Solve {l}{40x+60y=3800}{60x+100y=6000}

$$\left\{ \begin{array} { l } { 40 x + 60 y = 3800 } \\ { 60 x + 100 y = 6000 } \end{array} \right.$$

Solve for x, y

$$x=50$$

$$y=30$$

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.

$$40x+60y=3800,60x+100y=6000$$

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

$$40x+60y=3800$$

Subtract 60y from both sides of the equation.

$$40x=-60y+3800$$

Divide both sides by 40.

$$x=\frac{1}{40}\left(-60y+3800\right)$$

Multiply \frac{1}{40} times -60y+3800.

$$x=-\frac{3}{2}y+95$$

Substitute -\frac{3y}{2}+95 for x in the other equation, 60x+100y=6000.

$$60\left(-\frac{3}{2}y+95\right)+100y=6000$$

Multiply 60 times -\frac{3y}{2}+95.

$$-90y+5700+100y=6000$$

Add -90y to 100y.

$$10y+5700=6000$$

Subtract 5700 from both sides of the equation.

$$10y=300$$

Divide both sides by 10.

$$y=30$$

Substitute 30 for y in x=-\frac{3}{2}y+95. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-\frac{3}{2}\times 30+95$$

Multiply -\frac{3}{2} times 30.

$$x=-45+95$$

Add 95 to -45.

$$x=50$$

The system is now solved.

$$x=50,y=30$$

Steps Using Matrices

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

$$40x+60y=3800,60x+100y=6000$$

Write the equations in matrix form.

$$\left(\begin{matrix}40&60\\60&100\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}3800\\6000\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}40&60\\60&100\end{matrix}\right).

$$inverse(\left(\begin{matrix}40&60\\60&100\end{matrix}\right))\left(\begin{matrix}40&60\\60&100\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}40&60\\60&100\end{matrix}\right))\left(\begin{matrix}3800\\6000\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}40&60\\60&100\end{matrix}\right))\left(\begin{matrix}3800\\6000\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}40&60\\60&100\end{matrix}\right))\left(\begin{matrix}3800\\6000\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{100}{40\times 100-60\times 60}&-\frac{60}{40\times 100-60\times 60}\\-\frac{60}{40\times 100-60\times 60}&\frac{40}{40\times 100-60\times 60}\end{matrix}\right)\left(\begin{matrix}3800\\6000\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{4}&-\frac{3}{20}\\-\frac{3}{20}&\frac{1}{10}\end{matrix}\right)\left(\begin{matrix}3800\\6000\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{4}\times 3800-\frac{3}{20}\times 6000\\-\frac{3}{20}\times 3800+\frac{1}{10}\times 6000\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}50\\30\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=50,y=30$$

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.

$$40x+60y=3800,60x+100y=6000$$

To make 40x 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 40.

$$60\times 40x+60\times 60y=60\times 3800,40\times 60x+40\times 100y=40\times 6000$$

Simplify.

$$2400x+3600y=228000,2400x+4000y=240000$$

Subtract 2400x+4000y=240000 from 2400x+3600y=228000 by subtracting like terms on each side of the equal sign.

$$2400x-2400x+3600y-4000y=228000-240000$$

Add 2400x to -2400x. Terms 2400x and -2400x cancel out, leaving an equation with only one variable that can be solved.

$$3600y-4000y=228000-240000$$

Add 3600y to -4000y.

$$-400y=228000-240000$$

Add 228000 to -240000.

$$-400y=-12000$$

Divide both sides by -400.

$$y=30$$

Substitute 30 for y in 60x+100y=6000. Because the resulting equation contains only one variable, you can solve for x directly.

$$60x+100\times 30=6000$$

Multiply 100 times 30.

$$60x+3000=6000$$

Subtract 3000 from both sides of the equation.

$$60x=3000$$

Divide both sides by 60.

$$x=50$$

The system is now solved.

$$x=50,y=30$$