Solve {l}{(A+B)2-B=24}{=(2A+B)2^2-B=36}

$$\left. \begin{array} { l } { ( A + B ) 2 - B = 24 } \\ { = ( 2 A + B ) 2 ^ { 2 } - B = 36 } \end{array} \right.$$

Solve for A, B

$$A=-18$$

$$B=60$$

Steps Using Substitution

Consider the first equation. Use the distributive property to multiply A+B by 2.

$$2A+2B-B=24$$

Combine 2B and -B to get B.

$$2A+B=24$$

Consider the second equation. Calculate 2 to the power of 2 and get 4.

$$\left(2A+B\right)\times 4-B=36$$

Use the distributive property to multiply 2A+B by 4.

$$8A+4B-B=36$$

Combine 4B and -B to get 3B.

$$8A+3B=36$$

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.

$$2A+B=24,8A+3B=36$$

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

$$2A+B=24$$

Subtract B from both sides of the equation.

$$2A=-B+24$$

Divide both sides by 2.

$$A=\frac{1}{2}\left(-B+24\right)$$

Multiply \frac{1}{2} times -B+24.

$$A=-\frac{1}{2}B+12$$

Substitute -\frac{B}{2}+12 for A in the other equation, 8A+3B=36.

$$8\left(-\frac{1}{2}B+12\right)+3B=36$$

Multiply 8 times -\frac{B}{2}+12.

$$-4B+96+3B=36$$

Add -4B to 3B.

$$-B+96=36$$

Subtract 96 from both sides of the equation.

$$-B=-60$$

Divide both sides by -1.

$$B=60$$

Substitute 60 for B in A=-\frac{1}{2}B+12. Because the resulting equation contains only one variable, you can solve for A directly.

$$A=-\frac{1}{2}\times 60+12$$

Multiply -\frac{1}{2} times 60.

$$A=-30+12$$

Add 12 to -30.

$$A=-18$$

The system is now solved.

$$A=-18,B=60$$

Steps Using Matrices

Consider the first equation. Use the distributive property to multiply A+B by 2.

$$2A+2B-B=24$$

Combine 2B and -B to get B.

$$2A+B=24$$

Consider the second equation. Calculate 2 to the power of 2 and get 4.

$$\left(2A+B\right)\times 4-B=36$$

Use the distributive property to multiply 2A+B by 4.

$$8A+4B-B=36$$

Combine 4B and -B to get 3B.

$$8A+3B=36$$

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

$$2A+B=24,8A+3B=36$$

Write the equations in matrix form.

$$\left(\begin{matrix}2&1\\8&3\end{matrix}\right)\left(\begin{matrix}A\\B\end{matrix}\right)=\left(\begin{matrix}24\\36\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}2&1\\8&3\end{matrix}\right).

$$inverse(\left(\begin{matrix}2&1\\8&3\end{matrix}\right))\left(\begin{matrix}2&1\\8&3\end{matrix}\right)\left(\begin{matrix}A\\B\end{matrix}\right)=inverse(\left(\begin{matrix}2&1\\8&3\end{matrix}\right))\left(\begin{matrix}24\\36\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}A\\B\end{matrix}\right)=inverse(\left(\begin{matrix}2&1\\8&3\end{matrix}\right))\left(\begin{matrix}24\\36\end{matrix}\right)$$

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

$$\left(\begin{matrix}A\\B\end{matrix}\right)=inverse(\left(\begin{matrix}2&1\\8&3\end{matrix}\right))\left(\begin{matrix}24\\36\end{matrix}\right)$$

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.

$$\left(\begin{matrix}A\\B\end{matrix}\right)=\left(\begin{matrix}\frac{3}{2\times 3-8}&-\frac{1}{2\times 3-8}\\-\frac{8}{2\times 3-8}&\frac{2}{2\times 3-8}\end{matrix}\right)\left(\begin{matrix}24\\36\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}A\\B\end{matrix}\right)=\left(\begin{matrix}-\frac{3}{2}&\frac{1}{2}\\4&-1\end{matrix}\right)\left(\begin{matrix}24\\36\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}A\\B\end{matrix}\right)=\left(\begin{matrix}-\frac{3}{2}\times 24+\frac{1}{2}\times 36\\4\times 24-36\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}A\\B\end{matrix}\right)=\left(\begin{matrix}-18\\60\end{matrix}\right)$$

Extract the matrix elements A and B.

$$A=-18,B=60$$

Steps Using Elimination

Consider the first equation. Use the distributive property to multiply A+B by 2.

$$2A+2B-B=24$$

Combine 2B and -B to get B.

$$2A+B=24$$

Consider the second equation. Calculate 2 to the power of 2 and get 4.

$$\left(2A+B\right)\times 4-B=36$$

Use the distributive property to multiply 2A+B by 4.

$$8A+4B-B=36$$

Combine 4B and -B to get 3B.

$$8A+3B=36$$

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.

$$2A+B=24,8A+3B=36$$

To make 2A and 8A equal, multiply all terms on each side of the first equation by 8 and all terms on each side of the second by 2.

$$8\times 2A+8B=8\times 24,2\times 8A+2\times 3B=2\times 36$$

Simplify.

$$16A+8B=192,16A+6B=72$$

Subtract 16A+6B=72 from 16A+8B=192 by subtracting like terms on each side of the equal sign.

$$16A-16A+8B-6B=192-72$$

Add 16A to -16A. Terms 16A and -16A cancel out, leaving an equation with only one variable that can be solved.

$$8B-6B=192-72$$

Add 8B to -6B.

$$2B=192-72$$

Add 192 to -72.

$$2B=120$$

Divide both sides by 2.

$$B=60$$

Substitute 60 for B in 8A+3B=36. Because the resulting equation contains only one variable, you can solve for A directly.

$$8A+3\times 60=36$$

Multiply 3 times 60.

$$8A+180=36$$

Subtract 180 from both sides of the equation.

$$8A=-144$$

Divide both sides by 8.

$$A=-18$$

The system is now solved.

$$A=-18,B=60$$