Solve {l}{frac{40-V_{1}}{10}-3-frac{V_{1}}{40}-frac{V_{1}-V_{2}}{26}=0}{3+frac{V_{1}-V_{2}}{26}=0}

$$\left. \begin{array} { l } { \frac { 40 - V _ { 1 } } { 10 } - 3 - \frac { V _ { 1 } } { 40 } - \frac { V _ { 1 } - V _ { 2 } } { 26 } = 0 } \\ { 3 + \frac { V _ { 1 } - V _ { 2 } } { 26 } = 0 } \end{array} \right.$$

Solve for V_1, V_2

$$V_{1}=32$$

$$V_{2}=110$$

Steps Using Substitution

Consider the first equation. Multiply both sides of the equation by 520, the least common multiple of 10,40,26.

$$52\left(40-V_{1}\right)-1560-13V_{1}-20\left(V_{1}-V_{2}\right)=0$$

Use the distributive property to multiply 52 by 40-V_{1}.

$$2080-52V_{1}-1560-13V_{1}-20\left(V_{1}-V_{2}\right)=0$$

Subtract 1560 from 2080 to get 520.

$$520-52V_{1}-13V_{1}-20\left(V_{1}-V_{2}\right)=0$$

Combine -52V_{1} and -13V_{1} to get -65V_{1}.

$$520-65V_{1}-20\left(V_{1}-V_{2}\right)=0$$

Use the distributive property to multiply -20 by V_{1}-V_{2}.

$$520-65V_{1}-20V_{1}+20V_{2}=0$$

Combine -65V_{1} and -20V_{1} to get -85V_{1}.

$$520-85V_{1}+20V_{2}=0$$

Subtract 520 from both sides. Anything subtracted from zero gives its negation.

$$-85V_{1}+20V_{2}=-520$$

Consider the second equation. Multiply both sides of the equation by 26.

$$78+V_{1}-V_{2}=0$$

Subtract 78 from both sides. Anything subtracted from zero gives its negation.

$$V_{1}-V_{2}=-78$$

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.

$$-85V_{1}+20V_{2}=-520,V_{1}-V_{2}=-78$$

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

$$-85V_{1}+20V_{2}=-520$$

Subtract 20V_{2} from both sides of the equation.

$$-85V_{1}=-20V_{2}-520$$

Divide both sides by -85.

$$V_{1}=-\frac{1}{85}\left(-20V_{2}-520\right)$$

Multiply -\frac{1}{85} times -20V_{2}-520.

$$V_{1}=\frac{4}{17}V_{2}+\frac{104}{17}$$

Substitute \frac{104+4V_{2}}{17} for V_{1} in the other equation, V_{1}-V_{2}=-78.

$$\frac{4}{17}V_{2}+\frac{104}{17}-V_{2}=-78$$

Add \frac{4V_{2}}{17} to -V_{2}.

$$-\frac{13}{17}V_{2}+\frac{104}{17}=-78$$

Subtract \frac{104}{17} from both sides of the equation.

$$-\frac{13}{17}V_{2}=-\frac{1430}{17}$$

Divide both sides of the equation by -\frac{13}{17}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$V_{2}=110$$

Substitute 110 for V_{2} in V_{1}=\frac{4}{17}V_{2}+\frac{104}{17}. Because the resulting equation contains only one variable, you can solve for V_{1} directly.

$$V_{1}=\frac{4}{17}\times 110+\frac{104}{17}$$

Multiply \frac{4}{17} times 110.

$$V_{1}=\frac{440+104}{17}$$

Add \frac{104}{17} to \frac{440}{17} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$V_{1}=32$$

The system is now solved.

$$V_{1}=32,V_{2}=110$$

Steps Using Matrices

Consider the first equation. Multiply both sides of the equation by 520, the least common multiple of 10,40,26.

$$52\left(40-V_{1}\right)-1560-13V_{1}-20\left(V_{1}-V_{2}\right)=0$$

Use the distributive property to multiply 52 by 40-V_{1}.

$$2080-52V_{1}-1560-13V_{1}-20\left(V_{1}-V_{2}\right)=0$$

Subtract 1560 from 2080 to get 520.

$$520-52V_{1}-13V_{1}-20\left(V_{1}-V_{2}\right)=0$$

Combine -52V_{1} and -13V_{1} to get -65V_{1}.

$$520-65V_{1}-20\left(V_{1}-V_{2}\right)=0$$

Use the distributive property to multiply -20 by V_{1}-V_{2}.

$$520-65V_{1}-20V_{1}+20V_{2}=0$$

Combine -65V_{1} and -20V_{1} to get -85V_{1}.

$$520-85V_{1}+20V_{2}=0$$

Subtract 520 from both sides. Anything subtracted from zero gives its negation.

$$-85V_{1}+20V_{2}=-520$$

Consider the second equation. Multiply both sides of the equation by 26.

$$78+V_{1}-V_{2}=0$$

Subtract 78 from both sides. Anything subtracted from zero gives its negation.

