Solve {l}{77=k_{1}(5^2)+k_{2}(-1)^3}{32=k_{1}(-4)^2+k_{2}(2)^3}

$$\left\{ \begin{array} { l } { 77 = k _ { 1 } ( 5 ^ { 2 } ) + k _ { 2 } ( - 1 ) ^ { 3 } } \\ { 32 = k _ { 1 } ( - 4 ) ^ { 2 } + k _ { 2 } ( 2 ) ^ { 3 } } \end{array} \right.$$

Solve for k_1, k_2

$$k_{1}=3$$

$$k_{2}=-2$$

Steps Using Substitution

Consider the first equation. Calculate 5 to the power of 2 and get 25.

$$77=k_{1}\times 25+k_{2}\left(-1\right)^{3}$$

Calculate -1 to the power of 3 and get -1.

$$77=k_{1}\times 25+k_{2}\left(-1\right)$$

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

$$k_{1}\times 25+k_{2}\left(-1\right)=77$$

Consider the second equation. Calculate -4 to the power of 2 and get 16.

$$32=k_{1}\times 16+k_{2}\times 2^{3}$$

Calculate 2 to the power of 3 and get 8.

$$32=k_{1}\times 16+k_{2}\times 8$$

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

$$k_{1}\times 16+k_{2}\times 8=32$$

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.

$$25k_{1}-k_{2}=77,16k_{1}+8k_{2}=32$$

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

$$25k_{1}-k_{2}=77$$

Add k_{2} to both sides of the equation.

$$25k_{1}=k_{2}+77$$

Divide both sides by 25.

$$k_{1}=\frac{1}{25}\left(k_{2}+77\right)$$

Multiply \frac{1}{25} times k_{2}+77.

$$k_{1}=\frac{1}{25}k_{2}+\frac{77}{25}$$

Substitute \frac{77+k_{2}}{25} for k_{1} in the other equation, 16k_{1}+8k_{2}=32.

$$16\left(\frac{1}{25}k_{2}+\frac{77}{25}\right)+8k_{2}=32$$

Multiply 16 times \frac{77+k_{2}}{25}.

$$\frac{16}{25}k_{2}+\frac{1232}{25}+8k_{2}=32$$

Add \frac{16k_{2}}{25} to 8k_{2}.

$$\frac{216}{25}k_{2}+\frac{1232}{25}=32$$

Subtract \frac{1232}{25} from both sides of the equation.

$$\frac{216}{25}k_{2}=-\frac{432}{25}$$

Divide both sides of the equation by \frac{216}{25}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$k_{2}=-2$$

Substitute -2 for k_{2} in k_{1}=\frac{1}{25}k_{2}+\frac{77}{25}. Because the resulting equation contains only one variable, you can solve for k_{1} directly.

$$k_{1}=\frac{1}{25}\left(-2\right)+\frac{77}{25}$$

Multiply \frac{1}{25} times -2.

$$k_{1}=\frac{-2+77}{25}$$

Add \frac{77}{25} to -\frac{2}{25} 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.

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

Steps Using Matrices

Consider the first equation. Calculate 5 to the power of 2 and get 25.

$$77=k_{1}\times 25+k_{2}\left(-1\right)^{3}$$

Calculate -1 to the power of 3 and get -1.

$$77=k_{1}\times 25+k_{2}\left(-1\right)$$

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

$$k_{1}\times 25+k_{2}\left(-1\right)=77$$

Consider the second equation. Calculate -4 to the power of 2 and get 16.

$$32=k_{1}\times 16+k_{2}\times 2^{3}$$

Calculate 2 to the power of 3 and get 8.

$$32=k_{1}\times 16+k_{2}\times 8$$

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

$$k_{1}\times 16+k_{2}\times 8=32$$

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

$$25k_{1}-k_{2}=77,16k_{1}+8k_{2}=32$$

Write the equations in matrix form.

$$\left(\begin{matrix}25&-1\\16&8\end{matrix}\right)\left(\begin{matrix}k_{1}\\k_{2}\end{matrix}\right)=\left(\begin{matrix}77\\32\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}25&-1\\16&8\end{matrix}\right).

