Solve {l}{25c+22T=152000}{11c+12T=75000}

$$\left. \begin{array} { l } { 25 c + 22 T = 152000 } \\ { 11 c + 12 T = 75000 } \end{array} \right.$$

Solve for c, T

$$c=3000$$

$$T=3500$$

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.

$$25c+22T=152000,11c+12T=75000$$

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

$$25c+22T=152000$$

Subtract 22T from both sides of the equation.

$$25c=-22T+152000$$

Divide both sides by 25.

$$c=\frac{1}{25}\left(-22T+152000\right)$$

Multiply \frac{1}{25} times -22T+152000.

$$c=-\frac{22}{25}T+6080$$

Substitute -\frac{22T}{25}+6080 for c in the other equation, 11c+12T=75000.

$$11\left(-\frac{22}{25}T+6080\right)+12T=75000$$

Multiply 11 times -\frac{22T}{25}+6080.

$$-\frac{242}{25}T+66880+12T=75000$$

Add -\frac{242T}{25} to 12T.

$$\frac{58}{25}T+66880=75000$$

Subtract 66880 from both sides of the equation.

$$\frac{58}{25}T=8120$$

Divide both sides of the equation by \frac{58}{25}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$T=3500$$

Substitute 3500 for T in c=-\frac{22}{25}T+6080. Because the resulting equation contains only one variable, you can solve for c directly.

$$c=-\frac{22}{25}\times 3500+6080$$

Multiply -\frac{22}{25} times 3500.

$$c=-3080+6080$$

Add 6080 to -3080.

$$c=3000$$

The system is now solved.

$$c=3000,T=3500$$

Steps Using Matrices

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

$$25c+22T=152000,11c+12T=75000$$

Write the equations in matrix form.

$$\left(\begin{matrix}25&22\\11&12\end{matrix}\right)\left(\begin{matrix}c\\T\end{matrix}\right)=\left(\begin{matrix}152000\\75000\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}25&22\\11&12\end{matrix}\right).

$$inverse(\left(\begin{matrix}25&22\\11&12\end{matrix}\right))\left(\begin{matrix}25&22\\11&12\end{matrix}\right)\left(\begin{matrix}c\\T\end{matrix}\right)=inverse(\left(\begin{matrix}25&22\\11&12\end{matrix}\right))\left(\begin{matrix}152000\\75000\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}c\\T\end{matrix}\right)=inverse(\left(\begin{matrix}25&22\\11&12\end{matrix}\right))\left(\begin{matrix}152000\\75000\end{matrix}\right)$$

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

$$\left(\begin{matrix}c\\T\end{matrix}\right)=inverse(\left(\begin{matrix}25&22\\11&12\end{matrix}\right))\left(\begin{matrix}152000\\75000\end{matrix}\right)$$

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.

$$\left(\begin{matrix}c\\T\end{matrix}\right)=\left(\begin{matrix}\frac{12}{25\times 12-22\times 11}&-\frac{22}{25\times 12-22\times 11}\\-\frac{11}{25\times 12-22\times 11}&\frac{25}{25\times 12-22\times 11}\end{matrix}\right)\left(\begin{matrix}152000\\75000\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}c\\T\end{matrix}\right)=\left(\begin{matrix}\frac{6}{29}&-\frac{11}{29}\\-\frac{11}{58}&\frac{25}{58}\end{matrix}\right)\left(\begin{matrix}152000\\75000\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}c\\T\end{matrix}\right)=\left(\begin{matrix}\frac{6}{29}\times 152000-\frac{11}{29}\times 75000\\-\frac{11}{58}\times 152000+\frac{25}{58}\times 75000\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}c\\T\end{matrix}\right)=\left(\begin{matrix}3000\\3500\end{matrix}\right)$$

Extract the matrix elements c and T.

$$c=3000,T=3500$$

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.

$$25c+22T=152000,11c+12T=75000$$

To make 25c and 11c equal, multiply all terms on each side of the first equation by 11 and all terms on each side of the second by 25.

$$11\times 25c+11\times 22T=11\times 152000,25\times 11c+25\times 12T=25\times 75000$$

Simplify.

$$275c+242T=1672000,275c+300T=1875000$$

Subtract 275c+300T=1875000 from 275c+242T=1672000 by subtracting like terms on each side of the equal sign.

$$275c-275c+242T-300T=1672000-1875000$$

Add 275c to -275c. Terms 275c and -275c cancel out, leaving an equation with only one variable that can be solved.

$$242T-300T=1672000-1875000$$

Add 242T to -300T.

$$-58T=1672000-1875000$$

Add 1672000 to -1875000.

$$-58T=-203000$$

Divide both sides by -58.

$$T=3500$$

Substitute 3500 for T in 11c+12T=75000. Because the resulting equation contains only one variable, you can solve for c directly.

$$11c+12\times 3500=75000$$

Multiply 12 times 3500.

$$11c+42000=75000$$

Subtract 42000 from both sides of the equation.

$$11c=33000$$

Divide both sides by 11.

$$c=3000$$

The system is now solved.

$$c=3000,T=3500$$