Solve {c}{x+y=35}{35x+30y=1125}

$$\left. \begin{array} { c } { x + y = 35 } \\ { 35 x + 30 y = 1125 } \end{array} \right.$$

Solve for x, y

$$x=15$$

$$y=20$$

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.

$$x+y=35,35x+30y=1125$$

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

$$x+y=35$$

Subtract y from both sides of the equation.

$$x=-y+35$$

Substitute -y+35 for x in the other equation, 35x+30y=1125.

$$35\left(-y+35\right)+30y=1125$$

Multiply 35 times -y+35.

$$-35y+1225+30y=1125$$

Add -35y to 30y.

$$-5y+1225=1125$$

Subtract 1225 from both sides of the equation.

$$-5y=-100$$

Divide both sides by -5.

$$y=20$$

Substitute 20 for y in x=-y+35. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-20+35$$

Add 35 to -20.

$$x=15$$

The system is now solved.

$$x=15,y=20$$

Steps Using Matrices

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

$$x+y=35,35x+30y=1125$$

Write the equations in matrix form.

$$\left(\begin{matrix}1&1\\35&30\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}35\\1125\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&1\\35&30\end{matrix}\right).

$$inverse(\left(\begin{matrix}1&1\\35&30\end{matrix}\right))\left(\begin{matrix}1&1\\35&30\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\35&30\end{matrix}\right))\left(\begin{matrix}35\\1125\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\35&30\end{matrix}\right))\left(\begin{matrix}35\\1125\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\35&30\end{matrix}\right))\left(\begin{matrix}35\\1125\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{30}{30-35}&-\frac{1}{30-35}\\-\frac{35}{30-35}&\frac{1}{30-35}\end{matrix}\right)\left(\begin{matrix}35\\1125\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-6&\frac{1}{5}\\7&-\frac{1}{5}\end{matrix}\right)\left(\begin{matrix}35\\1125\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-6\times 35+\frac{1}{5}\times 1125\\7\times 35-\frac{1}{5}\times 1125\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}15\\20\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=15,y=20$$

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.

$$x+y=35,35x+30y=1125$$

To make x and 35x equal, multiply all terms on each side of the first equation by 35 and all terms on each side of the second by 1.

$$35x+35y=35\times 35,35x+30y=1125$$

Simplify.

$$35x+35y=1225,35x+30y=1125$$

Subtract 35x+30y=1125 from 35x+35y=1225 by subtracting like terms on each side of the equal sign.

$$35x-35x+35y-30y=1225-1125$$

Add 35x to -35x. Terms 35x and -35x cancel out, leaving an equation with only one variable that can be solved.

$$35y-30y=1225-1125$$

Add 35y to -30y.

$$5y=1225-1125$$

Add 1225 to -1125.

$$5y=100$$

Divide both sides by 5.

$$y=20$$

Substitute 20 for y in 35x+30y=1125. Because the resulting equation contains only one variable, you can solve for x directly.

$$35x+30\times 20=1125$$

Multiply 30 times 20.

$$35x+600=1125$$

Subtract 600 from both sides of the equation.

$$35x=525$$

Divide both sides by 35.

$$x=15$$

The system is now solved.

$$x=15,y=20$$