Solve {l}{frac{v_{1}-100}{1}+frac{v_{1}-v_{0}}{2}+12.5=0}{frac{v_{0}-v_{1}}{2}-12.5+frac{v_{0}}{4}+frac{v_{0}+50}{6}=0}

$$\left\{ \begin{array} { l } { \frac { v _ { 1 } - 100 } { 1 } + \frac { v _ { 1 } - v _ { 0 } } { 2 } + 12.5 = 0 } \\ { \frac { v _ { 0 } - v _ { 1 } } { 2 } - 12.5 + \frac { v _ { 0 } } { 4 } + \frac { v _ { 0 } + 50 } { 6 } = 0 } \end{array} \right.$$

Solve for v_1, v_0

$$v_{1} = \frac{1975}{27} = 73\frac{4}{27} \approx 73.148148148$$

$$v_{0} = \frac{400}{9} = 44\frac{4}{9} \approx 44.444444444$$

Steps Using Substitution

Consider the first equation. Multiply both sides of the equation by 2.

$$2\left(v_{1}-100\right)+v_{1}-v_{0}+25=0$$

Use the distributive property to multiply 2 by v_{1}-100.

$$2v_{1}-200+v_{1}-v_{0}+25=0$$

Combine 2v_{1} and v_{1} to get 3v_{1}.

$$3v_{1}-200-v_{0}+25=0$$

Add -200 and 25 to get -175.

$$3v_{1}-175-v_{0}=0$$

Add 175 to both sides. Anything plus zero gives itself.

$$3v_{1}-v_{0}=175$$

Consider the second equation. Multiply both sides of the equation by 12, the least common multiple of 2,4,6.

$$6\left(v_{0}-v_{1}\right)-150+3v_{0}+2\left(v_{0}+50\right)=0$$

Use the distributive property to multiply 6 by v_{0}-v_{1}.

$$6v_{0}-6v_{1}-150+3v_{0}+2\left(v_{0}+50\right)=0$$

Combine 6v_{0} and 3v_{0} to get 9v_{0}.

$$9v_{0}-6v_{1}-150+2\left(v_{0}+50\right)=0$$

Use the distributive property to multiply 2 by v_{0}+50.

$$9v_{0}-6v_{1}-150+2v_{0}+100=0$$

Combine 9v_{0} and 2v_{0} to get 11v_{0}.

$$11v_{0}-6v_{1}-150+100=0$$

Add -150 and 100 to get -50.

$$11v_{0}-6v_{1}-50=0$$

Add 50 to both sides. Anything plus zero gives itself.

$$11v_{0}-6v_{1}=50$$

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.

$$3v_{1}-v_{0}=175,-6v_{1}+11v_{0}=50$$

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

$$3v_{1}-v_{0}=175$$

Add v_{0} to both sides of the equation.

$$3v_{1}=v_{0}+175$$

Divide both sides by 3.

$$v_{1}=\frac{1}{3}\left(v_{0}+175\right)$$

Multiply \frac{1}{3} times v_{0}+175.

$$v_{1}=\frac{1}{3}v_{0}+\frac{175}{3}$$

Substitute \frac{175+v_{0}}{3} for v_{1} in the other equation, -6v_{1}+11v_{0}=50.

$$-6\left(\frac{1}{3}v_{0}+\frac{175}{3}\right)+11v_{0}=50$$

Multiply -6 times \frac{175+v_{0}}{3}.

$$-2v_{0}-350+11v_{0}=50$$

Add -2v_{0} to 11v_{0}.

$$9v_{0}-350=50$$

Add 350 to both sides of the equation.

$$9v_{0}=400$$

Divide both sides by 9.

$$v_{0}=\frac{400}{9}$$

Substitute \frac{400}{9} for v_{0} in v_{1}=\frac{1}{3}v_{0}+\frac{175}{3}. Because the resulting equation contains only one variable, you can solve for v_{1} directly.

$$v_{1}=\frac{1}{3}\times \frac{400}{9}+\frac{175}{3}$$

Multiply \frac{1}{3} times \frac{400}{9} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$v_{1}=\frac{400}{27}+\frac{175}{3}$$

Add \frac{175}{3} to \frac{400}{27} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$v_{1}=\frac{1975}{27}$$

The system is now solved.

$$v_{1}=\frac{1975}{27},v_{0}=\frac{400}{9}$$

Steps Using Matrices

Consider the first equation. Multiply both sides of the equation by 2.

$$2\left(v_{1}-100\right)+v_{1}-v_{0}+25=0$$

Use the distributive property to multiply 2 by v_{1}-100.

$$2v_{1}-200+v_{1}-v_{0}+25=0$$

Combine 2v_{1} and v_{1} to get 3v_{1}.

$$3v_{1}-200-v_{0}+25=0$$

Add -200 and 25 to get -175.

$$3v_{1}-175-v_{0}=0$$

Add 175 to both sides. Anything plus zero gives itself.

$$3v_{1}-v_{0}=175$$

Consider the second equation. Multiply both sides of the equation by 12, the least common multiple of 2,4,6.

$$6\left(v_{0}-v_{1}\right)-150+3v_{0}+2\left(v_{0}+50\right)=0$$

Use the distributive property to multiply 6 by v_{0}-v_{1}.

$$6v_{0}-6v_{1}-150+3v_{0}+2\left(v_{0}+50\right)=0$$

Combine 6v_{0} and 3v_{0} to get 9v_{0}.

$$9v_{0}-6v_{1}-150+2\left(v_{0}+50\right)=0$$

Use the distributive property to multiply 2 by v_{0}+50.

