Solve {l}{3c+4u=33}{6c+3u=36}

$$\left. \begin{array} { l } { 3 c + 4 u = 33 } \\ { 6 c + 3 u = 36 } \end{array} \right.$$

Solve for c, u

$$c=3$$

$$u=6$$

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.

$$3c+4u=33,6c+3u=36$$

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

$$3c+4u=33$$

Subtract 4u from both sides of the equation.

$$3c=-4u+33$$

Divide both sides by 3.

$$c=\frac{1}{3}\left(-4u+33\right)$$

Multiply \frac{1}{3} times -4u+33.

$$c=-\frac{4}{3}u+11$$

Substitute -\frac{4u}{3}+11 for c in the other equation, 6c+3u=36.

$$6\left(-\frac{4}{3}u+11\right)+3u=36$$

Multiply 6 times -\frac{4u}{3}+11.

$$-8u+66+3u=36$$

Add -8u to 3u.

$$-5u+66=36$$

Subtract 66 from both sides of the equation.

$$-5u=-30$$

Divide both sides by -5.

$$u=6$$

Substitute 6 for u in c=-\frac{4}{3}u+11. Because the resulting equation contains only one variable, you can solve for c directly.

$$c=-\frac{4}{3}\times 6+11$$

Multiply -\frac{4}{3} times 6.

$$c=-8+11$$

Add 11 to -8.

$$c=3$$

The system is now solved.

$$c=3,u=6$$

Steps Using Matrices

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

$$3c+4u=33,6c+3u=36$$

Write the equations in matrix form.

$$\left(\begin{matrix}3&4\\6&3\end{matrix}\right)\left(\begin{matrix}c\\u\end{matrix}\right)=\left(\begin{matrix}33\\36\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}3&4\\6&3\end{matrix}\right).

$$inverse(\left(\begin{matrix}3&4\\6&3\end{matrix}\right))\left(\begin{matrix}3&4\\6&3\end{matrix}\right)\left(\begin{matrix}c\\u\end{matrix}\right)=inverse(\left(\begin{matrix}3&4\\6&3\end{matrix}\right))\left(\begin{matrix}33\\36\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}c\\u\end{matrix}\right)=inverse(\left(\begin{matrix}3&4\\6&3\end{matrix}\right))\left(\begin{matrix}33\\36\end{matrix}\right)$$

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

$$\left(\begin{matrix}c\\u\end{matrix}\right)=inverse(\left(\begin{matrix}3&4\\6&3\end{matrix}\right))\left(\begin{matrix}33\\36\end{matrix}\right)$$

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.

$$\left(\begin{matrix}c\\u\end{matrix}\right)=\left(\begin{matrix}\frac{3}{3\times 3-4\times 6}&-\frac{4}{3\times 3-4\times 6}\\-\frac{6}{3\times 3-4\times 6}&\frac{3}{3\times 3-4\times 6}\end{matrix}\right)\left(\begin{matrix}33\\36\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}c\\u\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{5}&\frac{4}{15}\\\frac{2}{5}&-\frac{1}{5}\end{matrix}\right)\left(\begin{matrix}33\\36\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}c\\u\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{5}\times 33+\frac{4}{15}\times 36\\\frac{2}{5}\times 33-\frac{1}{5}\times 36\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}c\\u\end{matrix}\right)=\left(\begin{matrix}3\\6\end{matrix}\right)$$

Extract the matrix elements c and u.

$$c=3,u=6$$

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.

$$3c+4u=33,6c+3u=36$$

To make 3c and 6c equal, multiply all terms on each side of the first equation by 6 and all terms on each side of the second by 3.

$$6\times 3c+6\times 4u=6\times 33,3\times 6c+3\times 3u=3\times 36$$

Simplify.

$$18c+24u=198,18c+9u=108$$

Subtract 18c+9u=108 from 18c+24u=198 by subtracting like terms on each side of the equal sign.

$$18c-18c+24u-9u=198-108$$

Add 18c to -18c. Terms 18c and -18c cancel out, leaving an equation with only one variable that can be solved.

$$24u-9u=198-108$$

Add 24u to -9u.

$$15u=198-108$$

Add 198 to -108.

$$15u=90$$

Divide both sides by 15.

$$u=6$$

Substitute 6 for u in 6c+3u=36. Because the resulting equation contains only one variable, you can solve for c directly.

$$6c+3\times 6=36$$

Multiply 3 times 6.

$$6c+18=36$$

Subtract 18 from both sides of the equation.

$$6c=18$$

Divide both sides by 6.

$$c=3$$

The system is now solved.

$$c=3,u=6$$