Solve {c}{x+y=414}{55x+32y=22264}

$$\left. \begin{array} { c } { x + y = 414 } \\ { 55 x + 32 y = 22264 } \end{array} \right.$$

Solve for x, y

$$x=392$$

$$y=22$$

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.

$$x+y=414,55x+32y=22264$$

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

$$x+y=414$$

Subtract y from both sides of the equation.

$$x=-y+414$$

Substitute -y+414 for x in the other equation, 55x+32y=22264.

$$55\left(-y+414\right)+32y=22264$$

Multiply 55 times -y+414.

$$-55y+22770+32y=22264$$

Add -55y to 32y.

$$-23y+22770=22264$$

Subtract 22770 from both sides of the equation.

$$-23y=-506$$

Divide both sides by -23.

$$y=22$$

Substitute 22 for y in x=-y+414. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-22+414$$

Add 414 to -22.

$$x=392$$

The system is now solved.

$$x=392,y=22$$

Steps Using Matrices

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

$$x+y=414,55x+32y=22264$$

Write the equations in matrix form.

$$\left(\begin{matrix}1&1\\55&32\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}414\\22264\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&1\\55&32\end{matrix}\right).

$$inverse(\left(\begin{matrix}1&1\\55&32\end{matrix}\right))\left(\begin{matrix}1&1\\55&32\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\55&32\end{matrix}\right))\left(\begin{matrix}414\\22264\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\55&32\end{matrix}\right))\left(\begin{matrix}414\\22264\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\55&32\end{matrix}\right))\left(\begin{matrix}414\\22264\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{32}{32-55}&-\frac{1}{32-55}\\-\frac{55}{32-55}&\frac{1}{32-55}\end{matrix}\right)\left(\begin{matrix}414\\22264\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{32}{23}&\frac{1}{23}\\\frac{55}{23}&-\frac{1}{23}\end{matrix}\right)\left(\begin{matrix}414\\22264\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{32}{23}\times 414+\frac{1}{23}\times 22264\\\frac{55}{23}\times 414-\frac{1}{23}\times 22264\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}392\\22\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=392,y=22$$

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.

$$x+y=414,55x+32y=22264$$

To make x and 55x equal, multiply all terms on each side of the first equation by 55 and all terms on each side of the second by 1.

$$55x+55y=55\times 414,55x+32y=22264$$

Simplify.

$$55x+55y=22770,55x+32y=22264$$

Subtract 55x+32y=22264 from 55x+55y=22770 by subtracting like terms on each side of the equal sign.

$$55x-55x+55y-32y=22770-22264$$

Add 55x to -55x. Terms 55x and -55x cancel out, leaving an equation with only one variable that can be solved.

$$55y-32y=22770-22264$$

Add 55y to -32y.

$$23y=22770-22264$$

Add 22770 to -22264.

$$23y=506$$

Divide both sides by 23.

$$y=22$$

Substitute 22 for y in 55x+32y=22264. Because the resulting equation contains only one variable, you can solve for x directly.

$$55x+32\times 22=22264$$

Multiply 32 times 22.

$$55x+704=22264$$

Subtract 704 from both sides of the equation.

$$55x=21560$$

Divide both sides by 55.

$$x=392$$

The system is now solved.

$$x=392,y=22$$