Solve {l}{5x+20y=1100}{60x+80y=1600}

$$\left\{ \begin{array} { l } { 5 x + 20 y = 1100 } \\ { 60 x + 80 y = 1600 } \end{array} \right.$$

Solve for x, y

$$x=-70$$

$$y = \frac{145}{2} = 72\frac{1}{2} = 72.5$$

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.

$$5x+20y=1100,60x+80y=1600$$

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

$$5x+20y=1100$$

Subtract 20y from both sides of the equation.

$$5x=-20y+1100$$

Divide both sides by 5.

$$x=\frac{1}{5}\left(-20y+1100\right)$$

Multiply \frac{1}{5} times -20y+1100.

$$x=-4y+220$$

Substitute -4y+220 for x in the other equation, 60x+80y=1600.

$$60\left(-4y+220\right)+80y=1600$$

Multiply 60 times -4y+220.

$$-240y+13200+80y=1600$$

Add -240y to 80y.

$$-160y+13200=1600$$

Subtract 13200 from both sides of the equation.

$$-160y=-11600$$

Divide both sides by -160.

$$y=\frac{145}{2}$$

Substitute \frac{145}{2} for y in x=-4y+220. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-4\times \frac{145}{2}+220$$

Multiply -4 times \frac{145}{2}.

$$x=-290+220$$

Add 220 to -290.

$$x=-70$$

The system is now solved.

$$x=-70,y=\frac{145}{2}$$

Steps Using Matrices

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

$$5x+20y=1100,60x+80y=1600$$

Write the equations in matrix form.

$$\left(\begin{matrix}5&20\\60&80\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}1100\\1600\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}5&20\\60&80\end{matrix}\right).

$$inverse(\left(\begin{matrix}5&20\\60&80\end{matrix}\right))\left(\begin{matrix}5&20\\60&80\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}5&20\\60&80\end{matrix}\right))\left(\begin{matrix}1100\\1600\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}5&20\\60&80\end{matrix}\right))\left(\begin{matrix}1100\\1600\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}5&20\\60&80\end{matrix}\right))\left(\begin{matrix}1100\\1600\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{80}{5\times 80-20\times 60}&-\frac{20}{5\times 80-20\times 60}\\-\frac{60}{5\times 80-20\times 60}&\frac{5}{5\times 80-20\times 60}\end{matrix}\right)\left(\begin{matrix}1100\\1600\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{10}&\frac{1}{40}\\\frac{3}{40}&-\frac{1}{160}\end{matrix}\right)\left(\begin{matrix}1100\\1600\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{10}\times 1100+\frac{1}{40}\times 1600\\\frac{3}{40}\times 1100-\frac{1}{160}\times 1600\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-70\\\frac{145}{2}\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=-70,y=\frac{145}{2}$$

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.

$$5x+20y=1100,60x+80y=1600$$

To make 5x and 60x equal, multiply all terms on each side of the first equation by 60 and all terms on each side of the second by 5.

$$60\times 5x+60\times 20y=60\times 1100,5\times 60x+5\times 80y=5\times 1600$$

Simplify.

$$300x+1200y=66000,300x+400y=8000$$

Subtract 300x+400y=8000 from 300x+1200y=66000 by subtracting like terms on each side of the equal sign.

$$300x-300x+1200y-400y=66000-8000$$

Add 300x to -300x. Terms 300x and -300x cancel out, leaving an equation with only one variable that can be solved.

$$1200y-400y=66000-8000$$

Add 1200y to -400y.

$$800y=66000-8000$$

Add 66000 to -8000.

$$800y=58000$$

Divide both sides by 800.

$$y=\frac{145}{2}$$

Substitute \frac{145}{2} for y in 60x+80y=1600. Because the resulting equation contains only one variable, you can solve for x directly.

$$60x+80\times \frac{145}{2}=1600$$

Multiply 80 times \frac{145}{2}.

$$60x+5800=1600$$

Subtract 5800 from both sides of the equation.

$$60x=-4200$$

Divide both sides by 60.

$$x=-70$$

The system is now solved.

$$x=-70,y=\frac{145}{2}$$