Solve {l}{2=3k_{2}+k_{1}}{10=4k_{2}-2k_{1}}

$$\left\{ \begin{array} { l } { 2 = 3 k _ { 2 } + k _ { 1 } } \\ { 10 = 4 k _ { 2 } - 2 k _ { 1 } } \end{array} \right.$$

Solve for k_2, k_1

$$k_{2} = \frac{7}{5} = 1\frac{2}{5} = 1.4$$

$$k_{1} = -\frac{11}{5} = -2\frac{1}{5} = -2.2$$

Steps Using Substitution

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

$$3k_{2}+k_{1}=2$$

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

$$4k_{2}-2k_{1}=10$$

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.

$$3k_{2}+k_{1}=2,4k_{2}-2k_{1}=10$$

Choose one of the equations and solve it for k_{2} by isolating k_{2} on the left hand side of the equal sign.

$$3k_{2}+k_{1}=2$$

Subtract k_{1} from both sides of the equation.

$$3k_{2}=-k_{1}+2$$

Divide both sides by 3.

$$k_{2}=\frac{1}{3}\left(-k_{1}+2\right)$$

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

$$k_{2}=-\frac{1}{3}k_{1}+\frac{2}{3}$$

Substitute \frac{-k_{1}+2}{3} for k_{2} in the other equation, 4k_{2}-2k_{1}=10.

$$4\left(-\frac{1}{3}k_{1}+\frac{2}{3}\right)-2k_{1}=10$$

Multiply 4 times \frac{-k_{1}+2}{3}.

$$-\frac{4}{3}k_{1}+\frac{8}{3}-2k_{1}=10$$

Add -\frac{4k_{1}}{3} to -2k_{1}.

$$-\frac{10}{3}k_{1}+\frac{8}{3}=10$$

Subtract \frac{8}{3} from both sides of the equation.

$$-\frac{10}{3}k_{1}=\frac{22}{3}$$

Divide both sides of the equation by -\frac{10}{3}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$k_{1}=-\frac{11}{5}$$

Substitute -\frac{11}{5} for k_{1} in k_{2}=-\frac{1}{3}k_{1}+\frac{2}{3}. Because the resulting equation contains only one variable, you can solve for k_{2} directly.

$$k_{2}=-\frac{1}{3}\left(-\frac{11}{5}\right)+\frac{2}{3}$$

Multiply -\frac{1}{3} times -\frac{11}{5} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$k_{2}=\frac{11}{15}+\frac{2}{3}$$

Add \frac{2}{3} to \frac{11}{15} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$k_{2}=\frac{7}{5}$$

The system is now solved.

$$k_{2}=\frac{7}{5},k_{1}=-\frac{11}{5}$$

Steps Using Matrices

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

$$3k_{2}+k_{1}=2$$

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

$$4k_{2}-2k_{1}=10$$

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

$$3k_{2}+k_{1}=2,4k_{2}-2k_{1}=10$$

Write the equations in matrix form.

$$\left(\begin{matrix}3&1\\4&-2\end{matrix}\right)\left(\begin{matrix}k_{2}\\k_{1}\end{matrix}\right)=\left(\begin{matrix}2\\10\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}3&1\\4&-2\end{matrix}\right).

$$inverse(\left(\begin{matrix}3&1\\4&-2\end{matrix}\right))\left(\begin{matrix}3&1\\4&-2\end{matrix}\right)\left(\begin{matrix}k_{2}\\k_{1}\end{matrix}\right)=inverse(\left(\begin{matrix}3&1\\4&-2\end{matrix}\right))\left(\begin{matrix}2\\10\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}k_{2}\\k_{1}\end{matrix}\right)=inverse(\left(\begin{matrix}3&1\\4&-2\end{matrix}\right))\left(\begin{matrix}2\\10\end{matrix}\right)$$

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

$$\left(\begin{matrix}k_{2}\\k_{1}\end{matrix}\right)=inverse(\left(\begin{matrix}3&1\\4&-2\end{matrix}\right))\left(\begin{matrix}2\\10\end{matrix}\right)$$

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.

$$\left(\begin{matrix}k_{2}\\k_{1}\end{matrix}\right)=\left(\begin{matrix}-\frac{2}{3\left(-2\right)-4}&-\frac{1}{3\left(-2\right)-4}\\-\frac{4}{3\left(-2\right)-4}&\frac{3}{3\left(-2\right)-4}\end{matrix}\right)\left(\begin{matrix}2\\10\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}k_{2}\\k_{1}\end{matrix}\right)=\left(\begin{matrix}\frac{1}{5}&\frac{1}{10}\\\frac{2}{5}&-\frac{3}{10}\end{matrix}\right)\left(\begin{matrix}2\\10\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}k_{2}\\k_{1}\end{matrix}\right)=\left(\begin{matrix}\frac{1}{5}\times 2+\frac{1}{10}\times 10\\\frac{2}{5}\times 2-\frac{3}{10}\times 10\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}k_{2}\\k_{1}\end{matrix}\right)=\left(\begin{matrix}\frac{7}{5}\\-\frac{11}{5}\end{matrix}\right)$$

Extract the matrix elements k_{2} and k_{1}.

$$k_{2}=\frac{7}{5},k_{1}=-\frac{11}{5}$$

Steps Using Elimination

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

$$3k_{2}+k_{1}=2$$

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

$$4k_{2}-2k_{1}=10$$

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.

$$3k_{2}+k_{1}=2,4k_{2}-2k_{1}=10$$

To make 3k_{2} and 4k_{2} equal, multiply all terms on each side of the first equation by 4 and all terms on each side of the second by 3.

$$4\times 3k_{2}+4k_{1}=4\times 2,3\times 4k_{2}+3\left(-2\right)k_{1}=3\times 10$$

Simplify.

$$12k_{2}+4k_{1}=8,12k_{2}-6k_{1}=30$$

Subtract 12k_{2}-6k_{1}=30 from 12k_{2}+4k_{1}=8 by subtracting like terms on each side of the equal sign.

$$12k_{2}-12k_{2}+4k_{1}+6k_{1}=8-30$$

Add 12k_{2} to -12k_{2}. Terms 12k_{2} and -12k_{2} cancel out, leaving an equation with only one variable that can be solved.

$$4k_{1}+6k_{1}=8-30$$

Add 4k_{1} to 6k_{1}.

$$10k_{1}=8-30$$

Add 8 to -30.

$$10k_{1}=-22$$

Divide both sides by 10.

$$k_{1}=-\frac{11}{5}$$

Substitute -\frac{11}{5} for k_{1} in 4k_{2}-2k_{1}=10. Because the resulting equation contains only one variable, you can solve for k_{2} directly.

$$4k_{2}-2\left(-\frac{11}{5}\right)=10$$

Multiply -2 times -\frac{11}{5}.

$$4k_{2}+\frac{22}{5}=10$$

Subtract \frac{22}{5} from both sides of the equation.

$$4k_{2}=\frac{28}{5}$$

Divide both sides by 4.

$$k_{2}=\frac{7}{5}$$

The system is now solved.

$$k_{2}=\frac{7}{5},k_{1}=-\frac{11}{5}$$