Solve {l}{2300x+1200y=18800}{1955x+960y=18380}

$$\left\{ \begin{array} { l } { 2300 x + 1200 y = 18800 } \\ { 1955 x + 960 y = 18380 } \end{array} \right.$$

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.

$$2300x+1200y=18800,1955x+960y=18380$$

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, 1955x+960y=18380.

$$1955\left(-\frac{12}{23}y+\frac{188}{23}\right)+960y=18380$$

Multiply 1955 times \frac{-12y+188}{23}.

$$-1020y+15980+960y=18380$$

Add -1020y to 960y.

$$-60y+15980=18380$$

Subtract 15980 from both sides of the equation.

$$-60y=2400$$

Divide both sides by -60.

$$y=-40$$

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.

$$2300x+1200y=18800,1955x+960y=18380$$

Write the equations in matrix form.

$$\left(\begin{matrix}2300&1200\\1955&960\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}18800\\18380\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}2300&1200\\1955&960\end{matrix}\right).

$$inverse(\left(\begin{matrix}2300&1200\\1955&960\end{matrix}\right))\left(\begin{matrix}2300&1200\\1955&960\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}2300&1200\\1955&960\end{matrix}\right))\left(\begin{matrix}18800\\18380\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}2300&1200\\1955&960\end{matrix}\right))\left(\begin{matrix}18800\\18380\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}2300&1200\\1955&960\end{matrix}\right))\left(\begin{matrix}18800\\18380\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{960}{2300\times 960-1200\times 1955}&-\frac{1200}{2300\times 960-1200\times 1955}\\-\frac{1955}{2300\times 960-1200\times 1955}&\frac{2300}{2300\times 960-1200\times 1955}\end{matrix}\right)\left(\begin{matrix}18800\\18380\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{4}{575}&\frac{1}{115}\\\frac{17}{1200}&-\frac{1}{60}\end{matrix}\right)\left(\begin{matrix}18800\\18380\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{4}{575}\times 18800+\frac{1}{115}\times 18380\\\frac{17}{1200}\times 18800-\frac{1}{60}\times 18380\end{matrix}\right)$$

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.

$$2300x+1200y=18800,1955x+960y=18380$$

To make 2300x and 1955x equal, multiply all terms on each side of the first equation by 1955 and all terms on each side of the second by 2300.

$$1955\times 2300x+1955\times 1200y=1955\times 18800,2300\times 1955x+2300\times 960y=2300\times 18380$$

Simplify.

$$4496500x+2346000y=36754000,4496500x+2208000y=42274000$$

Subtract 4496500x+2208000y=42274000 from 4496500x+2346000y=36754000 by subtracting like terms on each side of the equal sign.

$$4496500x-4496500x+2346000y-2208000y=36754000-42274000$$

Add 4496500x to -4496500x. Terms 4496500x and -4496500x cancel out, leaving an equation with only one variable that can be solved.

$$2346000y-2208000y=36754000-42274000$$

Add 2346000y to -2208000y.

$$138000y=36754000-42274000$$

Add 36754000 to -42274000.

$$138000y=-5520000$$

Divide both sides by 138000.

$$y=-40$$

Substitute -40 for y in 1955x+960y=18380. Because the resulting equation contains only one variable, you can solve for x directly.

$$1955x+960\left(-40\right)=18380$$

Multiply 960 times -40.

$$1955x-38400=18380$$

Add 38400 to both sides of the equation.

$$1955x=56780$$

Divide both sides by 1955.

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

The system is now solved.

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