Solve {l}{frac{V_{1}-90}{500}+frac{V_{1}}{1000}+frac{V_{1}-V_{2}}{2000}=0}{frac{V_{2}-V_{1}}{2000}+frac{V_{1}}{750}+frac{V_{2}}{200}=0}

$$\left. \begin{array} { l } { \frac { V _ { 1 } - 90 } { 500 } + \frac { V _ { 1 } } { 1000 } + \frac { V _ { 1 } - V _ { 2 } } { 2000 } = 0 } \\ { \frac { V _ { 2 } - V _ { 1 } } { 2000 } + \frac { V _ { 1 } } { 750 } + \frac { V _ { 2 } } { 200 } = 0 } \end{array} \right.$$

Solve for V_1, V_2

$$V_{1} = \frac{2970}{59} = 50\frac{20}{59} \approx 50.338983051$$

$$V_{2} = -\frac{450}{59} = -7\frac{37}{59} \approx -7.627118644$$

Steps Using Substitution

Consider the first equation. Multiply both sides of the equation by 2000, the least common multiple of 500,1000,2000.

$$4\left(V_{1}-90\right)+2V_{1}+V_{1}-V_{2}=0$$

Use the distributive property to multiply 4 by V_{1}-90.

$$4V_{1}-360+2V_{1}+V_{1}-V_{2}=0$$

Combine 4V_{1} and 2V_{1} to get 6V_{1}.

$$6V_{1}-360+V_{1}-V_{2}=0$$

Combine 6V_{1} and V_{1} to get 7V_{1}.

$$7V_{1}-360-V_{2}=0$$

Add 360 to both sides. Anything plus zero gives itself.

$$7V_{1}-V_{2}=360$$

Consider the second equation. Multiply both sides of the equation by 6000, the least common multiple of 2000,750,200.

$$3\left(V_{2}-V_{1}\right)+8V_{1}+30V_{2}=0$$

Use the distributive property to multiply 3 by V_{2}-V_{1}.

$$3V_{2}-3V_{1}+8V_{1}+30V_{2}=0$$

Combine -3V_{1} and 8V_{1} to get 5V_{1}.

$$3V_{2}+5V_{1}+30V_{2}=0$$

Combine 3V_{2} and 30V_{2} to get 33V_{2}.

$$33V_{2}+5V_{1}=0$$

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.

$$7V_{1}-V_{2}=360,5V_{1}+33V_{2}=0$$

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

$$7V_{1}-V_{2}=360$$

Add V_{2} to both sides of the equation.

$$7V_{1}=V_{2}+360$$

Divide both sides by 7.

$$V_{1}=\frac{1}{7}\left(V_{2}+360\right)$$

Multiply \frac{1}{7} times V_{2}+360.

$$V_{1}=\frac{1}{7}V_{2}+\frac{360}{7}$$

Substitute \frac{360+V_{2}}{7} for V_{1} in the other equation, 5V_{1}+33V_{2}=0.

$$5\left(\frac{1}{7}V_{2}+\frac{360}{7}\right)+33V_{2}=0$$

Multiply 5 times \frac{360+V_{2}}{7}.

$$\frac{5}{7}V_{2}+\frac{1800}{7}+33V_{2}=0$$

Add \frac{5V_{2}}{7} to 33V_{2}.

$$\frac{236}{7}V_{2}+\frac{1800}{7}=0$$

Subtract \frac{1800}{7} from both sides of the equation.

$$\frac{236}{7}V_{2}=-\frac{1800}{7}$$

Divide both sides of the equation by \frac{236}{7}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$V_{2}=-\frac{450}{59}$$

Substitute -\frac{450}{59} for V_{2} in V_{1}=\frac{1}{7}V_{2}+\frac{360}{7}. Because the resulting equation contains only one variable, you can solve for V_{1} directly.

$$V_{1}=\frac{1}{7}\left(-\frac{450}{59}\right)+\frac{360}{7}$$

Multiply \frac{1}{7} times -\frac{450}{59} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$V_{1}=-\frac{450}{413}+\frac{360}{7}$$

Add \frac{360}{7} to -\frac{450}{413} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$V_{1}=\frac{2970}{59}$$

The system is now solved.

$$V_{1}=\frac{2970}{59},V_{2}=-\frac{450}{59}$$

Steps Using Matrices

Consider the first equation. Multiply both sides of the equation by 2000, the least common multiple of 500,1000,2000.

$$4\left(V_{1}-90\right)+2V_{1}+V_{1}-V_{2}=0$$

Use the distributive property to multiply 4 by V_{1}-90.

$$4V_{1}-360+2V_{1}+V_{1}-V_{2}=0$$

Combine 4V_{1} and 2V_{1} to get 6V_{1}.

$$6V_{1}-360+V_{1}-V_{2}=0$$

Combine 6V_{1} and V_{1} to get 7V_{1}.

$$7V_{1}-360-V_{2}=0$$

Add 360 to both sides. Anything plus zero gives itself.

$$7V_{1}-V_{2}=360$$

Consider the second equation. Multiply both sides of the equation by 6000, the least common multiple of 2000,750,200.

$$3\left(V_{2}-V_{1}\right)+8V_{1}+30V_{2}=0$$

Use the distributive property to multiply 3 by V_{2}-V_{1}.

$$3V_{2}-3V_{1}+8V_{1}+30V_{2}=0$$

Combine -3V_{1} and 8V_{1} to get 5V_{1}.

$$3V_{2}+5V_{1}+30V_{2}=0$$

Combine 3V_{2} and 30V_{2} to get 33V_{2}.

