Solve {l}{x+y=100}{140x+1000y=87000}

$$\left. \begin{array} { l } { x + y = 100 } \\ { 140 x + 1000 y = 87000 } \end{array} \right.$$

Solve for x, y

$$x = \frac{650}{43} = 15\frac{5}{43} \approx 15.11627907$$

$$y = \frac{3650}{43} = 84\frac{38}{43} \approx 84.88372093$$

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.

$$x+y=100,140x+1000y=87000$$

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, 140x+1000y=87000.

$$140\left(-y+100\right)+1000y=87000$$

Multiply 140 times -y+100.

$$-140y+14000+1000y=87000$$

Add -140y to 1000y.

$$860y+14000=87000$$

Subtract 14000 from both sides of the equation.

$$860y=73000$$

Divide both sides by 860.

$$y=\frac{3650}{43}$$

Substitute \frac{3650}{43} for y in x=-y+100. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-\frac{3650}{43}+100$$

Add 100 to -\frac{3650}{43}.

$$x=\frac{650}{43}$$

The system is now solved.

$$x=\frac{650}{43},y=\frac{3650}{43}$$

Steps Using Matrices

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

$$x+y=100,140x+1000y=87000$$

Write the equations in matrix form.

$$\left(\begin{matrix}1&1\\140&1000\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}100\\87000\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&1\\140&1000\end{matrix}\right).

$$inverse(\left(\begin{matrix}1&1\\140&1000\end{matrix}\right))\left(\begin{matrix}1&1\\140&1000\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\140&1000\end{matrix}\right))\left(\begin{matrix}100\\87000\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\140&1000\end{matrix}\right))\left(\begin{matrix}100\\87000\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\140&1000\end{matrix}\right))\left(\begin{matrix}100\\87000\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1000}{1000-140}&-\frac{1}{1000-140}\\-\frac{140}{1000-140}&\frac{1}{1000-140}\end{matrix}\right)\left(\begin{matrix}100\\87000\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{50}{43}&-\frac{1}{860}\\-\frac{7}{43}&\frac{1}{860}\end{matrix}\right)\left(\begin{matrix}100\\87000\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{50}{43}\times 100-\frac{1}{860}\times 87000\\-\frac{7}{43}\times 100+\frac{1}{860}\times 87000\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{650}{43}\\\frac{3650}{43}\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=\frac{650}{43},y=\frac{3650}{43}$$

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.

$$x+y=100,140x+1000y=87000$$

To make x and 140x equal, multiply all terms on each side of the first equation by 140 and all terms on each side of the second by 1.

$$140x+140y=140\times 100,140x+1000y=87000$$

Simplify.

$$140x+140y=14000,140x+1000y=87000$$

Subtract 140x+1000y=87000 from 140x+140y=14000 by subtracting like terms on each side of the equal sign.

$$140x-140x+140y-1000y=14000-87000$$

Add 140x to -140x. Terms 140x and -140x cancel out, leaving an equation with only one variable that can be solved.

$$140y-1000y=14000-87000$$

Add 140y to -1000y.

$$-860y=14000-87000$$

Add 14000 to -87000.

$$-860y=-73000$$

Divide both sides by -860.

$$y=\frac{3650}{43}$$

Substitute \frac{3650}{43} for y in 140x+1000y=87000. Because the resulting equation contains only one variable, you can solve for x directly.

$$140x+1000\times \frac{3650}{43}=87000$$

Multiply 1000 times \frac{3650}{43}.

$$140x+\frac{3650000}{43}=87000$$

Subtract \frac{3650000}{43} from both sides of the equation.

$$140x=\frac{91000}{43}$$

Divide both sides by 140.

$$x=\frac{650}{43}$$

The system is now solved.

$$x=\frac{650}{43},y=\frac{3650}{43}$$