Solve {l}{I_{1}(4+6+5)-I_{2}*6-12=0}{-I_{1}*6+I_{2}*12=0}

$$\left. \begin{array} { l } { I _ { 1 } ( 4 + 6 + 5 ) - I _ { 2 } \times 6 - 12 = 0 } \\ { - I _ { 1 } \times 6 + I _ { 2 } \times 12 = 0 } \end{array} \right.$$

Solve for I_1, I_2

$$I_{1}=1$$

$$I_{2}=\frac{1}{2}=0.5$$

Steps Using Substitution

Consider the first equation. Add 4 and 6 to get 10.

$$I_{1}\left(10+5\right)-I_{2}\times 6-12=0$$

Add 10 and 5 to get 15.

$$I_{1}\times 15-I_{2}\times 6-12=0$$

Add 12 to both sides. Anything plus zero gives itself.

$$I_{1}\times 15-I_{2}\times 6=12$$

Multiply -1 and 6 to get -6.

$$I_{1}\times 15-6I_{2}=12$$

Consider the second equation. Multiply -1 and 6 to get -6.

$$-6I_{1}+I_{2}\times 12=0$$

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.

$$15I_{1}-6I_{2}=12,-6I_{1}+12I_{2}=0$$

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

$$15I_{1}-6I_{2}=12$$

Add 6I_{2} to both sides of the equation.

$$15I_{1}=6I_{2}+12$$

Divide both sides by 15.

$$I_{1}=\frac{1}{15}\left(6I_{2}+12\right)$$

Multiply \frac{1}{15} times 12+6I_{2}.

$$I_{1}=\frac{2}{5}I_{2}+\frac{4}{5}$$

Substitute \frac{4+2I_{2}}{5} for I_{1} in the other equation, -6I_{1}+12I_{2}=0.

$$-6\left(\frac{2}{5}I_{2}+\frac{4}{5}\right)+12I_{2}=0$$

Multiply -6 times \frac{4+2I_{2}}{5}.

$$-\frac{12}{5}I_{2}-\frac{24}{5}+12I_{2}=0$$

Add -\frac{12I_{2}}{5} to 12I_{2}.

$$\frac{48}{5}I_{2}-\frac{24}{5}=0$$

Add \frac{24}{5} to both sides of the equation.

$$\frac{48}{5}I_{2}=\frac{24}{5}$$

Divide both sides of the equation by \frac{48}{5}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$I_{2}=\frac{1}{2}$$

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

$$I_{1}=\frac{2}{5}\times \frac{1}{2}+\frac{4}{5}$$

Multiply \frac{2}{5} times \frac{1}{2} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$I_{1}=\frac{1+4}{5}$$

Add \frac{4}{5} to \frac{1}{5} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$I_{1}=1$$

The system is now solved.

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

Steps Using Matrices

Consider the first equation. Add 4 and 6 to get 10.

$$I_{1}\left(10+5\right)-I_{2}\times 6-12=0$$

Add 10 and 5 to get 15.

$$I_{1}\times 15-I_{2}\times 6-12=0$$

Add 12 to both sides. Anything plus zero gives itself.

$$I_{1}\times 15-I_{2}\times 6=12$$

Multiply -1 and 6 to get -6.

$$I_{1}\times 15-6I_{2}=12$$

Consider the second equation. Multiply -1 and 6 to get -6.

$$-6I_{1}+I_{2}\times 12=0$$

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

$$15I_{1}-6I_{2}=12,-6I_{1}+12I_{2}=0$$

Write the equations in matrix form.

$$\left(\begin{matrix}15&-6\\-6&12\end{matrix}\right)\left(\begin{matrix}I_{1}\\I_{2}\end{matrix}\right)=\left(\begin{matrix}12\\0\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}15&-6\\-6&12\end{matrix}\right).

