Solve {l}{6I_{1}+6I_{2}=10}{6I_{1}+16I_{2}=30}

$$\left. \begin{array} { l } { 6 I _ { 1 } + 6 I _ { 2 } = 10 } \\ { 6 I _ { 1 } + 16 I _ { 2 } = 30 } \end{array} \right.$$

Solve for I_1, I_2

$$I_{1}=-\frac{1}{3}\approx -0.333333333$$

$$I_{2}=2$$

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.

$$6I_{1}+6I_{2}=10,6I_{1}+16I_{2}=30$$

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

$$6I_{1}+6I_{2}=10$$

Subtract 6I_{2} from both sides of the equation.

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

Divide both sides by 6.

$$I_{1}=\frac{1}{6}\left(-6I_{2}+10\right)$$

Multiply \frac{1}{6} times -6I_{2}+10.

$$I_{1}=-I_{2}+\frac{5}{3}$$

Substitute -I_{2}+\frac{5}{3} for I_{1} in the other equation, 6I_{1}+16I_{2}=30.

$$6\left(-I_{2}+\frac{5}{3}\right)+16I_{2}=30$$

Multiply 6 times -I_{2}+\frac{5}{3}.

$$-6I_{2}+10+16I_{2}=30$$

Add -6I_{2} to 16I_{2}.

$$10I_{2}+10=30$$

Subtract 10 from both sides of the equation.

$$10I_{2}=20$$

Divide both sides by 10.

$$I_{2}=2$$

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

$$I_{1}=-2+\frac{5}{3}$$

Add \frac{5}{3} to -2.

$$I_{1}=-\frac{1}{3}$$

The system is now solved.

$$I_{1}=-\frac{1}{3},I_{2}=2$$

Steps Using Matrices

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

$$6I_{1}+6I_{2}=10,6I_{1}+16I_{2}=30$$

Write the equations in matrix form.

$$\left(\begin{matrix}6&6\\6&16\end{matrix}\right)\left(\begin{matrix}I_{1}\\I_{2}\end{matrix}\right)=\left(\begin{matrix}10\\30\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}6&6\\6&16\end{matrix}\right).

$$inverse(\left(\begin{matrix}6&6\\6&16\end{matrix}\right))\left(\begin{matrix}6&6\\6&16\end{matrix}\right)\left(\begin{matrix}I_{1}\\I_{2}\end{matrix}\right)=inverse(\left(\begin{matrix}6&6\\6&16\end{matrix}\right))\left(\begin{matrix}10\\30\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}I_{1}\\I_{2}\end{matrix}\right)=inverse(\left(\begin{matrix}6&6\\6&16\end{matrix}\right))\left(\begin{matrix}10\\30\end{matrix}\right)$$

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

$$\left(\begin{matrix}I_{1}\\I_{2}\end{matrix}\right)=inverse(\left(\begin{matrix}6&6\\6&16\end{matrix}\right))\left(\begin{matrix}10\\30\end{matrix}\right)$$

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.

$$\left(\begin{matrix}I_{1}\\I_{2}\end{matrix}\right)=\left(\begin{matrix}\frac{16}{6\times 16-6\times 6}&-\frac{6}{6\times 16-6\times 6}\\-\frac{6}{6\times 16-6\times 6}&\frac{6}{6\times 16-6\times 6}\end{matrix}\right)\left(\begin{matrix}10\\30\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}I_{1}\\I_{2}\end{matrix}\right)=\left(\begin{matrix}\frac{4}{15}&-\frac{1}{10}\\-\frac{1}{10}&\frac{1}{10}\end{matrix}\right)\left(\begin{matrix}10\\30\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}I_{1}\\I_{2}\end{matrix}\right)=\left(\begin{matrix}\frac{4}{15}\times 10-\frac{1}{10}\times 30\\-\frac{1}{10}\times 10+\frac{1}{10}\times 30\end{matrix}\right)$$

Do the arithmetic.

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

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

$$I_{1}=-\frac{1}{3},I_{2}=2$$

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.

$$6I_{1}+6I_{2}=10,6I_{1}+16I_{2}=30$$

Subtract 6I_{1}+16I_{2}=30 from 6I_{1}+6I_{2}=10 by subtracting like terms on each side of the equal sign.

$$6I_{1}-6I_{1}+6I_{2}-16I_{2}=10-30$$

Add 6I_{1} to -6I_{1}. Terms 6I_{1} and -6I_{1} cancel out, leaving an equation with only one variable that can be solved.

$$6I_{2}-16I_{2}=10-30$$

Add 6I_{2} to -16I_{2}.

$$-10I_{2}=10-30$$

Add 10 to -30.

$$-10I_{2}=-20$$

Divide both sides by -10.

$$I_{2}=2$$

Substitute 2 for I_{2} in 6I_{1}+16I_{2}=30. Because the resulting equation contains only one variable, you can solve for I_{1} directly.

$$6I_{1}+16\times 2=30$$

Multiply 16 times 2.

$$6I_{1}+32=30$$

Subtract 32 from both sides of the equation.

$$6I_{1}=-2$$

Divide both sides by 6.

$$I_{1}=-\frac{1}{3}$$

The system is now solved.

$$I_{1}=-\frac{1}{3},I_{2}=2$$