Solve {l}{20x+10y=1100}{30x-20y=10}

$$\left. \begin{array} { l } { 20 x + 10 y = 1100 } \\ { 30 x - 20 y = 10 } \end{array} \right.$$

Solve for x, y

$$x = \frac{221}{7} = 31\frac{4}{7} \approx 31.571428571$$

$$y = \frac{328}{7} = 46\frac{6}{7} \approx 46.857142857$$

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.

$$20x+10y=1100,30x-20y=10$$

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

$$20x+10y=1100$$

Subtract 10y from both sides of the equation.

$$20x=-10y+1100$$

Divide both sides by 20.

$$x=\frac{1}{20}\left(-10y+1100\right)$$

Multiply \frac{1}{20} times -10y+1100.

$$x=-\frac{1}{2}y+55$$

Substitute -\frac{y}{2}+55 for x in the other equation, 30x-20y=10.

$$30\left(-\frac{1}{2}y+55\right)-20y=10$$

Multiply 30 times -\frac{y}{2}+55.

$$-15y+1650-20y=10$$

Add -15y to -20y.

$$-35y+1650=10$$

Subtract 1650 from both sides of the equation.

$$-35y=-1640$$

Divide both sides by -35.

$$y=\frac{328}{7}$$

Substitute \frac{328}{7} for y in x=-\frac{1}{2}y+55. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-\frac{1}{2}\times \frac{328}{7}+55$$

Multiply -\frac{1}{2} times \frac{328}{7} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$x=-\frac{164}{7}+55$$

Add 55 to -\frac{164}{7}.

$$x=\frac{221}{7}$$

The system is now solved.

$$x=\frac{221}{7},y=\frac{328}{7}$$

Steps Using Matrices

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

$$20x+10y=1100,30x-20y=10$$

Write the equations in matrix form.

$$\left(\begin{matrix}20&10\\30&-20\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}1100\\10\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}20&10\\30&-20\end{matrix}\right).

$$inverse(\left(\begin{matrix}20&10\\30&-20\end{matrix}\right))\left(\begin{matrix}20&10\\30&-20\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}20&10\\30&-20\end{matrix}\right))\left(\begin{matrix}1100\\10\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}20&10\\30&-20\end{matrix}\right))\left(\begin{matrix}1100\\10\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}20&10\\30&-20\end{matrix}\right))\left(\begin{matrix}1100\\10\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{20}{20\left(-20\right)-10\times 30}&-\frac{10}{20\left(-20\right)-10\times 30}\\-\frac{30}{20\left(-20\right)-10\times 30}&\frac{20}{20\left(-20\right)-10\times 30}\end{matrix}\right)\left(\begin{matrix}1100\\10\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{35}&\frac{1}{70}\\\frac{3}{70}&-\frac{1}{35}\end{matrix}\right)\left(\begin{matrix}1100\\10\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{35}\times 1100+\frac{1}{70}\times 10\\\frac{3}{70}\times 1100-\frac{1}{35}\times 10\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{221}{7}\\\frac{328}{7}\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=\frac{221}{7},y=\frac{328}{7}$$

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.

$$20x+10y=1100,30x-20y=10$$

To make 20x 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 20.

$$30\times 20x+30\times 10y=30\times 1100,20\times 30x+20\left(-20\right)y=20\times 10$$

Simplify.

$$600x+300y=33000,600x-400y=200$$

Subtract 600x-400y=200 from 600x+300y=33000 by subtracting like terms on each side of the equal sign.

$$600x-600x+300y+400y=33000-200$$

Add 600x to -600x. Terms 600x and -600x cancel out, leaving an equation with only one variable that can be solved.

$$300y+400y=33000-200$$

Add 300y to 400y.

$$700y=33000-200$$

Add 33000 to -200.

$$700y=32800$$

Divide both sides by 700.

$$y=\frac{328}{7}$$

Substitute \frac{328}{7} for y in 30x-20y=10. Because the resulting equation contains only one variable, you can solve for x directly.

$$30x-20\times \frac{328}{7}=10$$

Multiply -20 times \frac{328}{7}.

$$30x-\frac{6560}{7}=10$$

Add \frac{6560}{7} to both sides of the equation.

$$30x=\frac{6630}{7}$$

Divide both sides by 30.

$$x=\frac{221}{7}$$

The system is now solved.

$$x=\frac{221}{7},y=\frac{328}{7}$$