$$V_{1}-V_{2}=-78$$

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

$$-85V_{1}+20V_{2}=-520,V_{1}-V_{2}=-78$$

Write the equations in matrix form.

$$\left(\begin{matrix}-85&20\\1&-1\end{matrix}\right)\left(\begin{matrix}V_{1}\\V_{2}\end{matrix}\right)=\left(\begin{matrix}-520\\-78\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}-85&20\\1&-1\end{matrix}\right).

$$inverse(\left(\begin{matrix}-85&20\\1&-1\end{matrix}\right))\left(\begin{matrix}-85&20\\1&-1\end{matrix}\right)\left(\begin{matrix}V_{1}\\V_{2}\end{matrix}\right)=inverse(\left(\begin{matrix}-85&20\\1&-1\end{matrix}\right))\left(\begin{matrix}-520\\-78\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}V_{1}\\V_{2}\end{matrix}\right)=inverse(\left(\begin{matrix}-85&20\\1&-1\end{matrix}\right))\left(\begin{matrix}-520\\-78\end{matrix}\right)$$

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

$$\left(\begin{matrix}V_{1}\\V_{2}\end{matrix}\right)=inverse(\left(\begin{matrix}-85&20\\1&-1\end{matrix}\right))\left(\begin{matrix}-520\\-78\end{matrix}\right)$$

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.

$$\left(\begin{matrix}V_{1}\\V_{2}\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{-85\left(-1\right)-20}&-\frac{20}{-85\left(-1\right)-20}\\-\frac{1}{-85\left(-1\right)-20}&-\frac{85}{-85\left(-1\right)-20}\end{matrix}\right)\left(\begin{matrix}-520\\-78\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}V_{1}\\V_{2}\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{65}&-\frac{4}{13}\\-\frac{1}{65}&-\frac{17}{13}\end{matrix}\right)\left(\begin{matrix}-520\\-78\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}V_{1}\\V_{2}\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{65}\left(-520\right)-\frac{4}{13}\left(-78\right)\\-\frac{1}{65}\left(-520\right)-\frac{17}{13}\left(-78\right)\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}V_{1}\\V_{2}\end{matrix}\right)=\left(\begin{matrix}32\\110\end{matrix}\right)$$

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

$$V_{1}=32,V_{2}=110$$

Steps Using Elimination

Consider the first equation. Multiply both sides of the equation by 520, the least common multiple of 10,40,26.

$$52\left(40-V_{1}\right)-1560-13V_{1}-20\left(V_{1}-V_{2}\right)=0$$

Use the distributive property to multiply 52 by 40-V_{1}.

$$2080-52V_{1}-1560-13V_{1}-20\left(V_{1}-V_{2}\right)=0$$

Subtract 1560 from 2080 to get 520.

$$520-52V_{1}-13V_{1}-20\left(V_{1}-V_{2}\right)=0$$

Combine -52V_{1} and -13V_{1} to get -65V_{1}.

$$520-65V_{1}-20\left(V_{1}-V_{2}\right)=0$$

Use the distributive property to multiply -20 by V_{1}-V_{2}.

$$520-65V_{1}-20V_{1}+20V_{2}=0$$

Combine -65V_{1} and -20V_{1} to get -85V_{1}.

$$520-85V_{1}+20V_{2}=0$$

Subtract 520 from both sides. Anything subtracted from zero gives its negation.

$$-85V_{1}+20V_{2}=-520$$

Consider the second equation. Multiply both sides of the equation by 26.

$$78+V_{1}-V_{2}=0$$

Subtract 78 from both sides. Anything subtracted from zero gives its negation.

$$V_{1}-V_{2}=-78$$

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.

$$-85V_{1}+20V_{2}=-520,V_{1}-V_{2}=-78$$

To make -85V_{1} and V_{1} equal, multiply all terms on each side of the first equation by 1 and all terms on each side of the second by -85.

$$-85V_{1}+20V_{2}=-520,-85V_{1}-85\left(-1\right)V_{2}=-85\left(-78\right)$$

Simplify.

$$-85V_{1}+20V_{2}=-520,-85V_{1}+85V_{2}=6630$$

Subtract -85V_{1}+85V_{2}=6630 from -85V_{1}+20V_{2}=-520 by subtracting like terms on each side of the equal sign.

$$-85V_{1}+85V_{1}+20V_{2}-85V_{2}=-520-6630$$

Add -85V_{1} to 85V_{1}. Terms -85V_{1} and 85V_{1} cancel out, leaving an equation with only one variable that can be solved.

$$20V_{2}-85V_{2}=-520-6630$$

Add 20V_{2} to -85V_{2}.

$$-65V_{2}=-520-6630$$

Add -520 to -6630.

$$-65V_{2}=-7150$$

Divide both sides by -65.

$$V_{2}=110$$

Substitute 110 for V_{2} in V_{1}-V_{2}=-78. Because the resulting equation contains only one variable, you can solve for V_{1} directly.

$$V_{1}-110=-78$$

Add 110 to both sides of the equation.

$$V_{1}=32$$

The system is now solved.

$$V_{1}=32,V_{2}=110$$