Solve {l}{200=10k+b}{150=15k+b}

$$\left\{ \begin{array} { l } { 200 = 10 k + b } \\ { 150 = 15 k + b } \end{array} \right.$$

Solve for k, b

$$k=-10$$

$$b=300$$

Steps Using Substitution

Consider the first equation. Swap sides so that all variable terms are on the left hand side.

$$10k+b=200$$

Consider the second equation. Swap sides so that all variable terms are on the left hand side.

$$15k+b=150$$

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.

$$10k+b=200,15k+b=150$$

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

$$10k+b=200$$

Subtract b from both sides of the equation.

$$10k=-b+200$$

Divide both sides by 10.

$$k=\frac{1}{10}\left(-b+200\right)$$

Multiply \frac{1}{10} times -b+200.

$$k=-\frac{1}{10}b+20$$

Substitute -\frac{b}{10}+20 for k in the other equation, 15k+b=150.

$$15\left(-\frac{1}{10}b+20\right)+b=150$$

Multiply 15 times -\frac{b}{10}+20.

$$-\frac{3}{2}b+300+b=150$$

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

$$-\frac{1}{2}b+300=150$$

Subtract 300 from both sides of the equation.

$$-\frac{1}{2}b=-150$$

Multiply both sides by -2.

$$b=300$$

Substitute 300 for b in k=-\frac{1}{10}b+20. Because the resulting equation contains only one variable, you can solve for k directly.

$$k=-\frac{1}{10}\times 300+20$$

Multiply -\frac{1}{10} times 300.

$$k=-30+20$$

Add 20 to -30.

$$k=-10$$

The system is now solved.

$$k=-10,b=300$$

Steps Using Matrices

Consider the first equation. Swap sides so that all variable terms are on the left hand side.

$$10k+b=200$$

Consider the second equation. Swap sides so that all variable terms are on the left hand side.

$$15k+b=150$$

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

$$10k+b=200,15k+b=150$$

Write the equations in matrix form.

$$\left(\begin{matrix}10&1\\15&1\end{matrix}\right)\left(\begin{matrix}k\\b\end{matrix}\right)=\left(\begin{matrix}200\\150\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}10&1\\15&1\end{matrix}\right).

$$inverse(\left(\begin{matrix}10&1\\15&1\end{matrix}\right))\left(\begin{matrix}10&1\\15&1\end{matrix}\right)\left(\begin{matrix}k\\b\end{matrix}\right)=inverse(\left(\begin{matrix}10&1\\15&1\end{matrix}\right))\left(\begin{matrix}200\\150\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}k\\b\end{matrix}\right)=inverse(\left(\begin{matrix}10&1\\15&1\end{matrix}\right))\left(\begin{matrix}200\\150\end{matrix}\right)$$

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

$$\left(\begin{matrix}k\\b\end{matrix}\right)=inverse(\left(\begin{matrix}10&1\\15&1\end{matrix}\right))\left(\begin{matrix}200\\150\end{matrix}\right)$$

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.

$$\left(\begin{matrix}k\\b\end{matrix}\right)=\left(\begin{matrix}\frac{1}{10-15}&-\frac{1}{10-15}\\-\frac{15}{10-15}&\frac{10}{10-15}\end{matrix}\right)\left(\begin{matrix}200\\150\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}k\\b\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{5}&\frac{1}{5}\\3&-2\end{matrix}\right)\left(\begin{matrix}200\\150\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}k\\b\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{5}\times 200+\frac{1}{5}\times 150\\3\times 200-2\times 150\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}k\\b\end{matrix}\right)=\left(\begin{matrix}-10\\300\end{matrix}\right)$$

Extract the matrix elements k and b.

$$k=-10,b=300$$

Steps Using Elimination

Consider the first equation. Swap sides so that all variable terms are on the left hand side.

$$10k+b=200$$

Consider the second equation. Swap sides so that all variable terms are on the left hand side.

$$15k+b=150$$

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.

$$10k+b=200,15k+b=150$$

Subtract 15k+b=150 from 10k+b=200 by subtracting like terms on each side of the equal sign.

$$10k-15k+b-b=200-150$$

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

$$10k-15k=200-150$$

Add 10k to -15k.

$$-5k=200-150$$

Add 200 to -150.

$$-5k=50$$

Divide both sides by -5.

$$k=-10$$

Substitute -10 for k in 15k+b=150. Because the resulting equation contains only one variable, you can solve for b directly.

$$15\left(-10\right)+b=150$$

Multiply 15 times -10.

$$-150+b=150$$

Add 150 to both sides of the equation.

$$b=300$$

The system is now solved.

$$k=-10,b=300$$