$$33V_{2}+5V_{1}=0$$

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

$$7V_{1}-V_{2}=360,5V_{1}+33V_{2}=0$$

Write the equations in matrix form.

$$\left(\begin{matrix}7&-1\\5&33\end{matrix}\right)\left(\begin{matrix}V_{1}\\V_{2}\end{matrix}\right)=\left(\begin{matrix}360\\0\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}7&-1\\5&33\end{matrix}\right).

$$inverse(\left(\begin{matrix}7&-1\\5&33\end{matrix}\right))\left(\begin{matrix}7&-1\\5&33\end{matrix}\right)\left(\begin{matrix}V_{1}\\V_{2}\end{matrix}\right)=inverse(\left(\begin{matrix}7&-1\\5&33\end{matrix}\right))\left(\begin{matrix}360\\0\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}V_{1}\\V_{2}\end{matrix}\right)=inverse(\left(\begin{matrix}7&-1\\5&33\end{matrix}\right))\left(\begin{matrix}360\\0\end{matrix}\right)$$

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

$$\left(\begin{matrix}V_{1}\\V_{2}\end{matrix}\right)=inverse(\left(\begin{matrix}7&-1\\5&33\end{matrix}\right))\left(\begin{matrix}360\\0\end{matrix}\right)$$

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.

$$\left(\begin{matrix}V_{1}\\V_{2}\end{matrix}\right)=\left(\begin{matrix}\frac{33}{7\times 33-\left(-5\right)}&-\frac{-1}{7\times 33-\left(-5\right)}\\-\frac{5}{7\times 33-\left(-5\right)}&\frac{7}{7\times 33-\left(-5\right)}\end{matrix}\right)\left(\begin{matrix}360\\0\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}V_{1}\\V_{2}\end{matrix}\right)=\left(\begin{matrix}\frac{33}{236}&\frac{1}{236}\\-\frac{5}{236}&\frac{7}{236}\end{matrix}\right)\left(\begin{matrix}360\\0\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}V_{1}\\V_{2}\end{matrix}\right)=\left(\begin{matrix}\frac{33}{236}\times 360\\-\frac{5}{236}\times 360\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}V_{1}\\V_{2}\end{matrix}\right)=\left(\begin{matrix}\frac{2970}{59}\\-\frac{450}{59}\end{matrix}\right)$$

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

$$V_{1}=\frac{2970}{59},V_{2}=-\frac{450}{59}$$

Steps Using Elimination

Consider the first equation. Multiply both sides of the equation by 2000, the least common multiple of 500,1000,2000.

$$4\left(V_{1}-90\right)+2V_{1}+V_{1}-V_{2}=0$$

Use the distributive property to multiply 4 by V_{1}-90.

$$4V_{1}-360+2V_{1}+V_{1}-V_{2}=0$$

Combine 4V_{1} and 2V_{1} to get 6V_{1}.

$$6V_{1}-360+V_{1}-V_{2}=0$$

Combine 6V_{1} and V_{1} to get 7V_{1}.

$$7V_{1}-360-V_{2}=0$$

Add 360 to both sides. Anything plus zero gives itself.

$$7V_{1}-V_{2}=360$$

Consider the second equation. Multiply both sides of the equation by 6000, the least common multiple of 2000,750,200.

$$3\left(V_{2}-V_{1}\right)+8V_{1}+30V_{2}=0$$

Use the distributive property to multiply 3 by V_{2}-V_{1}.

$$3V_{2}-3V_{1}+8V_{1}+30V_{2}=0$$

Combine -3V_{1} and 8V_{1} to get 5V_{1}.

$$3V_{2}+5V_{1}+30V_{2}=0$$

Combine 3V_{2} and 30V_{2} to get 33V_{2}.

$$33V_{2}+5V_{1}=0$$

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.

$$7V_{1}-V_{2}=360,5V_{1}+33V_{2}=0$$

To make 7V_{1} and 5V_{1} equal, multiply all terms on each side of the first equation by 5 and all terms on each side of the second by 7.

$$5\times 7V_{1}+5\left(-1\right)V_{2}=5\times 360,7\times 5V_{1}+7\times 33V_{2}=0$$

Simplify.

$$35V_{1}-5V_{2}=1800,35V_{1}+231V_{2}=0$$

Subtract 35V_{1}+231V_{2}=0 from 35V_{1}-5V_{2}=1800 by subtracting like terms on each side of the equal sign.

$$35V_{1}-35V_{1}-5V_{2}-231V_{2}=1800$$

Add 35V_{1} to -35V_{1}. Terms 35V_{1} and -35V_{1} cancel out, leaving an equation with only one variable that can be solved.

$$-5V_{2}-231V_{2}=1800$$

Add -5V_{2} to -231V_{2}.

$$-236V_{2}=1800$$

Divide both sides by -236.

$$V_{2}=-\frac{450}{59}$$

Substitute -\frac{450}{59} for V_{2} in 5V_{1}+33V_{2}=0. Because the resulting equation contains only one variable, you can solve for V_{1} directly.

$$5V_{1}+33\left(-\frac{450}{59}\right)=0$$

Multiply 33 times -\frac{450}{59}.

$$5V_{1}-\frac{14850}{59}=0$$

Add \frac{14850}{59} to both sides of the equation.

$$5V_{1}=\frac{14850}{59}$$

Divide both sides by 5.

$$V_{1}=\frac{2970}{59}$$

The system is now solved.

$$V_{1}=\frac{2970}{59},V_{2}=-\frac{450}{59}$$