$$inverse(\left(\begin{matrix}25&-1\\16&8\end{matrix}\right))\left(\begin{matrix}25&-1\\16&8\end{matrix}\right)\left(\begin{matrix}k_{1}\\k_{2}\end{matrix}\right)=inverse(\left(\begin{matrix}25&-1\\16&8\end{matrix}\right))\left(\begin{matrix}77\\32\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}k_{1}\\k_{2}\end{matrix}\right)=inverse(\left(\begin{matrix}25&-1\\16&8\end{matrix}\right))\left(\begin{matrix}77\\32\end{matrix}\right)$$

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

$$\left(\begin{matrix}k_{1}\\k_{2}\end{matrix}\right)=inverse(\left(\begin{matrix}25&-1\\16&8\end{matrix}\right))\left(\begin{matrix}77\\32\end{matrix}\right)$$

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.

$$\left(\begin{matrix}k_{1}\\k_{2}\end{matrix}\right)=\left(\begin{matrix}\frac{8}{25\times 8-\left(-16\right)}&-\frac{-1}{25\times 8-\left(-16\right)}\\-\frac{16}{25\times 8-\left(-16\right)}&\frac{25}{25\times 8-\left(-16\right)}\end{matrix}\right)\left(\begin{matrix}77\\32\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}k_{1}\\k_{2}\end{matrix}\right)=\left(\begin{matrix}\frac{1}{27}&\frac{1}{216}\\-\frac{2}{27}&\frac{25}{216}\end{matrix}\right)\left(\begin{matrix}77\\32\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}k_{1}\\k_{2}\end{matrix}\right)=\left(\begin{matrix}\frac{1}{27}\times 77+\frac{1}{216}\times 32\\-\frac{2}{27}\times 77+\frac{25}{216}\times 32\end{matrix}\right)$$

Do the arithmetic.

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

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

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

Steps Using Elimination

Consider the first equation. Calculate 5 to the power of 2 and get 25.

$$77=k_{1}\times 25+k_{2}\left(-1\right)^{3}$$

Calculate -1 to the power of 3 and get -1.

$$77=k_{1}\times 25+k_{2}\left(-1\right)$$

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

$$k_{1}\times 25+k_{2}\left(-1\right)=77$$

Consider the second equation. Calculate -4 to the power of 2 and get 16.

$$32=k_{1}\times 16+k_{2}\times 2^{3}$$

Calculate 2 to the power of 3 and get 8.

$$32=k_{1}\times 16+k_{2}\times 8$$

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

$$k_{1}\times 16+k_{2}\times 8=32$$

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.

$$25k_{1}-k_{2}=77,16k_{1}+8k_{2}=32$$

To make 25k_{1} and 16k_{1} equal, multiply all terms on each side of the first equation by 16 and all terms on each side of the second by 25.

$$16\times 25k_{1}+16\left(-1\right)k_{2}=16\times 77,25\times 16k_{1}+25\times 8k_{2}=25\times 32$$

Simplify.

$$400k_{1}-16k_{2}=1232,400k_{1}+200k_{2}=800$$

Subtract 400k_{1}+200k_{2}=800 from 400k_{1}-16k_{2}=1232 by subtracting like terms on each side of the equal sign.

$$400k_{1}-400k_{1}-16k_{2}-200k_{2}=1232-800$$

Add 400k_{1} to -400k_{1}. Terms 400k_{1} and -400k_{1} cancel out, leaving an equation with only one variable that can be solved.

$$-16k_{2}-200k_{2}=1232-800$$

Add -16k_{2} to -200k_{2}.

$$-216k_{2}=1232-800$$

Add 1232 to -800.

$$-216k_{2}=432$$

Divide both sides by -216.

$$k_{2}=-2$$

Substitute -2 for k_{2} in 16k_{1}+8k_{2}=32. Because the resulting equation contains only one variable, you can solve for k_{1} directly.

$$16k_{1}+8\left(-2\right)=32$$

Multiply 8 times -2.

$$16k_{1}-16=32$$

Add 16 to both sides of the equation.

$$16k_{1}=48$$

Divide both sides by 16.

$$k_{1}=3$$

The system is now solved.

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