Solve {l}{x+y=2400}{2x=3y}

$$\left\{ \begin{array} { l } { x + y = 2400 } \\ { 2 x = 3 y } \end{array} \right.$$

Solve for x, y

$$x=1440$$

$$y=960$$

Steps Using Substitution

Consider the second equation. Subtract 3y from both sides.

$$2x-3y=0$$

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.

$$x+y=2400,2x-3y=0$$

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

$$x+y=2400$$

Subtract y from both sides of the equation.

$$x=-y+2400$$

Substitute -y+2400 for x in the other equation, 2x-3y=0.

$$2\left(-y+2400\right)-3y=0$$

Multiply 2 times -y+2400.

$$-2y+4800-3y=0$$

Add -2y to -3y.

$$-5y+4800=0$$

Subtract 4800 from both sides of the equation.

$$-5y=-4800$$

Divide both sides by -5.

$$y=960$$

Substitute 960 for y in x=-y+2400. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-960+2400$$

Add 2400 to -960.

$$x=1440$$

The system is now solved.

$$x=1440,y=960$$

Steps Using Matrices

Consider the second equation. Subtract 3y from both sides.

$$2x-3y=0$$

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

$$x+y=2400,2x-3y=0$$

Write the equations in matrix form.

$$\left(\begin{matrix}1&1\\2&-3\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}2400\\0\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&1\\2&-3\end{matrix}\right).

$$inverse(\left(\begin{matrix}1&1\\2&-3\end{matrix}\right))\left(\begin{matrix}1&1\\2&-3\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\2&-3\end{matrix}\right))\left(\begin{matrix}2400\\0\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\2&-3\end{matrix}\right))\left(\begin{matrix}2400\\0\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\2&-3\end{matrix}\right))\left(\begin{matrix}2400\\0\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{3}{-3-2}&-\frac{1}{-3-2}\\-\frac{2}{-3-2}&\frac{1}{-3-2}\end{matrix}\right)\left(\begin{matrix}2400\\0\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{3}{5}&\frac{1}{5}\\\frac{2}{5}&-\frac{1}{5}\end{matrix}\right)\left(\begin{matrix}2400\\0\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{3}{5}\times 2400\\\frac{2}{5}\times 2400\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}1440\\960\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=1440,y=960$$

Steps Using Elimination

Consider the second equation. Subtract 3y from both sides.

$$2x-3y=0$$

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.

$$x+y=2400,2x-3y=0$$

To make x and 2x equal, multiply all terms on each side of the first equation by 2 and all terms on each side of the second by 1.

$$2x+2y=2\times 2400,2x-3y=0$$

Simplify.

$$2x+2y=4800,2x-3y=0$$

Subtract 2x-3y=0 from 2x+2y=4800 by subtracting like terms on each side of the equal sign.

$$2x-2x+2y+3y=4800$$

Add 2x to -2x. Terms 2x and -2x cancel out, leaving an equation with only one variable that can be solved.

$$2y+3y=4800$$

Add 2y to 3y.

$$5y=4800$$

Divide both sides by 5.

$$y=960$$

Substitute 960 for y in 2x-3y=0. Because the resulting equation contains only one variable, you can solve for x directly.

$$2x-3\times 960=0$$

Multiply -3 times 960.

$$2x-2880=0$$

Add 2880 to both sides of the equation.

$$2x=2880$$

Divide both sides by 2.

$$x=1440$$

The system is now solved.

$$x=1440,y=960$$