Solve {l}{x+y=130}{45x+35y=5250}

$$\left. \begin{array} { l } { x + y = 130 } \\ { 45 x + 35 y = 5250 } \end{array} \right.$$

Solve for x, y

$$x=70$$

$$y=60$$

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.

$$x+y=130,45x+35y=5250$$

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

$$x+y=130$$

Subtract y from both sides of the equation.

$$x=-y+130$$

Substitute -y+130 for x in the other equation, 45x+35y=5250.

$$45\left(-y+130\right)+35y=5250$$

Multiply 45 times -y+130.

$$-45y+5850+35y=5250$$

Add -45y to 35y.

$$-10y+5850=5250$$

Subtract 5850 from both sides of the equation.

$$-10y=-600$$

Divide both sides by -10.

$$y=60$$

Substitute 60 for y in x=-y+130. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-60+130$$

Add 130 to -60.

$$x=70$$

The system is now solved.

$$x=70,y=60$$

Steps Using Matrices

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

$$x+y=130,45x+35y=5250$$

Write the equations in matrix form.

$$\left(\begin{matrix}1&1\\45&35\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}130\\5250\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&1\\45&35\end{matrix}\right).

$$inverse(\left(\begin{matrix}1&1\\45&35\end{matrix}\right))\left(\begin{matrix}1&1\\45&35\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\45&35\end{matrix}\right))\left(\begin{matrix}130\\5250\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\45&35\end{matrix}\right))\left(\begin{matrix}130\\5250\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\45&35\end{matrix}\right))\left(\begin{matrix}130\\5250\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{35}{35-45}&-\frac{1}{35-45}\\-\frac{45}{35-45}&\frac{1}{35-45}\end{matrix}\right)\left(\begin{matrix}130\\5250\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{7}{2}&\frac{1}{10}\\\frac{9}{2}&-\frac{1}{10}\end{matrix}\right)\left(\begin{matrix}130\\5250\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{7}{2}\times 130+\frac{1}{10}\times 5250\\\frac{9}{2}\times 130-\frac{1}{10}\times 5250\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}70\\60\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=70,y=60$$

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.

$$x+y=130,45x+35y=5250$$

To make x and 45x equal, multiply all terms on each side of the first equation by 45 and all terms on each side of the second by 1.

$$45x+45y=45\times 130,45x+35y=5250$$

Simplify.

$$45x+45y=5850,45x+35y=5250$$

Subtract 45x+35y=5250 from 45x+45y=5850 by subtracting like terms on each side of the equal sign.

$$45x-45x+45y-35y=5850-5250$$

Add 45x to -45x. Terms 45x and -45x cancel out, leaving an equation with only one variable that can be solved.

$$45y-35y=5850-5250$$

Add 45y to -35y.

$$10y=5850-5250$$

Add 5850 to -5250.

$$10y=600$$

Divide both sides by 10.

$$y=60$$

Substitute 60 for y in 45x+35y=5250. Because the resulting equation contains only one variable, you can solve for x directly.

$$45x+35\times 60=5250$$

Multiply 35 times 60.

$$45x+2100=5250$$

Subtract 2100 from both sides of the equation.

$$45x=3150$$

Divide both sides by 45.

$$x=70$$

The system is now solved.

$$x=70,y=60$$