$$inverse(\left(\begin{matrix}15&-6\\-6&12\end{matrix}\right))\left(\begin{matrix}15&-6\\-6&12\end{matrix}\right)\left(\begin{matrix}I_{1}\\I_{2}\end{matrix}\right)=inverse(\left(\begin{matrix}15&-6\\-6&12\end{matrix}\right))\left(\begin{matrix}12\\0\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}I_{1}\\I_{2}\end{matrix}\right)=inverse(\left(\begin{matrix}15&-6\\-6&12\end{matrix}\right))\left(\begin{matrix}12\\0\end{matrix}\right)$$

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

$$\left(\begin{matrix}I_{1}\\I_{2}\end{matrix}\right)=inverse(\left(\begin{matrix}15&-6\\-6&12\end{matrix}\right))\left(\begin{matrix}12\\0\end{matrix}\right)$$

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.

$$\left(\begin{matrix}I_{1}\\I_{2}\end{matrix}\right)=\left(\begin{matrix}\frac{12}{15\times 12-\left(-6\left(-6\right)\right)}&-\frac{-6}{15\times 12-\left(-6\left(-6\right)\right)}\\-\frac{-6}{15\times 12-\left(-6\left(-6\right)\right)}&\frac{15}{15\times 12-\left(-6\left(-6\right)\right)}\end{matrix}\right)\left(\begin{matrix}12\\0\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}I_{1}\\I_{2}\end{matrix}\right)=\left(\begin{matrix}\frac{1}{12}&\frac{1}{24}\\\frac{1}{24}&\frac{5}{48}\end{matrix}\right)\left(\begin{matrix}12\\0\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}I_{1}\\I_{2}\end{matrix}\right)=\left(\begin{matrix}\frac{1}{12}\times 12\\\frac{1}{24}\times 12\end{matrix}\right)$$

Do the arithmetic.

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

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

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

Steps Using Elimination

Consider the first equation. Add 4 and 6 to get 10.

$$I_{1}\left(10+5\right)-I_{2}\times 6-12=0$$

Add 10 and 5 to get 15.

$$I_{1}\times 15-I_{2}\times 6-12=0$$

Add 12 to both sides. Anything plus zero gives itself.

$$I_{1}\times 15-I_{2}\times 6=12$$

Multiply -1 and 6 to get -6.

$$I_{1}\times 15-6I_{2}=12$$

Consider the second equation. Multiply -1 and 6 to get -6.

$$-6I_{1}+I_{2}\times 12=0$$

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.

$$15I_{1}-6I_{2}=12,-6I_{1}+12I_{2}=0$$

To make 15I_{1} and -6I_{1} equal, multiply all terms on each side of the first equation by -6 and all terms on each side of the second by 15.

$$-6\times 15I_{1}-6\left(-6\right)I_{2}=-6\times 12,15\left(-6\right)I_{1}+15\times 12I_{2}=0$$

Simplify.

$$-90I_{1}+36I_{2}=-72,-90I_{1}+180I_{2}=0$$

Subtract -90I_{1}+180I_{2}=0 from -90I_{1}+36I_{2}=-72 by subtracting like terms on each side of the equal sign.

$$-90I_{1}+90I_{1}+36I_{2}-180I_{2}=-72$$

Add -90I_{1} to 90I_{1}. Terms -90I_{1} and 90I_{1} cancel out, leaving an equation with only one variable that can be solved.

$$36I_{2}-180I_{2}=-72$$

Add 36I_{2} to -180I_{2}.

$$-144I_{2}=-72$$

Divide both sides by -144.

$$I_{2}=\frac{1}{2}$$

Substitute \frac{1}{2} for I_{2} in -6I_{1}+12I_{2}=0. Because the resulting equation contains only one variable, you can solve for I_{1} directly.

$$-6I_{1}+12\times \frac{1}{2}=0$$

Multiply 12 times \frac{1}{2}.

$$-6I_{1}+6=0$$

Subtract 6 from both sides of the equation.

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

Divide both sides by -6.

$$I_{1}=1$$

The system is now solved.

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