Solve {l}{7=5700I_{1}+4700I_{2}}{5=4700I_{1}+10300I_{2}}

$$\left. \begin{array} { l } { 7 = 5700 I _ { 1 } + 4700 I _ { 2 } } \\ { 5 = 4700 I _ { 1 } + 10300 I _ { 2 } } \end{array} \right.$$

Solve for I_1, I_2

$$I_{1}=\frac{243}{183100}\approx 0.001327144$$

$$I_{2}=-\frac{11}{91550}\approx -0.000120153$$

Steps Using Substitution

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

$$5700I_{1}+4700I_{2}=7$$

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

$$4700I_{1}+10300I_{2}=5$$

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.

$$5700I_{1}+4700I_{2}=7,4700I_{1}+10300I_{2}=5$$

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

$$5700I_{1}+4700I_{2}=7$$

Subtract 4700I_{2} from both sides of the equation.

$$5700I_{1}=-4700I_{2}+7$$

Divide both sides by 5700.

$$I_{1}=\frac{1}{5700}\left(-4700I_{2}+7\right)$$

Multiply \frac{1}{5700} times -4700I_{2}+7.

$$I_{1}=-\frac{47}{57}I_{2}+\frac{7}{5700}$$

Substitute -\frac{47I_{2}}{57}+\frac{7}{5700} for I_{1} in the other equation, 4700I_{1}+10300I_{2}=5.

$$4700\left(-\frac{47}{57}I_{2}+\frac{7}{5700}\right)+10300I_{2}=5$$

Multiply 4700 times -\frac{47I_{2}}{57}+\frac{7}{5700}.

$$-\frac{220900}{57}I_{2}+\frac{329}{57}+10300I_{2}=5$$

Add -\frac{220900I_{2}}{57} to 10300I_{2}.

$$\frac{366200}{57}I_{2}+\frac{329}{57}=5$$

Subtract \frac{329}{57} from both sides of the equation.

$$\frac{366200}{57}I_{2}=-\frac{44}{57}$$

Divide both sides of the equation by \frac{366200}{57}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$I_{2}=-\frac{11}{91550}$$

Substitute -\frac{11}{91550} for I_{2} in I_{1}=-\frac{47}{57}I_{2}+\frac{7}{5700}. Because the resulting equation contains only one variable, you can solve for I_{1} directly.

$$I_{1}=-\frac{47}{57}\left(-\frac{11}{91550}\right)+\frac{7}{5700}$$

Multiply -\frac{47}{57} times -\frac{11}{91550} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$I_{1}=\frac{517}{5218350}+\frac{7}{5700}$$

Add \frac{7}{5700} to \frac{517}{5218350} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$I_{1}=\frac{243}{183100}$$

The system is now solved.

$$I_{1}=\frac{243}{183100},I_{2}=-\frac{11}{91550}$$

Steps Using Matrices

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

$$5700I_{1}+4700I_{2}=7$$

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

$$4700I_{1}+10300I_{2}=5$$

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

$$5700I_{1}+4700I_{2}=7,4700I_{1}+10300I_{2}=5$$

Write the equations in matrix form.

$$\left(\begin{matrix}5700&4700\\4700&10300\end{matrix}\right)\left(\begin{matrix}I_{1}\\I_{2}\end{matrix}\right)=\left(\begin{matrix}7\\5\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}5700&4700\\4700&10300\end{matrix}\right).

$$inverse(\left(\begin{matrix}5700&4700\\4700&10300\end{matrix}\right))\left(\begin{matrix}5700&4700\\4700&10300\end{matrix}\right)\left(\begin{matrix}I_{1}\\I_{2}\end{matrix}\right)=inverse(\left(\begin{matrix}5700&4700\\4700&10300\end{matrix}\right))\left(\begin{matrix}7\\5\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}I_{1}\\I_{2}\end{matrix}\right)=inverse(\left(\begin{matrix}5700&4700\\4700&10300\end{matrix}\right))\left(\begin{matrix}7\\5\end{matrix}\right)$$

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

$$\left(\begin{matrix}I_{1}\\I_{2}\end{matrix}\right)=inverse(\left(\begin{matrix}5700&4700\\4700&10300\end{matrix}\right))\left(\begin{matrix}7\\5\end{matrix}\right)$$

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.

$$\left(\begin{matrix}I_{1}\\I_{2}\end{matrix}\right)=\left(\begin{matrix}\frac{10300}{5700\times 10300-4700\times 4700}&-\frac{4700}{5700\times 10300-4700\times 4700}\\-\frac{4700}{5700\times 10300-4700\times 4700}&\frac{5700}{5700\times 10300-4700\times 4700}\end{matrix}\right)\left(\begin{matrix}7\\5\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}I_{1}\\I_{2}\end{matrix}\right)=\left(\begin{matrix}\frac{103}{366200}&-\frac{47}{366200}\\-\frac{47}{366200}&\frac{57}{366200}\end{matrix}\right)\left(\begin{matrix}7\\5\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}I_{1}\\I_{2}\end{matrix}\right)=\left(\begin{matrix}\frac{103}{366200}\times 7-\frac{47}{366200}\times 5\\-\frac{47}{366200}\times 7+\frac{57}{366200}\times 5\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}I_{1}\\I_{2}\end{matrix}\right)=\left(\begin{matrix}\frac{243}{183100}\\-\frac{11}{91550}\end{matrix}\right)$$

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

$$I_{1}=\frac{243}{183100},I_{2}=-\frac{11}{91550}$$

Steps Using Elimination

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

$$5700I_{1}+4700I_{2}=7$$

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

$$4700I_{1}+10300I_{2}=5$$

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.

$$5700I_{1}+4700I_{2}=7,4700I_{1}+10300I_{2}=5$$

To make 5700I_{1} and 4700I_{1} equal, multiply all terms on each side of the first equation by 4700 and all terms on each side of the second by 5700.

$$4700\times 5700I_{1}+4700\times 4700I_{2}=4700\times 7,5700\times 4700I_{1}+5700\times 10300I_{2}=5700\times 5$$

Simplify.

$$26790000I_{1}[phone]I_{2}=32900,26790000I_{1}[phone]I_{2}=28500$$

Subtract 26790000I_{1}[phone]I_{2}=28500 from 26790000I_{1}[phone]I_{2}=32900 by subtracting like terms on each side of the equal sign.

$$26790000I_{1}-26790000I_{1}[phone]I_{2}-58710000I_{2}=32900-28500$$

Add 26790000I_{1} to -26790000I_{1}. Terms 26790000I_{1} and -26790000I_{1} cancel out, leaving an equation with only one variable that can be solved.

$$22090000I_{2}-58710000I_{2}=32900-28500$$

Add 22090000I_{2} to -58710000I_{2}.

$$-36620000I_{2}=32900-28500$$

Add 32900 to -28500.

$$-36620000I_{2}=4400$$

Divide both sides by -36620000.

$$I_{2}=-\frac{11}{91550}$$

Substitute -\frac{11}{91550} for I_{2} in 4700I_{1}+10300I_{2}=5. Because the resulting equation contains only one variable, you can solve for I_{1} directly.

$$4700I_{1}+10300\left(-\frac{11}{91550}\right)=5$$

Multiply 10300 times -\frac{11}{91550}.

$$4700I_{1}-\frac{2266}{1831}=5$$

Add \frac{2266}{1831} to both sides of the equation.

$$4700I_{1}=\frac{11421}{1831}$$

Divide both sides by 4700.

$$I_{1}=\frac{243}{183100}$$

The system is now solved.

$$I_{1}=\frac{243}{183100},I_{2}=-\frac{11}{91550}$$