Solve {l}{2.5I_{1}-5I_{2}=10}{-2I_{1}+6I_{2}=20}

$$\left. \begin{array} { l } { 2.5 I _ { 1 } - 5 I _ { 2 } = 10 } \\ { - 2 I _ { 1 } + 6 I _ { 2 } = 20 } \end{array} \right.$$

Solve for I_1, I_2

$$I_{1}=32$$

$$I_{2}=14$$

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.

$$2.5I_{1}-5I_{2}=10,-2I_{1}+6I_{2}=20$$

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

$$2.5I_{1}-5I_{2}=10$$

Add 5I_{2} to both sides of the equation.

$$2.5I_{1}=5I_{2}+10$$

Divide both sides of the equation by 2.5, which is the same as multiplying both sides by the reciprocal of the fraction.

$$I_{1}=0.4\left(5I_{2}+10\right)$$

Multiply 0.4 times 10+5I_{2}.

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

Substitute 4+2I_{2} for I_{1} in the other equation, -2I_{1}+6I_{2}=20.

$$-2\left(2I_{2}+4\right)+6I_{2}=20$$

Multiply -2 times 4+2I_{2}.

$$-4I_{2}-8+6I_{2}=20$$

Add -4I_{2} to 6I_{2}.

$$2I_{2}-8=20$$

Add 8 to both sides of the equation.

$$2I_{2}=28$$

Divide both sides by 2.

$$I_{2}=14$$

Substitute 14 for I_{2} in I_{1}=2I_{2}+4. Because the resulting equation contains only one variable, you can solve for I_{1} directly.

$$I_{1}=2\times 14+4$$

Multiply 2 times 14.

$$I_{1}=28+4$$

Add 4 to 28.

$$I_{1}=32$$

The system is now solved.

$$I_{1}=32,I_{2}=14$$

Steps Using Matrices

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

$$2.5I_{1}-5I_{2}=10,-2I_{1}+6I_{2}=20$$

Write the equations in matrix form.

$$\left(\begin{matrix}2.5&-5\\-2&6\end{matrix}\right)\left(\begin{matrix}I_{1}\\I_{2}\end{matrix}\right)=\left(\begin{matrix}10\\20\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}2.5&-5\\-2&6\end{matrix}\right).

$$inverse(\left(\begin{matrix}2.5&-5\\-2&6\end{matrix}\right))\left(\begin{matrix}2.5&-5\\-2&6\end{matrix}\right)\left(\begin{matrix}I_{1}\\I_{2}\end{matrix}\right)=inverse(\left(\begin{matrix}2.5&-5\\-2&6\end{matrix}\right))\left(\begin{matrix}10\\20\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}I_{1}\\I_{2}\end{matrix}\right)=inverse(\left(\begin{matrix}2.5&-5\\-2&6\end{matrix}\right))\left(\begin{matrix}10\\20\end{matrix}\right)$$

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

$$\left(\begin{matrix}I_{1}\\I_{2}\end{matrix}\right)=inverse(\left(\begin{matrix}2.5&-5\\-2&6\end{matrix}\right))\left(\begin{matrix}10\\20\end{matrix}\right)$$

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.

$$\left(\begin{matrix}I_{1}\\I_{2}\end{matrix}\right)=\left(\begin{matrix}\frac{6}{2.5\times 6-\left(-5\left(-2\right)\right)}&-\frac{-5}{2.5\times 6-\left(-5\left(-2\right)\right)}\\-\frac{-2}{2.5\times 6-\left(-5\left(-2\right)\right)}&\frac{2.5}{2.5\times 6-\left(-5\left(-2\right)\right)}\end{matrix}\right)\left(\begin{matrix}10\\20\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}I_{1}\\I_{2}\end{matrix}\right)=\left(\begin{matrix}\frac{6}{5}&1\\\frac{2}{5}&\frac{1}{2}\end{matrix}\right)\left(\begin{matrix}10\\20\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}I_{1}\\I_{2}\end{matrix}\right)=\left(\begin{matrix}\frac{6}{5}\times 10+20\\\frac{2}{5}\times 10+\frac{1}{2}\times 20\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}I_{1}\\I_{2}\end{matrix}\right)=\left(\begin{matrix}32\\14\end{matrix}\right)$$

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

$$I_{1}=32,I_{2}=14$$

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.

$$2.5I_{1}-5I_{2}=10,-2I_{1}+6I_{2}=20$$

To make \frac{5I_{1}}{2} and -2I_{1} equal, multiply all terms on each side of the first equation by -2 and all terms on each side of the second by 2.5.

$$-2\times 2.5I_{1}-2\left(-5\right)I_{2}=-2\times 10,2.5\left(-2\right)I_{1}+2.5\times 6I_{2}=2.5\times 20$$

Simplify.

$$-5I_{1}+10I_{2}=-20,-5I_{1}+15I_{2}=50$$

Subtract -5I_{1}+15I_{2}=50 from -5I_{1}+10I_{2}=-20 by subtracting like terms on each side of the equal sign.

$$-5I_{1}+5I_{1}+10I_{2}-15I_{2}=-20-50$$

Add -5I_{1} to 5I_{1}. Terms -5I_{1} and 5I_{1} cancel out, leaving an equation with only one variable that can be solved.

$$10I_{2}-15I_{2}=-20-50$$

Add 10I_{2} to -15I_{2}.

$$-5I_{2}=-20-50$$

Add -20 to -50.

$$-5I_{2}=-70$$

Divide both sides by -5.

$$I_{2}=14$$

Substitute 14 for I_{2} in -2I_{1}+6I_{2}=20. Because the resulting equation contains only one variable, you can solve for I_{1} directly.

$$-2I_{1}+6\times 14=20$$

Multiply 6 times 14.

$$-2I_{1}+84=20$$

Subtract 84 from both sides of the equation.

$$-2I_{1}=-64$$

Divide both sides by -2.

$$I_{1}=32$$

The system is now solved.

$$I_{1}=32,I_{2}=14$$