Solve {l}{10x+20y=4000}{20x+10y=3500}

$$\left\{ \begin{array} { l } { 10 x + 20 y = 4000 } \\ { 20 x + 10 y = 3500 } \end{array} \right.$$

Solve for x, y

$$x=100$$

$$y=150$$

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.

$$10x+20y=4000,20x+10y=3500$$

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

$$10x+20y=4000$$

Subtract 20y from both sides of the equation.

$$10x=-20y+4000$$

Divide both sides by 10.

$$x=\frac{1}{10}\left(-20y+4000\right)$$

Multiply \frac{1}{10} times -20y+4000.

$$x=-2y+400$$

Substitute -2y+400 for x in the other equation, 20x+10y=3500.

$$20\left(-2y+400\right)+10y=3500$$

Multiply 20 times -2y+400.

$$-40y+8000+10y=3500$$

Add -40y to 10y.

$$-30y+8000=3500$$

Subtract 8000 from both sides of the equation.

$$-30y=-4500$$

Divide both sides by -30.

$$y=150$$

Substitute 150 for y in x=-2y+400. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-2\times 150+400$$

Multiply -2 times 150.

$$x=-300+400$$

Add 400 to -300.

$$x=100$$

The system is now solved.

$$x=100,y=150$$

Steps Using Matrices

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

$$10x+20y=4000,20x+10y=3500$$

Write the equations in matrix form.

$$\left(\begin{matrix}10&20\\20&10\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}4000\\3500\end{matrix}\right)$$

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

$$inverse(\left(\begin{matrix}10&20\\20&10\end{matrix}\right))\left(\begin{matrix}10&20\\20&10\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}10&20\\20&10\end{matrix}\right))\left(\begin{matrix}4000\\3500\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}10&20\\20&10\end{matrix}\right))\left(\begin{matrix}4000\\3500\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}10&20\\20&10\end{matrix}\right))\left(\begin{matrix}4000\\3500\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{10}{10\times 10-20\times 20}&-\frac{20}{10\times 10-20\times 20}\\-\frac{20}{10\times 10-20\times 20}&\frac{10}{10\times 10-20\times 20}\end{matrix}\right)\left(\begin{matrix}4000\\3500\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{30}&\frac{1}{15}\\\frac{1}{15}&-\frac{1}{30}\end{matrix}\right)\left(\begin{matrix}4000\\3500\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{30}\times 4000+\frac{1}{15}\times 3500\\\frac{1}{15}\times 4000-\frac{1}{30}\times 3500\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}100\\150\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=100,y=150$$

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.

$$10x+20y=4000,20x+10y=3500$$

To make 10x and 20x equal, multiply all terms on each side of the first equation by 20 and all terms on each side of the second by 10.

$$20\times 10x+20\times 20y=20\times 4000,10\times 20x+10\times 10y=10\times 3500$$

Simplify.

$$200x+400y=80000,200x+100y=35000$$

Subtract 200x+100y=35000 from 200x+400y=80000 by subtracting like terms on each side of the equal sign.

$$200x-200x+400y-100y=80000-35000$$

Add 200x to -200x. Terms 200x and -200x cancel out, leaving an equation with only one variable that can be solved.

$$400y-100y=80000-35000$$

Add 400y to -100y.

$$300y=80000-35000$$

Add 80000 to -35000.

$$300y=45000$$

Divide both sides by 300.

$$y=150$$

Substitute 150 for y in 20x+10y=3500. Because the resulting equation contains only one variable, you can solve for x directly.

$$20x+10\times 150=3500$$

Multiply 10 times 150.

$$20x+1500=3500$$

Subtract 1500 from both sides of the equation.

$$20x=2000$$

Divide both sides by 20.

$$x=100$$

The system is now solved.

$$x=100,y=150$$