Solve {l}{x+y=100}{17x+45y=6/5(5x+35y)}

$$\left\{ \begin{array} { l } { x + y = 100 } \\ { 17 x + 45 y = \frac { 6 } { 5 } ( 5 x + 35 y ) } \end{array} \right.$$

Solve for x, y

$$x = -\frac{75}{2} = -37\frac{1}{2} = -37.5$$

$$y = \frac{275}{2} = 137\frac{1}{2} = 137.5$$

Steps Using Substitution

Consider the second equation. Use the distributive property to multiply \frac{6}{5} by 5x+35y.

$$17x+45y=6x+42y$$

Subtract 6x from both sides.

$$17x+45y-6x=42y$$

Combine 17x and -6x to get 11x.

$$11x+45y=42y$$

Subtract 42y from both sides.

$$11x+45y-42y=0$$

Combine 45y and -42y to get 3y.

$$11x+3y=0$$

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.

$$x+y=100,11x+3y=0$$

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

$$x+y=100$$

Subtract y from both sides of the equation.

$$x=-y+100$$

Substitute -y+100 for x in the other equation, 11x+3y=0.

$$11\left(-y+100\right)+3y=0$$

Multiply 11 times -y+100.

$$-11y+1100+3y=0$$

Add -11y to 3y.

$$-8y+1100=0$$

Subtract 1100 from both sides of the equation.

$$-8y=-1100$$

Divide both sides by -8.

$$y=\frac{275}{2}$$

Substitute \frac{275}{2} for y in x=-y+100. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-\frac{275}{2}+100$$

Add 100 to -\frac{275}{2}.

$$x=-\frac{75}{2}$$

The system is now solved.

$$x=-\frac{75}{2},y=\frac{275}{2}$$

Steps Using Matrices

Consider the second equation. Use the distributive property to multiply \frac{6}{5} by 5x+35y.

$$17x+45y=6x+42y$$

Subtract 6x from both sides.

$$17x+45y-6x=42y$$

Combine 17x and -6x to get 11x.

$$11x+45y=42y$$

Subtract 42y from both sides.

$$11x+45y-42y=0$$

Combine 45y and -42y to get 3y.

$$11x+3y=0$$

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

$$x+y=100,11x+3y=0$$

Write the equations in matrix form.

$$\left(\begin{matrix}1&1\\11&3\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}100\\0\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&1\\11&3\end{matrix}\right).

$$inverse(\left(\begin{matrix}1&1\\11&3\end{matrix}\right))\left(\begin{matrix}1&1\\11&3\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\11&3\end{matrix}\right))\left(\begin{matrix}100\\0\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\11&3\end{matrix}\right))\left(\begin{matrix}100\\0\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\11&3\end{matrix}\right))\left(\begin{matrix}100\\0\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{3}{3-11}&-\frac{1}{3-11}\\-\frac{11}{3-11}&\frac{1}{3-11}\end{matrix}\right)\left(\begin{matrix}100\\0\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{3}{8}&\frac{1}{8}\\\frac{11}{8}&-\frac{1}{8}\end{matrix}\right)\left(\begin{matrix}100\\0\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{3}{8}\times 100\\\frac{11}{8}\times 100\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{75}{2}\\\frac{275}{2}\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=-\frac{75}{2},y=\frac{275}{2}$$

Steps Using Elimination

Consider the second equation. Use the distributive property to multiply \frac{6}{5} by 5x+35y.

$$17x+45y=6x+42y$$

Subtract 6x from both sides.

$$17x+45y-6x=42y$$

Combine 17x and -6x to get 11x.

$$11x+45y=42y$$

Subtract 42y from both sides.

$$11x+45y-42y=0$$

Combine 45y and -42y to get 3y.

$$11x+3y=0$$

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.

$$x+y=100,11x+3y=0$$

To make x and 11x equal, multiply all terms on each side of the first equation by 11 and all terms on each side of the second by 1.

$$11x+11y=11\times 100,11x+3y=0$$

Simplify.

$$11x+11y=1100,11x+3y=0$$

Subtract 11x+3y=0 from 11x+11y=1100 by subtracting like terms on each side of the equal sign.

$$11x-11x+11y-3y=1100$$

Add 11x to -11x. Terms 11x and -11x cancel out, leaving an equation with only one variable that can be solved.

$$11y-3y=1100$$

Add 11y to -3y.

$$8y=1100$$

Divide both sides by 8.

$$y=\frac{275}{2}$$

Substitute \frac{275}{2} for y in 11x+3y=0. Because the resulting equation contains only one variable, you can solve for x directly.

$$11x+3\times \frac{275}{2}=0$$

Multiply 3 times \frac{275}{2}.

$$11x+\frac{825}{2}=0$$

Subtract \frac{825}{2} from both sides of the equation.

$$11x=-\frac{825}{2}$$

Divide both sides by 11.

$$x=-\frac{75}{2}$$

The system is now solved.

$$x=-\frac{75}{2},y=\frac{275}{2}$$