Solve {l}{a+3k=500}{22a+k=600}

$$\left. \begin{array}{l}{ a + 3 k = 500 }\\{ 22 a + k = 600 }\end{array} \right.$$

Solve for a, k

$$a=20$$

$$k=160$$

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.

$$a+3k=500,22a+k=600$$

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

$$a+3k=500$$

Subtract 3k from both sides of the equation.

$$a=-3k+500$$

Substitute -3k+500 for a in the other equation, 22a+k=600.

$$22\left(-3k+500\right)+k=600$$

Multiply 22 times -3k+500.

$$-66k+11000+k=600$$

Add -66k to k.

$$-65k+11000=600$$

Subtract 11000 from both sides of the equation.

$$-65k=-10400$$

Divide both sides by -65.

$$k=160$$

Substitute 160 for k in a=-3k+500. Because the resulting equation contains only one variable, you can solve for a directly.

$$a=-3\times 160+500$$

Multiply -3 times 160.

$$a=-480+500$$

Add 500 to -480.

$$a=20$$

The system is now solved.

$$a=20,k=160$$

Steps Using Matrices

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

$$a+3k=500,22a+k=600$$

Write the equations in matrix form.

$$\left(\begin{matrix}1&3\\22&1\end{matrix}\right)\left(\begin{matrix}a\\k\end{matrix}\right)=\left(\begin{matrix}500\\600\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&3\\22&1\end{matrix}\right).

$$inverse(\left(\begin{matrix}1&3\\22&1\end{matrix}\right))\left(\begin{matrix}1&3\\22&1\end{matrix}\right)\left(\begin{matrix}a\\k\end{matrix}\right)=inverse(\left(\begin{matrix}1&3\\22&1\end{matrix}\right))\left(\begin{matrix}500\\600\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}a\\k\end{matrix}\right)=inverse(\left(\begin{matrix}1&3\\22&1\end{matrix}\right))\left(\begin{matrix}500\\600\end{matrix}\right)$$

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

$$\left(\begin{matrix}a\\k\end{matrix}\right)=inverse(\left(\begin{matrix}1&3\\22&1\end{matrix}\right))\left(\begin{matrix}500\\600\end{matrix}\right)$$

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.

$$\left(\begin{matrix}a\\k\end{matrix}\right)=\left(\begin{matrix}\frac{1}{1-3\times 22}&-\frac{3}{1-3\times 22}\\-\frac{22}{1-3\times 22}&\frac{1}{1-3\times 22}\end{matrix}\right)\left(\begin{matrix}500\\600\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}a\\k\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{65}&\frac{3}{65}\\\frac{22}{65}&-\frac{1}{65}\end{matrix}\right)\left(\begin{matrix}500\\600\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}a\\k\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{65}\times 500+\frac{3}{65}\times 600\\\frac{22}{65}\times 500-\frac{1}{65}\times 600\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}a\\k\end{matrix}\right)=\left(\begin{matrix}20\\160\end{matrix}\right)$$

Extract the matrix elements a and k.

$$a=20,k=160$$

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.

$$a+3k=500,22a+k=600$$

To make a and 22a equal, multiply all terms on each side of the first equation by 22 and all terms on each side of the second by 1.

$$22a+22\times 3k=22\times 500,22a+k=600$$

Simplify.

$$22a+66k=11000,22a+k=600$$

Subtract 22a+k=600 from 22a+66k=11000 by subtracting like terms on each side of the equal sign.

$$22a-22a+66k-k=11000-600$$

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

$$66k-k=11000-600$$

Add 66k to -k.

$$65k=11000-600$$

Add 11000 to -600.

$$65k=10400$$

Divide both sides by 65.

$$k=160$$

Substitute 160 for k in 22a+k=600. Because the resulting equation contains only one variable, you can solve for a directly.

$$22a+160=600$$

Subtract 160 from both sides of the equation.

$$22a=440$$

Divide both sides by 22.

$$a=20$$

The system is now solved.

$$a=20,k=160$$