050-1800$$

Add -4050 to -1800.

$$900y=-5850$$

Divide both sides by 900.

$$y=-\frac{13}{2}$$

Substitute -\frac{13}{2} for y in -45x+30y=30. Because the resulting equation contains only one variable, you can solve for x directly.

$$-45x+30\left(-\frac{13}{2}\right)=30$$

Multiply 30 times -\frac{13}{2}.

$$-45x-195=30$$

Add 195 to both sides of the equation.

$$-45x=225$$

Divide both sides by -45.

$$x=-5$$

The system is now solved.

$$x=-5,y=-\frac{13}{2}$$