Solve {l}{2300x+1200y=18800}{2443.75x+1200y=22975}

$$\left. \begin{array} { l } { 2300 x + 1200 y = 18800 } \\ { 2443.75 x + 1200 y = 22975 } \end{array} \right.$$

Solve for x, y

$$x = \frac{668}{23} = 29\frac{1}{23} \approx 29.043478261$$

$$y=-40$$

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.

$$2300x+1200y=18800,2443.75x+1200y=22975$$

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

$$2300x+1200y=18800$$

Subtract 1200y from both sides of the equation.

$$2300x=-1200y+18800$$

Divide both sides by 2300.

$$x=\frac{1}{2300}\left(-1200y+18800\right)$$

Multiply \frac{1}{2300} times -1200y+18800.

$$x=-\frac{12}{23}y+\frac{188}{23}$$

Substitute \frac{-12y+188}{23} for x in the other equation, 2443.75x+1200y=22975.

$$2443.75\left(-\frac{12}{23}y+\frac{188}{23}\right)+1200y=22975$$

Multiply 2443.75 times \frac{-12y+188}{23}.

$$-1275y+19975+1200y=22975$$

Add -1275y to 1200y.

$$-75y+19975=22975$$

Subtract 19975 from both sides of the equation.

$$-75y=3000$$

Divide both sides by -75.

$$y=-40$$

Substitute -40 for y in x=-\frac{12}{23}y+\frac{188}{23}. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-\frac{12}{23}\left(-40\right)+\frac{188}{23}$$

Multiply -\frac{12}{23} times -40.

$$x=\frac{480+188}{23}$$

Add \frac{188}{23} to \frac{480}{23} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{668}{23}$$

The system is now solved.

$$x=\frac{668}{23},y=-40$$

Steps Using Matrices

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

$$2300x+1200y=18800,2443.75x+1200y=22975$$

Write the equations in matrix form.

$$\left(\begin{matrix}2300&1200\\2443.75&1200\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}18800\\22975\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}2300&1200\\2443.75&1200\end{matrix}\right).

$$inverse(\left(\begin{matrix}2300&1200\\2443.75&1200\end{matrix}\right))\left(\begin{matrix}2300&1200\\2443.75&1200\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}2300&1200\\2443.75&1200\end{matrix}\right))\left(\begin{matrix}18800\\22975\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}2300&1200\\2443.75&1200\end{matrix}\right))\left(\begin{matrix}18800\\22975\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}2300&1200\\2443.75&1200\end{matrix}\right))\left(\begin{matrix}18800\\22975\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1200}{2300\times 1200-1200\times 2443.75}&-\frac{1200}{2300\times 1200-1200\times 2443.75}\\-\frac{2443.75}{2300\times 1200-1200\times 2443.75}&\frac{2300}{2300\times 1200-1200\times 2443.75}\end{matrix}\right)\left(\begin{matrix}18800\\22975\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{4}{575}&\frac{4}{575}\\\frac{17}{1200}&-\frac{1}{75}\end{matrix}\right)\left(\begin{matrix}18800\\22975\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{4}{575}\times 18800+\frac{4}{575}\times 22975\\\frac{17}{1200}\times 18800-\frac{1}{75}\times 22975\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{668}{23}\\-40\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=\frac{668}{23},y=-40$$

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.

$$2300x+1200y=18800,2443.75x+1200y=22975$$

Subtract 2443.75x+1200y=22975 from 2300x+1200y=18800 by subtracting like terms on each side of the equal sign.

$$2300x-2443.75x+1200y-1200y=18800-22975$$

Add 1200y to -1200y. Terms 1200y and -1200y cancel out, leaving an equation with only one variable that can be solved.

$$2300x-2443.75x=18800-22975$$

Add 2300x to -\frac{9775x}{4}.

$$-143.75x=18800-22975$$

Add 18800 to -22975.

$$-143.75x=-4175$$

Divide both sides of the equation by -143.75, which is the same as multiplying both sides by the reciprocal of the fraction.

$$x=\frac{668}{23}$$

Substitute \frac{668}{23} for x in 2443.75x+1200y=22975. Because the resulting equation contains only one variable, you can solve for y directly.

$$2443.75\times \frac{668}{23}+1200y=22975$$

Multiply 2443.75 times \frac{668}{23} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$70975+1200y=22975$$

Subtract 70975 from both sides of the equation.

$$1200y=-48000$$

Divide both sides by 1200.

$$y=-40$$

The system is now solved.

$$x=\frac{668}{23},y=-40$$