Solve {l}{3s+2c=57}{3s+12c=147}

$$\left. \begin{array} { l } { 3 s + 2 c = 57 } \\ { 3 s + 12 c = 147 } \end{array} \right.$$

Solve for s, c

$$s=13$$

$$c=9$$

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.

$$3s+2c=57,3s+12c=147$$

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

$$3s+2c=57$$

Subtract 2c from both sides of the equation.

$$3s=-2c+57$$

Divide both sides by 3.

$$s=\frac{1}{3}\left(-2c+57\right)$$

Multiply \frac{1}{3} times -2c+57.

$$s=-\frac{2}{3}c+19$$

Substitute -\frac{2c}{3}+19 for s in the other equation, 3s+12c=147.

$$3\left(-\frac{2}{3}c+19\right)+12c=147$$

Multiply 3 times -\frac{2c}{3}+19.

$$-2c+57+12c=147$$

Add -2c to 12c.

$$10c+57=147$$

Subtract 57 from both sides of the equation.

$$10c=90$$

Divide both sides by 10.

$$c=9$$

Substitute 9 for c in s=-\frac{2}{3}c+19. Because the resulting equation contains only one variable, you can solve for s directly.

$$s=-\frac{2}{3}\times 9+19$$

Multiply -\frac{2}{3} times 9.

$$s=-6+19$$

Add 19 to -6.

$$s=13$$

The system is now solved.

$$s=13,c=9$$

Steps Using Matrices

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

$$3s+2c=57,3s+12c=147$$

Write the equations in matrix form.

$$\left(\begin{matrix}3&2\\3&12\end{matrix}\right)\left(\begin{matrix}s\\c\end{matrix}\right)=\left(\begin{matrix}57\\147\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}3&2\\3&12\end{matrix}\right).

$$inverse(\left(\begin{matrix}3&2\\3&12\end{matrix}\right))\left(\begin{matrix}3&2\\3&12\end{matrix}\right)\left(\begin{matrix}s\\c\end{matrix}\right)=inverse(\left(\begin{matrix}3&2\\3&12\end{matrix}\right))\left(\begin{matrix}57\\147\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}s\\c\end{matrix}\right)=inverse(\left(\begin{matrix}3&2\\3&12\end{matrix}\right))\left(\begin{matrix}57\\147\end{matrix}\right)$$

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

$$\left(\begin{matrix}s\\c\end{matrix}\right)=inverse(\left(\begin{matrix}3&2\\3&12\end{matrix}\right))\left(\begin{matrix}57\\147\end{matrix}\right)$$

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.

$$\left(\begin{matrix}s\\c\end{matrix}\right)=\left(\begin{matrix}\frac{12}{3\times 12-2\times 3}&-\frac{2}{3\times 12-2\times 3}\\-\frac{3}{3\times 12-2\times 3}&\frac{3}{3\times 12-2\times 3}\end{matrix}\right)\left(\begin{matrix}57\\147\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}s\\c\end{matrix}\right)=\left(\begin{matrix}\frac{2}{5}&-\frac{1}{15}\\-\frac{1}{10}&\frac{1}{10}\end{matrix}\right)\left(\begin{matrix}57\\147\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}s\\c\end{matrix}\right)=\left(\begin{matrix}\frac{2}{5}\times 57-\frac{1}{15}\times 147\\-\frac{1}{10}\times 57+\frac{1}{10}\times 147\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}s\\c\end{matrix}\right)=\left(\begin{matrix}13\\9\end{matrix}\right)$$

Extract the matrix elements s and c.

$$s=13,c=9$$

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.

$$3s+2c=57,3s+12c=147$$

Subtract 3s+12c=147 from 3s+2c=57 by subtracting like terms on each side of the equal sign.

$$3s-3s+2c-12c=57-147$$

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

$$2c-12c=57-147$$

Add 2c to -12c.

$$-10c=57-147$$

Add 57 to -147.

$$-10c=-90$$

Divide both sides by -10.

$$c=9$$

Substitute 9 for c in 3s+12c=147. Because the resulting equation contains only one variable, you can solve for s directly.

$$3s+12\times 9=147$$

Multiply 12 times 9.

$$3s+108=147$$

Subtract 108 from both sides of the equation.

$$3s=39$$

Divide both sides by 3.

$$s=13$$

The system is now solved.

$$s=13,c=9$$