Solve {l}{3w-2u=12}{w=u+60}

$$\left. \begin{array}{l}{ 3 w - 2 u = 12 }\\{ w = u + 60 }\end{array} \right.$$

Solve for w, u

$$w=-108$$

$$u=-168$$

Steps Using Substitution

Consider the second equation. Subtract u from both sides.

$$w-u=60$$

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.

$$3w-2u=12,w-u=60$$

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

$$3w-2u=12$$

Add 2u to both sides of the equation.

$$3w=2u+12$$

Divide both sides by 3.

$$w=\frac{1}{3}\left(2u+12\right)$$

Multiply \frac{1}{3} times 12+2u.

$$w=\frac{2}{3}u+4$$

Substitute \frac{2u}{3}+4 for w in the other equation, w-u=60.

$$\frac{2}{3}u+4-u=60$$

Add \frac{2u}{3} to -u.

$$-\frac{1}{3}u+4=60$$

Subtract 4 from both sides of the equation.

$$-\frac{1}{3}u=56$$

Multiply both sides by -3.

$$u=-168$$

Substitute -168 for u in w=\frac{2}{3}u+4. Because the resulting equation contains only one variable, you can solve for w directly.

$$w=\frac{2}{3}\left(-168\right)+4$$

Multiply \frac{2}{3} times -168.

$$w=-112+4$$

Add 4 to -112.

$$w=-108$$

The system is now solved.

$$w=-108,u=-168$$

Steps Using Matrices

Consider the second equation. Subtract u from both sides.

$$w-u=60$$

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

$$3w-2u=12,w-u=60$$

Write the equations in matrix form.

$$\left(\begin{matrix}3&-2\\1&-1\end{matrix}\right)\left(\begin{matrix}w\\u\end{matrix}\right)=\left(\begin{matrix}12\\60\end{matrix}\right)$$

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

$$inverse(\left(\begin{matrix}3&-2\\1&-1\end{matrix}\right))\left(\begin{matrix}3&-2\\1&-1\end{matrix}\right)\left(\begin{matrix}w\\u\end{matrix}\right)=inverse(\left(\begin{matrix}3&-2\\1&-1\end{matrix}\right))\left(\begin{matrix}12\\60\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}w\\u\end{matrix}\right)=inverse(\left(\begin{matrix}3&-2\\1&-1\end{matrix}\right))\left(\begin{matrix}12\\60\end{matrix}\right)$$

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

$$\left(\begin{matrix}w\\u\end{matrix}\right)=inverse(\left(\begin{matrix}3&-2\\1&-1\end{matrix}\right))\left(\begin{matrix}12\\60\end{matrix}\right)$$

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.

$$\left(\begin{matrix}w\\u\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{3\left(-1\right)-\left(-2\right)}&-\frac{-2}{3\left(-1\right)-\left(-2\right)}\\-\frac{1}{3\left(-1\right)-\left(-2\right)}&\frac{3}{3\left(-1\right)-\left(-2\right)}\end{matrix}\right)\left(\begin{matrix}12\\60\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}w\\u\end{matrix}\right)=\left(\begin{matrix}1&-2\\1&-3\end{matrix}\right)\left(\begin{matrix}12\\60\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}w\\u\end{matrix}\right)=\left(\begin{matrix}12-2\times 60\\12-3\times 60\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}w\\u\end{matrix}\right)=\left(\begin{matrix}-108\\-168\end{matrix}\right)$$

Extract the matrix elements w and u.

$$w=-108,u=-168$$

Steps Using Elimination

Consider the second equation. Subtract u from both sides.

$$w-u=60$$

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.

$$3w-2u=12,w-u=60$$

To make 3w and w equal, multiply all terms on each side of the first equation by 1 and all terms on each side of the second by 3.

$$3w-2u=12,3w+3\left(-1\right)u=3\times 60$$

Simplify.

$$3w-2u=12,3w-3u=180$$

Subtract 3w-3u=180 from 3w-2u=12 by subtracting like terms on each side of the equal sign.

$$3w-3w-2u+3u=12-180$$

Add 3w to -3w. Terms 3w and -3w cancel out, leaving an equation with only one variable that can be solved.

$$-2u+3u=12-180$$

Add -2u to 3u.

$$u=12-180$$

Add 12 to -180.

$$u=-168$$

Substitute -168 for u in w-u=60. Because the resulting equation contains only one variable, you can solve for w directly.

$$w-\left(-168\right)=60$$

Subtract 168 from both sides of the equation.

$$w=-108$$

The system is now solved.

$$w=-108,u=-168$$