Solve {l}{4=k_{1}+k_{2}}{5=3k_{1}+frac{k_{2}}{5}}

$$\left\{ \begin{array} { l } { 4 = k _ { 1 } + k _ { 2 } } \\ { 5 = 3 k _ { 1 } + \frac { k _ { 2 } } { 5 } } \end{array} \right.$$

Solve for k_1, k_2

$$k_{1} = \frac{3}{2} = 1\frac{1}{2} = 1.5$$

$$k_{2} = \frac{5}{2} = 2\frac{1}{2} = 2.5$$

Steps Using Substitution

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

$$k_{1}+k_{2}=4$$

Consider the second equation. Multiply both sides of the equation by 5.

$$25=15k_{1}+k_{2}$$

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

$$15k_{1}+k_{2}=25$$

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.

$$k_{1}+k_{2}=4,15k_{1}+k_{2}=25$$

Choose one of the equations and solve it for k_{1} by isolating k_{1} on the left hand side of the equal sign.

$$k_{1}+k_{2}=4$$

Subtract k_{2} from both sides of the equation.

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

Substitute -k_{2}+4 for k_{1} in the other equation, 15k_{1}+k_{2}=25.

$$15\left(-k_{2}+4\right)+k_{2}=25$$

Multiply 15 times -k_{2}+4.

$$-15k_{2}+60+k_{2}=25$$

Add -15k_{2} to k_{2}.

$$-14k_{2}+60=25$$

Subtract 60 from both sides of the equation.

$$-14k_{2}=-35$$

Divide both sides by -14.

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

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

$$k_{1}=-\frac{5}{2}+4$$

Add 4 to -\frac{5}{2}.

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

The system is now solved.

$$k_{1}=\frac{3}{2},k_{2}=\frac{5}{2}$$

Steps Using Matrices

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

$$k_{1}+k_{2}=4$$

Consider the second equation. Multiply both sides of the equation by 5.

$$25=15k_{1}+k_{2}$$

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

$$15k_{1}+k_{2}=25$$

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

$$k_{1}+k_{2}=4,15k_{1}+k_{2}=25$$

Write the equations in matrix form.

$$\left(\begin{matrix}1&1\\15&1\end{matrix}\right)\left(\begin{matrix}k_{1}\\k_{2}\end{matrix}\right)=\left(\begin{matrix}4\\25\end{matrix}\right)$$

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

$$inverse(\left(\begin{matrix}1&1\\15&1\end{matrix}\right))\left(\begin{matrix}1&1\\15&1\end{matrix}\right)\left(\begin{matrix}k_{1}\\k_{2}\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\15&1\end{matrix}\right))\left(\begin{matrix}4\\25\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}k_{1}\\k_{2}\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\15&1\end{matrix}\right))\left(\begin{matrix}4\\25\end{matrix}\right)$$

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

$$\left(\begin{matrix}k_{1}\\k_{2}\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\15&1\end{matrix}\right))\left(\begin{matrix}4\\25\end{matrix}\right)$$

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.

$$\left(\begin{matrix}k_{1}\\k_{2}\end{matrix}\right)=\left(\begin{matrix}\frac{1}{1-15}&-\frac{1}{1-15}\\-\frac{15}{1-15}&\frac{1}{1-15}\end{matrix}\right)\left(\begin{matrix}4\\25\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}k_{1}\\k_{2}\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{14}&\frac{1}{14}\\\frac{15}{14}&-\frac{1}{14}\end{matrix}\right)\left(\begin{matrix}4\\25\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}k_{1}\\k_{2}\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{14}\times 4+\frac{1}{14}\times 25\\\frac{15}{14}\times 4-\frac{1}{14}\times 25\end{matrix}\right)$$

Do the arithmetic.

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

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

$$k_{1}=\frac{3}{2},k_{2}=\frac{5}{2}$$

Steps Using Elimination

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

$$k_{1}+k_{2}=4$$

Consider the second equation. Multiply both sides of the equation by 5.

$$25=15k_{1}+k_{2}$$

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

$$15k_{1}+k_{2}=25$$

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.

$$k_{1}+k_{2}=4,15k_{1}+k_{2}=25$$

Subtract 15k_{1}+k_{2}=25 from k_{1}+k_{2}=4 by subtracting like terms on each side of the equal sign.

$$k_{1}-15k_{1}+k_{2}-k_{2}=4-25$$

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

$$k_{1}-15k_{1}=4-25$$

Add k_{1} to -15k_{1}.

$$-14k_{1}=4-25$$

Add 4 to -25.

$$-14k_{1}=-21$$

Divide both sides by -14.

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

Substitute \frac{3}{2} for k_{1} in 15k_{1}+k_{2}=25. Because the resulting equation contains only one variable, you can solve for k_{2} directly.

$$15\times \frac{3}{2}+k_{2}=25$$

Multiply 15 times \frac{3}{2}.

$$\frac{45}{2}+k_{2}=25$$

Subtract \frac{45}{2} from both sides of the equation.

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

The system is now solved.

$$k_{1}=\frac{3}{2},k_{2}=\frac{5}{2}$$