Solve {l}{(A+B)3-B=6}{(2A+B)3^2-B=42}

$$\left. \begin{array} { l } { ( A + B ) 3 - B = 6 } \\ { ( 2 A + B ) 3 ^ { 2 } - B = 42 } \end{array} \right.$$

Solve for A, B

$$A=3$$

$$B = -\frac{3}{2} = -1\frac{1}{2} = -1.5$$

Steps Using Substitution

Consider the first equation. Use the distributive property to multiply A+B by 3.

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

Combine 3B and -B to get 2B.

$$3A+2B=6$$

Consider the second equation. Calculate 3 to the power of 2 and get 9.

$$\left(2A+B\right)\times 9-B=42$$

Use the distributive property to multiply 2A+B by 9.

$$18A+9B-B=42$$

Combine 9B and -B to get 8B.

$$18A+8B=42$$

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.

$$3A+2B=6,18A+8B=42$$

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

$$3A+2B=6$$

Subtract 2B from both sides of the equation.

$$3A=-2B+6$$

Divide both sides by 3.

$$A=\frac{1}{3}\left(-2B+6\right)$$

Multiply \frac{1}{3} times -2B+6.

$$A=-\frac{2}{3}B+2$$

Substitute -\frac{2B}{3}+2 for A in the other equation, 18A+8B=42.

$$18\left(-\frac{2}{3}B+2\right)+8B=42$$

Multiply 18 times -\frac{2B}{3}+2.

$$-12B+36+8B=42$$

Add -12B to 8B.

$$-4B+36=42$$

Subtract 36 from both sides of the equation.

$$-4B=6$$

Divide both sides by -4.

$$B=-\frac{3}{2}$$

Substitute -\frac{3}{2} for B in A=-\frac{2}{3}B+2. Because the resulting equation contains only one variable, you can solve for A directly.

$$A=-\frac{2}{3}\left(-\frac{3}{2}\right)+2$$

Multiply -\frac{2}{3} times -\frac{3}{2} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$A=1+2$$

Add 2 to 1.

$$A=3$$

The system is now solved.

$$A=3,B=-\frac{3}{2}$$

Steps Using Matrices

Consider the first equation. Use the distributive property to multiply A+B by 3.

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

Combine 3B and -B to get 2B.

$$3A+2B=6$$

Consider the second equation. Calculate 3 to the power of 2 and get 9.

$$\left(2A+B\right)\times 9-B=42$$

Use the distributive property to multiply 2A+B by 9.

$$18A+9B-B=42$$

Combine 9B and -B to get 8B.

$$18A+8B=42$$

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

$$3A+2B=6,18A+8B=42$$

Write the equations in matrix form.

$$\left(\begin{matrix}3&2\\18&8\end{matrix}\right)\left(\begin{matrix}A\\B\end{matrix}\right)=\left(\begin{matrix}6\\42\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}3&2\\18&8\end{matrix}\right).

$$inverse(\left(\begin{matrix}3&2\\18&8\end{matrix}\right))\left(\begin{matrix}3&2\\18&8\end{matrix}\right)\left(\begin{matrix}A\\B\end{matrix}\right)=inverse(\left(\begin{matrix}3&2\\18&8\end{matrix}\right))\left(\begin{matrix}6\\42\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}A\\B\end{matrix}\right)=inverse(\left(\begin{matrix}3&2\\18&8\end{matrix}\right))\left(\begin{matrix}6\\42\end{matrix}\right)$$

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

$$\left(\begin{matrix}A\\B\end{matrix}\right)=inverse(\left(\begin{matrix}3&2\\18&8\end{matrix}\right))\left(\begin{matrix}6\\42\end{matrix}\right)$$

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.

$$\left(\begin{matrix}A\\B\end{matrix}\right)=\left(\begin{matrix}\frac{8}{3\times 8-2\times 18}&-\frac{2}{3\times 8-2\times 18}\\-\frac{18}{3\times 8-2\times 18}&\frac{3}{3\times 8-2\times 18}\end{matrix}\right)\left(\begin{matrix}6\\42\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}A\\B\end{matrix}\right)=\left(\begin{matrix}-\frac{2}{3}&\frac{1}{6}\\\frac{3}{2}&-\frac{1}{4}\end{matrix}\right)\left(\begin{matrix}6\\42\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}A\\B\end{matrix}\right)=\left(\begin{matrix}-\frac{2}{3}\times 6+\frac{1}{6}\times 42\\\frac{3}{2}\times 6-\frac{1}{4}\times 42\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}A\\B\end{matrix}\right)=\left(\begin{matrix}3\\-\frac{3}{2}\end{matrix}\right)$$

Extract the matrix elements A and B.

$$A=3,B=-\frac{3}{2}$$

Steps Using Elimination

Consider the first equation. Use the distributive property to multiply A+B by 3.

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

Combine 3B and -B to get 2B.

$$3A+2B=6$$

Consider the second equation. Calculate 3 to the power of 2 and get 9.

$$\left(2A+B\right)\times 9-B=42$$

Use the distributive property to multiply 2A+B by 9.

$$18A+9B-B=42$$

Combine 9B and -B to get 8B.

$$18A+8B=42$$

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.

$$3A+2B=6,18A+8B=42$$

To make 3A and 18A equal, multiply all terms on each side of the first equation by 18 and all terms on each side of the second by 3.

$$18\times 3A+18\times 2B=18\times 6,3\times 18A+3\times 8B=3\times 42$$

Simplify.

$$54A+36B=108,54A+24B=126$$

Subtract 54A+24B=126 from 54A+36B=108 by subtracting like terms on each side of the equal sign.

$$54A-54A+36B-24B=108-126$$

Add 54A to -54A. Terms 54A and -54A cancel out, leaving an equation with only one variable that can be solved.

$$36B-24B=108-126$$

Add 36B to -24B.

$$12B=108-126$$

Add 108 to -126.

$$12B=-18$$

Divide both sides by 12.

$$B=-\frac{3}{2}$$

Substitute -\frac{3}{2} for B in 18A+8B=42. Because the resulting equation contains only one variable, you can solve for A directly.

$$18A+8\left(-\frac{3}{2}\right)=42$$

Multiply 8 times -\frac{3}{2}.

$$18A-12=42$$

Add 12 to both sides of the equation.

$$18A=54$$

Divide both sides by 18.

$$A=3$$

The system is now solved.

$$A=3,B=-\frac{3}{2}$$