Solve {l}{11a+55d=132}{2a+13d=30}

$$\left\{ \begin{array} { l } { 11 a + 55 d = 132 } \\ { 2 a + 13 d = 30 } \end{array} \right.$$

Solve for a, d

$$a=2$$

$$d=2$$

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.

$$11a+55d=132,2a+13d=30$$

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

$$11a+55d=132$$

Subtract 55d from both sides of the equation.

$$11a=-55d+132$$

Divide both sides by 11.

$$a=\frac{1}{11}\left(-55d+132\right)$$

Multiply \frac{1}{11} times -55d+132.

$$a=-5d+12$$

Substitute -5d+12 for a in the other equation, 2a+13d=30.

$$2\left(-5d+12\right)+13d=30$$

Multiply 2 times -5d+12.

$$-10d+24+13d=30$$

Add -10d to 13d.

$$3d+24=30$$

Subtract 24 from both sides of the equation.

$$3d=6$$

Divide both sides by 3.

$$d=2$$

Substitute 2 for d in a=-5d+12. Because the resulting equation contains only one variable, you can solve for a directly.

$$a=-5\times 2+12$$

Multiply -5 times 2.

$$a=-10+12$$

Add 12 to -10.

$$a=2$$

The system is now solved.

$$a=2,d=2$$

Steps Using Matrices

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

$$11a+55d=132,2a+13d=30$$

Write the equations in matrix form.

$$\left(\begin{matrix}11&55\\2&13\end{matrix}\right)\left(\begin{matrix}a\\d\end{matrix}\right)=\left(\begin{matrix}132\\30\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}11&55\\2&13\end{matrix}\right).

$$inverse(\left(\begin{matrix}11&55\\2&13\end{matrix}\right))\left(\begin{matrix}11&55\\2&13\end{matrix}\right)\left(\begin{matrix}a\\d\end{matrix}\right)=inverse(\left(\begin{matrix}11&55\\2&13\end{matrix}\right))\left(\begin{matrix}132\\30\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}a\\d\end{matrix}\right)=inverse(\left(\begin{matrix}11&55\\2&13\end{matrix}\right))\left(\begin{matrix}132\\30\end{matrix}\right)$$

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

$$\left(\begin{matrix}a\\d\end{matrix}\right)=inverse(\left(\begin{matrix}11&55\\2&13\end{matrix}\right))\left(\begin{matrix}132\\30\end{matrix}\right)$$

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.

$$\left(\begin{matrix}a\\d\end{matrix}\right)=\left(\begin{matrix}\frac{13}{11\times 13-55\times 2}&-\frac{55}{11\times 13-55\times 2}\\-\frac{2}{11\times 13-55\times 2}&\frac{11}{11\times 13-55\times 2}\end{matrix}\right)\left(\begin{matrix}132\\30\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}a\\d\end{matrix}\right)=\left(\begin{matrix}\frac{13}{33}&-\frac{5}{3}\\-\frac{2}{33}&\frac{1}{3}\end{matrix}\right)\left(\begin{matrix}132\\30\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}a\\d\end{matrix}\right)=\left(\begin{matrix}\frac{13}{33}\times 132-\frac{5}{3}\times 30\\-\frac{2}{33}\times 132+\frac{1}{3}\times 30\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}a\\d\end{matrix}\right)=\left(\begin{matrix}2\\2\end{matrix}\right)$$

Extract the matrix elements a and d.

$$a=2,d=2$$

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.

$$11a+55d=132,2a+13d=30$$

To make 11a and 2a equal, multiply all terms on each side of the first equation by 2 and all terms on each side of the second by 11.

$$2\times 11a+2\times 55d=2\times 132,11\times 2a+11\times 13d=11\times 30$$

Simplify.

$$22a+110d=264,22a+143d=330$$

Subtract 22a+143d=330 from 22a+110d=264 by subtracting like terms on each side of the equal sign.

$$22a-22a+110d-143d=264-330$$

Add 22a to -22a. Terms 22a and -22a cancel out, leaving an equation with only one variable that can be solved.

$$110d-143d=264-330$$

Add 110d to -143d.

$$-33d=264-330$$

Add 264 to -330.

$$-33d=-66$$

Divide both sides by -33.

$$d=2$$

Substitute 2 for d in 2a+13d=30. Because the resulting equation contains only one variable, you can solve for a directly.

$$2a+13\times 2=30$$

Multiply 13 times 2.

$$2a+26=30$$

Subtract 26 from both sides of the equation.

$$2a=4$$

Divide both sides by 2.

$$a=2$$

The system is now solved.

$$a=2,d=2$$