Solve {l}{60=10k_{1}+100k_{2}}{225=25k_{1}+k_{2}(25)^2}

$$\left\{ \begin{array} { l } { 60 = 10 k _ { 1 } + 100 k _ { 2 } } \\ { 225 = 25 k _ { 1 } + k _ { 2 } ( 25 ) ^ { 2 } } \end{array} \right.$$

Solve for k_1, k_2

$$k_{1}=4$$

$$k_{2}=\frac{1}{5}=0.2$$

Steps Using Substitution

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

$$10k_{1}+100k_{2}=60$$

Consider the second equation. Calculate 25 to the power of 2 and get 625.

$$225=25k_{1}+k_{2}\times 625$$

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

$$25k_{1}+k_{2}\times 625=225$$

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.

$$10k_{1}+100k_{2}=60,25k_{1}+625k_{2}=225$$

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

$$10k_{1}+100k_{2}=60$$

Subtract 100k_{2} from both sides of the equation.

$$10k_{1}=-100k_{2}+60$$

Divide both sides by 10.

$$k_{1}=\frac{1}{10}\left(-100k_{2}+60\right)$$

Multiply \frac{1}{10} times -100k_{2}+60.

$$k_{1}=-10k_{2}+6$$

Substitute -10k_{2}+6 for k_{1} in the other equation, 25k_{1}+625k_{2}=225.

$$25\left(-10k_{2}+6\right)+625k_{2}=225$$

Multiply 25 times -10k_{2}+6.

$$-250k_{2}+150+625k_{2}=225$$

Add -250k_{2} to 625k_{2}.

$$375k_{2}+150=225$$

Subtract 150 from both sides of the equation.

$$375k_{2}=75$$

Divide both sides by 375.

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

Substitute \frac{1}{5} for k_{2} in k_{1}=-10k_{2}+6. Because the resulting equation contains only one variable, you can solve for k_{1} directly.

$$k_{1}=-10\times \frac{1}{5}+6$$

Multiply -10 times \frac{1}{5}.

$$k_{1}=-2+6$$

Add 6 to -2.

$$k_{1}=4$$

The system is now solved.

$$k_{1}=4,k_{2}=\frac{1}{5}$$

Steps Using Matrices

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

$$10k_{1}+100k_{2}=60$$

Consider the second equation. Calculate 25 to the power of 2 and get 625.

$$225=25k_{1}+k_{2}\times 625$$

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

$$25k_{1}+k_{2}\times 625=225$$

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

$$10k_{1}+100k_{2}=60,25k_{1}+625k_{2}=225$$

Write the equations in matrix form.

$$\left(\begin{matrix}10&100\\25&625\end{matrix}\right)\left(\begin{matrix}k_{1}\\k_{2}\end{matrix}\right)=\left(\begin{matrix}60\\225\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}10&100\\25&625\end{matrix}\right).

$$inverse(\left(\begin{matrix}10&100\\25&625\end{matrix}\right))\left(\begin{matrix}10&100\\25&625\end{matrix}\right)\left(\begin{matrix}k_{1}\\k_{2}\end{matrix}\right)=inverse(\left(\begin{matrix}10&100\\25&625\end{matrix}\right))\left(\begin{matrix}60\\225\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}k_{1}\\k_{2}\end{matrix}\right)=inverse(\left(\begin{matrix}10&100\\25&625\end{matrix}\right))\left(\begin{matrix}60\\225\end{matrix}\right)$$

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

$$\left(\begin{matrix}k_{1}\\k_{2}\end{matrix}\right)=inverse(\left(\begin{matrix}10&100\\25&625\end{matrix}\right))\left(\begin{matrix}60\\225\end{matrix}\right)$$

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.

$$\left(\begin{matrix}k_{1}\\k_{2}\end{matrix}\right)=\left(\begin{matrix}\frac{625}{10\times 625-100\times 25}&-\frac{100}{10\times 625-100\times 25}\\-\frac{25}{10\times 625-100\times 25}&\frac{10}{10\times 625-100\times 25}\end{matrix}\right)\left(\begin{matrix}60\\225\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}k_{1}\\k_{2}\end{matrix}\right)=\left(\begin{matrix}\frac{1}{6}&-\frac{2}{75}\\-\frac{1}{150}&\frac{1}{375}\end{matrix}\right)\left(\begin{matrix}60\\225\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}k_{1}\\k_{2}\end{matrix}\right)=\left(\begin{matrix}\frac{1}{6}\times 60-\frac{2}{75}\times 225\\-\frac{1}{150}\times 60+\frac{1}{375}\times 225\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}k_{1}\\k_{2}\end{matrix}\right)=\left(\begin{matrix}4\\\frac{1}{5}\end{matrix}\right)$$

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

$$k_{1}=4,k_{2}=\frac{1}{5}$$

Steps Using Elimination

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

$$10k_{1}+100k_{2}=60$$

Consider the second equation. Calculate 25 to the power of 2 and get 625.

$$225=25k_{1}+k_{2}\times 625$$

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

$$25k_{1}+k_{2}\times 625=225$$

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.

$$10k_{1}+100k_{2}=60,25k_{1}+625k_{2}=225$$

To make 10k_{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 10.

$$25\times 10k_{1}+25\times 100k_{2}=25\times 60,10\times 25k_{1}+10\times 625k_{2}=10\times 225$$

Simplify.

$$250k_{1}+2500k_{2}=1500,250k_{1}+6250k_{2}=2250$$

Subtract 250k_{1}+6250k_{2}=2250 from 250k_{1}+2500k_{2}=1500 by subtracting like terms on each side of the equal sign.

$$250k_{1}-250k_{1}+2500k_{2}-6250k_{2}=1500-2250$$

Add 250k_{1} to -250k_{1}. Terms 250k_{1} and -250k_{1} cancel out, leaving an equation with only one variable that can be solved.

$$2500k_{2}-6250k_{2}=1500-2250$$

Add 2500k_{2} to -6250k_{2}.

$$-3750k_{2}=1500-2250$$

Add 1500 to -2250.

$$-3750k_{2}=-750$$

Divide both sides by -3750.

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

Substitute \frac{1}{5} for k_{2} in 25k_{1}+625k_{2}=225. Because the resulting equation contains only one variable, you can solve for k_{1} directly.

$$25k_{1}+625\times \frac{1}{5}=225$$

Multiply 625 times \frac{1}{5}.

$$25k_{1}+125=225$$

Subtract 125 from both sides of the equation.

$$25k_{1}=100$$

Divide both sides by 25.

$$k_{1}=4$$

The system is now solved.

$$k_{1}=4,k_{2}=\frac{1}{5}$$