Solve {l}{9k_{1}+8k_{2}=59}{25k_{1}+k_{2}=79}

$$\left. \begin{array} { l } { 9 k _ { 1 } + 8 k _ { 2 } = 59 } \\ { 25 k _ { 1 } + k _ { 2 } = 79 } \end{array} \right.$$

Solve for k_1, k_2

$$k_{1}=3$$

$$k_{2}=4$$

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.

$$9k_{1}+8k_{2}=59,25k_{1}+k_{2}=79$$

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

$$9k_{1}+8k_{2}=59$$

Subtract 8k_{2} from both sides of the equation.

$$9k_{1}=-8k_{2}+59$$

Divide both sides by 9.

$$k_{1}=\frac{1}{9}\left(-8k_{2}+59\right)$$

Multiply \frac{1}{9} times -8k_{2}+59.

$$k_{1}=-\frac{8}{9}k_{2}+\frac{59}{9}$$

Substitute \frac{-8k_{2}+59}{9} for k_{1} in the other equation, 25k_{1}+k_{2}=79.

$$25\left(-\frac{8}{9}k_{2}+\frac{59}{9}\right)+k_{2}=79$$

Multiply 25 times \frac{-8k_{2}+59}{9}.

$$-\frac{200}{9}k_{2}+\frac{1475}{9}+k_{2}=79$$

Add -\frac{200k_{2}}{9} to k_{2}.

$$-\frac{191}{9}k_{2}+\frac{1475}{9}=79$$

Subtract \frac{1475}{9} from both sides of the equation.

$$-\frac{191}{9}k_{2}=-\frac{764}{9}$$

Divide both sides of the equation by -\frac{191}{9}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$k_{2}=4$$

Substitute 4 for k_{2} in k_{1}=-\frac{8}{9}k_{2}+\frac{59}{9}. Because the resulting equation contains only one variable, you can solve for k_{1} directly.

$$k_{1}=-\frac{8}{9}\times 4+\frac{59}{9}$$

Multiply -\frac{8}{9} times 4.

$$k_{1}=\frac{-32+59}{9}$$

Add \frac{59}{9} to -\frac{32}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$k_{1}=3$$

The system is now solved.

$$k_{1}=3,k_{2}=4$$

Steps Using Matrices

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

$$9k_{1}+8k_{2}=59,25k_{1}+k_{2}=79$$

Write the equations in matrix form.

$$\left(\begin{matrix}9&8\\25&1\end{matrix}\right)\left(\begin{matrix}k_{1}\\k_{2}\end{matrix}\right)=\left(\begin{matrix}59\\79\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}9&8\\25&1\end{matrix}\right).

$$inverse(\left(\begin{matrix}9&8\\25&1\end{matrix}\right))\left(\begin{matrix}9&8\\25&1\end{matrix}\right)\left(\begin{matrix}k_{1}\\k_{2}\end{matrix}\right)=inverse(\left(\begin{matrix}9&8\\25&1\end{matrix}\right))\left(\begin{matrix}59\\79\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}k_{1}\\k_{2}\end{matrix}\right)=inverse(\left(\begin{matrix}9&8\\25&1\end{matrix}\right))\left(\begin{matrix}59\\79\end{matrix}\right)$$

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

$$\left(\begin{matrix}k_{1}\\k_{2}\end{matrix}\right)=inverse(\left(\begin{matrix}9&8\\25&1\end{matrix}\right))\left(\begin{matrix}59\\79\end{matrix}\right)$$

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.

$$\left(\begin{matrix}k_{1}\\k_{2}\end{matrix}\right)=\left(\begin{matrix}\frac{1}{9-8\times 25}&-\frac{8}{9-8\times 25}\\-\frac{25}{9-8\times 25}&\frac{9}{9-8\times 25}\end{matrix}\right)\left(\begin{matrix}59\\79\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}k_{1}\\k_{2}\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{191}&\frac{8}{191}\\\frac{25}{191}&-\frac{9}{191}\end{matrix}\right)\left(\begin{matrix}59\\79\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}k_{1}\\k_{2}\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{191}\times 59+\frac{8}{191}\times 79\\\frac{25}{191}\times 59-\frac{9}{191}\times 79\end{matrix}\right)$$

Do the arithmetic.

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

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

$$k_{1}=3,k_{2}=4$$

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.

$$9k_{1}+8k_{2}=59,25k_{1}+k_{2}=79$$

To make 9k_{1} and 25k_{1} equal, multiply all terms on each side of the first equation by 25 and all terms on each side of the second by 9.

$$25\times 9k_{1}+25\times 8k_{2}=25\times 59,9\times 25k_{1}+9k_{2}=9\times 79$$

Simplify.

$$225k_{1}+200k_{2}=1475,225k_{1}+9k_{2}=711$$

Subtract 225k_{1}+9k_{2}=711 from 225k_{1}+200k_{2}=1475 by subtracting like terms on each side of the equal sign.

$$225k_{1}-225k_{1}+200k_{2}-9k_{2}=1475-711$$

Add 225k_{1} to -225k_{1}. Terms 225k_{1} and -225k_{1} cancel out, leaving an equation with only one variable that can be solved.

$$200k_{2}-9k_{2}=1475-711$$

Add 200k_{2} to -9k_{2}.

$$191k_{2}=1475-711$$

Add 1475 to -711.

$$191k_{2}=764$$

Divide both sides by 191.

$$k_{2}=4$$

Substitute 4 for k_{2} in 25k_{1}+k_{2}=79. Because the resulting equation contains only one variable, you can solve for k_{1} directly.

$$25k_{1}+4=79$$

Subtract 4 from both sides of the equation.

$$25k_{1}=75$$

Divide both sides by 25.

$$k_{1}=3$$

The system is now solved.

$$k_{1}=3,k_{2}=4$$