Solve {l}{48x+40y=1200}{120x+80y=2800}

$$\left\{ \begin{array} { l } { 48 x + 40 y = 1200 } \\ { 120 x + 80 y = 2800 } \end{array} \right.$$

Solve for x, y

$$x = \frac{50}{3} = 16\frac{2}{3} \approx 16.666666667$$

$$y=10$$

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.

$$48x+40y=1200,120x+80y=2800$$

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

$$48x+40y=1200$$

Subtract 40y from both sides of the equation.

$$48x=-40y+1200$$

Divide both sides by 48.

$$x=\frac{1}{48}\left(-40y+1200\right)$$

Multiply \frac{1}{48} times -40y+1200.

$$x=-\frac{5}{6}y+25$$

Substitute -\frac{5y}{6}+25 for x in the other equation, 120x+80y=2800.

$$120\left(-\frac{5}{6}y+25\right)+80y=2800$$

Multiply 120 times -\frac{5y}{6}+25.

$$-100y+3000+80y=2800$$

Add -100y to 80y.

$$-20y+3000=2800$$

Subtract 3000 from both sides of the equation.

$$-20y=-200$$

Divide both sides by -20.

$$y=10$$

Substitute 10 for y in x=-\frac{5}{6}y+25. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-\frac{5}{6}\times 10+25$$

Multiply -\frac{5}{6} times 10.

$$x=-\frac{25}{3}+25$$

Add 25 to -\frac{25}{3}.

$$x=\frac{50}{3}$$

The system is now solved.

$$x=\frac{50}{3},y=10$$

Steps Using Matrices

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

$$48x+40y=1200,120x+80y=2800$$

Write the equations in matrix form.

$$\left(\begin{matrix}48&40\\120&80\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}1200\\2800\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}48&40\\120&80\end{matrix}\right).

$$inverse(\left(\begin{matrix}48&40\\120&80\end{matrix}\right))\left(\begin{matrix}48&40\\120&80\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}48&40\\120&80\end{matrix}\right))\left(\begin{matrix}1200\\2800\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}48&40\\120&80\end{matrix}\right))\left(\begin{matrix}1200\\2800\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}48&40\\120&80\end{matrix}\right))\left(\begin{matrix}1200\\2800\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{80}{48\times 80-40\times 120}&-\frac{40}{48\times 80-40\times 120}\\-\frac{120}{48\times 80-40\times 120}&\frac{48}{48\times 80-40\times 120}\end{matrix}\right)\left(\begin{matrix}1200\\2800\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{12}&\frac{1}{24}\\\frac{1}{8}&-\frac{1}{20}\end{matrix}\right)\left(\begin{matrix}1200\\2800\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{12}\times 1200+\frac{1}{24}\times 2800\\\frac{1}{8}\times 1200-\frac{1}{20}\times 2800\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{50}{3}\\10\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=\frac{50}{3},y=10$$

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.

$$48x+40y=1200,120x+80y=2800$$

To make 48x and 120x equal, multiply all terms on each side of the first equation by 120 and all terms on each side of the second by 48.

$$120\times 48x+120\times 40y=120\times 1200,48\times 120x+48\times 80y=48\times 2800$$

Simplify.

$$5760x+4800y=144000,5760x+3840y=134400$$

Subtract 5760x+3840y=134400 from 5760x+4800y=144000 by subtracting like terms on each side of the equal sign.

$$5760x-5760x+4800y-3840y=144000-134400$$

Add 5760x to -5760x. Terms 5760x and -5760x cancel out, leaving an equation with only one variable that can be solved.

$$4800y-3840y=144000-134400$$

Add 4800y to -3840y.

$$960y=144000-134400$$

Add 144000 to -134400.

$$960y=9600$$

Divide both sides by 960.

$$y=10$$

Substitute 10 for y in 120x+80y=2800. Because the resulting equation contains only one variable, you can solve for x directly.

$$120x+80\times 10=2800$$

Multiply 80 times 10.

$$120x+800=2800$$

Subtract 800 from both sides of the equation.

$$120x=2000$$

Divide both sides by 120.

$$x=\frac{50}{3}$$

The system is now solved.

$$x=\frac{50}{3},y=10$$