Solve {l}{10I_{1}+2I_{1}+4I_{1}-4I_{2}=15}{4(I_{2}-I_{1})=40I_{2}}

$$\left. \begin{array}{l}{ 10 I _ { 1 } + 2 I _ { 1 } + 4 I _ { 1 } - 4 I _ { 2 } = 15 }\\{ 4 ( I _ { 2 } - I _ { 1 } ) = 40 I _ { 2 } }\end{array} \right.$$

Solve for I_1, I_2

$$I_{1}=\frac{135}{148}\approx 0.912162162$$

$$I_{2}=-\frac{15}{148}\approx -0.101351351$$

Steps Using Substitution

Consider the first equation. Combine 10I_{1} and 2I_{1} to get 12I_{1}.

$$12I_{1}+4I_{1}-4I_{2}=15$$

Combine 12I_{1} and 4I_{1} to get 16I_{1}.

$$16I_{1}-4I_{2}=15$$

Consider the second equation. Use the distributive property to multiply 4 by I_{2}-I_{1}.

$$4I_{2}-4I_{1}=40I_{2}$$

Subtract 40I_{2} from both sides.

$$4I_{2}-4I_{1}-40I_{2}=0$$

Combine 4I_{2} and -40I_{2} to get -36I_{2}.

$$-36I_{2}-4I_{1}=0$$

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.

$$16I_{1}-4I_{2}=15,-4I_{1}-36I_{2}=0$$

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

$$16I_{1}-4I_{2}=15$$

Add 4I_{2} to both sides of the equation.

$$16I_{1}=4I_{2}+15$$

Divide both sides by 16.

$$I_{1}=\frac{1}{16}\left(4I_{2}+15\right)$$

Multiply \frac{1}{16} times 4I_{2}+15.

$$I_{1}=\frac{1}{4}I_{2}+\frac{15}{16}$$

Substitute \frac{I_{2}}{4}+\frac{15}{16} for I_{1} in the other equation, -4I_{1}-36I_{2}=0.

$$-4\left(\frac{1}{4}I_{2}+\frac{15}{16}\right)-36I_{2}=0$$

Multiply -4 times \frac{I_{2}}{4}+\frac{15}{16}.

$$-I_{2}-\frac{15}{4}-36I_{2}=0$$

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

$$-37I_{2}-\frac{15}{4}=0$$

Add \frac{15}{4} to both sides of the equation.

$$-37I_{2}=\frac{15}{4}$$

Divide both sides by -37.

$$I_{2}=-\frac{15}{148}$$

Substitute -\frac{15}{148} for I_{2} in I_{1}=\frac{1}{4}I_{2}+\frac{15}{16}. Because the resulting equation contains only one variable, you can solve for I_{1} directly.

$$I_{1}=\frac{1}{4}\left(-\frac{15}{148}\right)+\frac{15}{16}$$

Multiply \frac{1}{4} times -\frac{15}{148} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$I_{1}=-\frac{15}{592}+\frac{15}{16}$$

Add \frac{15}{16} to -\frac{15}{592} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$I_{1}=\frac{135}{148}$$

The system is now solved.

$$I_{1}=\frac{135}{148},I_{2}=-\frac{15}{148}$$

Steps Using Matrices

Consider the first equation. Combine 10I_{1} and 2I_{1} to get 12I_{1}.

$$12I_{1}+4I_{1}-4I_{2}=15$$

Combine 12I_{1} and 4I_{1} to get 16I_{1}.

$$16I_{1}-4I_{2}=15$$

Consider the second equation. Use the distributive property to multiply 4 by I_{2}-I_{1}.

$$4I_{2}-4I_{1}=40I_{2}$$

Subtract 40I_{2} from both sides.

$$4I_{2}-4I_{1}-40I_{2}=0$$

Combine 4I_{2} and -40I_{2} to get -36I_{2}.

$$-36I_{2}-4I_{1}=0$$

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

$$16I_{1}-4I_{2}=15,-4I_{1}-36I_{2}=0$$

Write the equations in matrix form.

$$\left(\begin{matrix}16&-4\\-4&-36\end{matrix}\right)\left(\begin{matrix}I_{1}\\I_{2}\end{matrix}\right)=\left(\begin{matrix}15\\0\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}16&-4\\-4&-36\end{matrix}\right).

