Solve {l}{B=500-2A}{(A+4B)-(2A+B)=600-500}

$$\left. \begin{array} { l } { B = 500 - 2 A } \\ { ( A + 4 B ) - ( 2 A + B ) = 600 - 500 } \end{array} \right.$$

Solve for B, A

$$B=100$$

$$A=200$$

Steps Using Substitution

Consider the first equation. Add 2A to both sides.

$$B+2A=500$$

Consider the second equation. To find the opposite of 2A+B, find the opposite of each term.

$$A+4B-2A-B=600-500$$

Combine A and -2A to get -A.

$$-A+4B-B=600-500$$

Combine 4B and -B to get 3B.

$$-A+3B=600-500$$

Subtract 500 from 600 to get 100.

$$-A+3B=100$$

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.

$$B+2A=500,3B-A=100$$

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

$$B+2A=500$$

Subtract 2A from both sides of the equation.

$$B=-2A+500$$

Substitute -2A+500 for B in the other equation, 3B-A=100.

$$3\left(-2A+500\right)-A=100$$

Multiply 3 times -2A+500.

$$-6A+1500-A=100$$

Add -6A to -A.

$$-7A+1500=100$$

Subtract 1500 from both sides of the equation.

$$-7A=-1400$$

Divide both sides by -7.

$$A=200$$

Substitute 200 for A in B=-2A+500. Because the resulting equation contains only one variable, you can solve for B directly.

$$B=-2\times 200+500$$

Multiply -2 times 200.

$$B=-400+500$$

Add 500 to -400.

$$B=100$$

The system is now solved.

$$B=100,A=200$$

Steps Using Matrices

Consider the first equation. Add 2A to both sides.

$$B+2A=500$$

Consider the second equation. To find the opposite of 2A+B, find the opposite of each term.

$$A+4B-2A-B=600-500$$

Combine A and -2A to get -A.

$$-A+4B-B=600-500$$

Combine 4B and -B to get 3B.

$$-A+3B=600-500$$

Subtract 500 from 600 to get 100.

$$-A+3B=100$$

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

$$B+2A=500,3B-A=100$$

Write the equations in matrix form.

$$\left(\begin{matrix}1&2\\3&-1\end{matrix}\right)\left(\begin{matrix}B\\A\end{matrix}\right)=\left(\begin{matrix}500\\100\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&2\\3&-1\end{matrix}\right).

$$inverse(\left(\begin{matrix}1&2\\3&-1\end{matrix}\right))\left(\begin{matrix}1&2\\3&-1\end{matrix}\right)\left(\begin{matrix}B\\A\end{matrix}\right)=inverse(\left(\begin{matrix}1&2\\3&-1\end{matrix}\right))\left(\begin{matrix}500\\100\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}B\\A\end{matrix}\right)=inverse(\left(\begin{matrix}1&2\\3&-1\end{matrix}\right))\left(\begin{matrix}500\\100\end{matrix}\right)$$

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

$$\left(\begin{matrix}B\\A\end{matrix}\right)=inverse(\left(\begin{matrix}1&2\\3&-1\end{matrix}\right))\left(\begin{matrix}500\\100\end{matrix}\right)$$

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.

$$\left(\begin{matrix}B\\A\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{-1-2\times 3}&-\frac{2}{-1-2\times 3}\\-\frac{3}{-1-2\times 3}&\frac{1}{-1-2\times 3}\end{matrix}\right)\left(\begin{matrix}500\\100\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}B\\A\end{matrix}\right)=\left(\begin{matrix}\frac{1}{7}&\frac{2}{7}\\\frac{3}{7}&-\frac{1}{7}\end{matrix}\right)\left(\begin{matrix}500\\100\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}B\\A\end{matrix}\right)=\left(\begin{matrix}\frac{1}{7}\times 500+\frac{2}{7}\times 100\\\frac{3}{7}\times 500-\frac{1}{7}\times 100\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}B\\A\end{matrix}\right)=\left(\begin{matrix}100\\200\end{matrix}\right)$$

Extract the matrix elements B and A.

$$B=100,A=200$$

Steps Using Elimination

Consider the first equation. Add 2A to both sides.

$$B+2A=500$$

Consider the second equation. To find the opposite of 2A+B, find the opposite of each term.

$$A+4B-2A-B=600-500$$

Combine A and -2A to get -A.

$$-A+4B-B=600-500$$

Combine 4B and -B to get 3B.

$$-A+3B=600-500$$

Subtract 500 from 600 to get 100.

$$-A+3B=100$$

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.

$$B+2A=500,3B-A=100$$

To make B and 3B equal, multiply all terms on each side of the first equation by 3 and all terms on each side of the second by 1.

$$3B+3\times 2A=3\times 500,3B-A=100$$

Simplify.

$$3B+6A=1500,3B-A=100$$

Subtract 3B-A=100 from 3B+6A=1500 by subtracting like terms on each side of the equal sign.

$$3B-3B+6A+A=1500-100$$

Add 3B to -3B. Terms 3B and -3B cancel out, leaving an equation with only one variable that can be solved.

$$6A+A=1500-100$$

Add 6A to A.

$$7A=1500-100$$

Add 1500 to -100.

$$7A=1400$$

Divide both sides by 7.

$$A=200$$

Substitute 200 for A in 3B-A=100. Because the resulting equation contains only one variable, you can solve for B directly.

$$3B-200=100$$

Add 200 to both sides of the equation.

$$3B=300$$

Divide both sides by 3.

$$B=100$$

The system is now solved.

$$B=100,A=200$$