$$inverse(\left(\begin{matrix}16&-4\\-4&-36\end{matrix}\right))\left(\begin{matrix}16&-4\\-4&-36\end{matrix}\right)\left(\begin{matrix}I_{1}\\I_{2}\end{matrix}\right)=inverse(\left(\begin{matrix}16&-4\\-4&-36\end{matrix}\right))\left(\begin{matrix}15\\0\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}I_{1}\\I_{2}\end{matrix}\right)=inverse(\left(\begin{matrix}16&-4\\-4&-36\end{matrix}\right))\left(\begin{matrix}15\\0\end{matrix}\right)$$

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

$$\left(\begin{matrix}I_{1}\\I_{2}\end{matrix}\right)=inverse(\left(\begin{matrix}16&-4\\-4&-36\end{matrix}\right))\left(\begin{matrix}15\\0\end{matrix}\right)$$

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.

$$\left(\begin{matrix}I_{1}\\I_{2}\end{matrix}\right)=\left(\begin{matrix}-\frac{36}{16\left(-36\right)-\left(-4\left(-4\right)\right)}&-\frac{-4}{16\left(-36\right)-\left(-4\left(-4\right)\right)}\\-\frac{-4}{16\left(-36\right)-\left(-4\left(-4\right)\right)}&\frac{16}{16\left(-36\right)-\left(-4\left(-4\right)\right)}\end{matrix}\right)\left(\begin{matrix}15\\0\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}I_{1}\\I_{2}\end{matrix}\right)=\left(\begin{matrix}\frac{9}{148}&-\frac{1}{148}\\-\frac{1}{148}&-\frac{1}{37}\end{matrix}\right)\left(\begin{matrix}15\\0\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}I_{1}\\I_{2}\end{matrix}\right)=\left(\begin{matrix}\frac{9}{148}\times 15\\-\frac{1}{148}\times 15\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}I_{1}\\I_{2}\end{matrix}\right)=\left(\begin{matrix}\frac{135}{148}\\-\frac{15}{148}\end{matrix}\right)$$

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

$$I_{1}=\frac{135}{148},I_{2}=-\frac{15}{148}$$

Steps Using Elimination

Consider the first equation. Combine 10I_{1} and 2I_{1} to get 12I_{1}.

$$12I_{1}+4I_{1}-4I_{2}=15$$

Combine 12I_{1} and 4I_{1} to get 16I_{1}.

$$16I_{1}-4I_{2}=15$$

Consider the second equation. Use the distributive property to multiply 4 by I_{2}-I_{1}.

$$4I_{2}-4I_{1}=40I_{2}$$

Subtract 40I_{2} from both sides.

$$4I_{2}-4I_{1}-40I_{2}=0$$

Combine 4I_{2} and -40I_{2} to get -36I_{2}.

$$-36I_{2}-4I_{1}=0$$

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.

$$16I_{1}-4I_{2}=15,-4I_{1}-36I_{2}=0$$

To make 16I_{1} and -4I_{1} equal, multiply all terms on each side of the first equation by -4 and all terms on each side of the second by 16.

$$-4\times 16I_{1}-4\left(-4\right)I_{2}=-4\times 15,16\left(-4\right)I_{1}+16\left(-36\right)I_{2}=0$$

Simplify.

$$-64I_{1}+16I_{2}=-60,-64I_{1}-576I_{2}=0$$

Subtract -64I_{1}-576I_{2}=0 from -64I_{1}+16I_{2}=-60 by subtracting like terms on each side of the equal sign.

$$-64I_{1}+64I_{1}+16I_{2}+576I_{2}=-60$$

Add -64I_{1} to 64I_{1}. Terms -64I_{1} and 64I_{1} cancel out, leaving an equation with only one variable that can be solved.

$$16I_{2}+576I_{2}=-60$$

Add 16I_{2} to 576I_{2}.

$$592I_{2}=-60$$

Divide both sides by 592.

$$I_{2}=-\frac{15}{148}$$

Substitute -\frac{15}{148} for I_{2} in -4I_{1}-36I_{2}=0. Because the resulting equation contains only one variable, you can solve for I_{1} directly.

$$-4I_{1}-36\left(-\frac{15}{148}\right)=0$$

Multiply -36 times -\frac{15}{148}.

$$-4I_{1}+\frac{135}{37}=0$$

Subtract \frac{135}{37} from both sides of the equation.

$$-4I_{1}=-\frac{135}{37}$$

Divide both sides by -4.

$$I_{1}=\frac{135}{148}$$

The system is now solved.

$$I_{1}=\frac{135}{148},I_{2}=-\frac{15}{148}$$