$$9v_{0}-6v_{1}-150+2v_{0}+100=0$$

Combine 9v_{0} and 2v_{0} to get 11v_{0}.

$$11v_{0}-6v_{1}-150+100=0$$

Add -150 and 100 to get -50.

$$11v_{0}-6v_{1}-50=0$$

Add 50 to both sides. Anything plus zero gives itself.

$$11v_{0}-6v_{1}=50$$

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

$$3v_{1}-v_{0}=175,-6v_{1}+11v_{0}=50$$

Write the equations in matrix form.

$$\left(\begin{matrix}3&-1\\-6&11\end{matrix}\right)\left(\begin{matrix}v_{1}\\v_{0}\end{matrix}\right)=\left(\begin{matrix}175\\50\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}3&-1\\-6&11\end{matrix}\right).

$$inverse(\left(\begin{matrix}3&-1\\-6&11\end{matrix}\right))\left(\begin{matrix}3&-1\\-6&11\end{matrix}\right)\left(\begin{matrix}v_{1}\\v_{0}\end{matrix}\right)=inverse(\left(\begin{matrix}3&-1\\-6&11\end{matrix}\right))\left(\begin{matrix}175\\50\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}v_{1}\\v_{0}\end{matrix}\right)=inverse(\left(\begin{matrix}3&-1\\-6&11\end{matrix}\right))\left(\begin{matrix}175\\50\end{matrix}\right)$$

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

$$\left(\begin{matrix}v_{1}\\v_{0}\end{matrix}\right)=inverse(\left(\begin{matrix}3&-1\\-6&11\end{matrix}\right))\left(\begin{matrix}175\\50\end{matrix}\right)$$

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.

$$\left(\begin{matrix}v_{1}\\v_{0}\end{matrix}\right)=\left(\begin{matrix}\frac{11}{3\times 11-\left(-\left(-6\right)\right)}&-\frac{-1}{3\times 11-\left(-\left(-6\right)\right)}\\-\frac{-6}{3\times 11-\left(-\left(-6\right)\right)}&\frac{3}{3\times 11-\left(-\left(-6\right)\right)}\end{matrix}\right)\left(\begin{matrix}175\\50\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}v_{1}\\v_{0}\end{matrix}\right)=\left(\begin{matrix}\frac{11}{27}&\frac{1}{27}\\\frac{2}{9}&\frac{1}{9}\end{matrix}\right)\left(\begin{matrix}175\\50\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}v_{1}\\v_{0}\end{matrix}\right)=\left(\begin{matrix}\frac{11}{27}\times 175+\frac{1}{27}\times 50\\\frac{2}{9}\times 175+\frac{1}{9}\times 50\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}v_{1}\\v_{0}\end{matrix}\right)=\left(\begin{matrix}\frac{1975}{27}\\\frac{400}{9}\end{matrix}\right)$$

Extract the matrix elements v_{1} and v_{0}.

$$v_{1}=\frac{1975}{27},v_{0}=\frac{400}{9}$$

Steps Using Elimination

Consider the first equation. Multiply both sides of the equation by 2.

$$2\left(v_{1}-100\right)+v_{1}-v_{0}+25=0$$

Use the distributive property to multiply 2 by v_{1}-100.

$$2v_{1}-200+v_{1}-v_{0}+25=0$$

Combine 2v_{1} and v_{1} to get 3v_{1}.

$$3v_{1}-200-v_{0}+25=0$$

Add -200 and 25 to get -175.

$$3v_{1}-175-v_{0}=0$$

Add 175 to both sides. Anything plus zero gives itself.

$$3v_{1}-v_{0}=175$$

Consider the second equation. Multiply both sides of the equation by 12, the least common multiple of 2,4,6.

$$6\left(v_{0}-v_{1}\right)-150+3v_{0}+2\left(v_{0}+50\right)=0$$

Use the distributive property to multiply 6 by v_{0}-v_{1}.

$$6v_{0}-6v_{1}-150+3v_{0}+2\left(v_{0}+50\right)=0$$

Combine 6v_{0} and 3v_{0} to get 9v_{0}.

$$9v_{0}-6v_{1}-150+2\left(v_{0}+50\right)=0$$

Use the distributive property to multiply 2 by v_{0}+50.

$$9v_{0}-6v_{1}-150+2v_{0}+100=0$$

Combine 9v_{0} and 2v_{0} to get 11v_{0}.

$$11v_{0}-6v_{1}-150+100=0$$

Add -150 and 100 to get -50.

$$11v_{0}-6v_{1}-50=0$$

Add 50 to both sides. Anything plus zero gives itself.

$$11v_{0}-6v_{1}=50$$

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.

$$3v_{1}-v_{0}=175,-6v_{1}+11v_{0}=50$$

To make 3v_{1} and -6v_{1} equal, multiply all terms on each side of the first equation by -6 and all terms on each side of the second by 3.

$$-6\times 3v_{1}-6\left(-1\right)v_{0}=-6\times 175,3\left(-6\right)v_{1}+3\times 11v_{0}=3\times 50$$

Simplify.

$$-18v_{1}+6v_{0}=-1050,-18v_{1}+33v_{0}=150$$

Subtract -18v_{1}+33v_{0}=150 from -18v_{1}+6v_{0}=-1050 by subtracting like terms on each side of the equal sign.

$$-18v_{1}+18v_{1}+6v_{0}-33v_{0}=-1050-150$$

Add -18v_{1} to 18v_{1}. Terms -18v_{1} and 18v_{1} cancel out, leaving an equation with only one variable that can be solved.

$$6v_{0}-33v_{0}=-1050-150$$

Add 6v_{0} to -33v_{0}.

$$-27v_{0}=-1050-150$$

Add -1050 to -150.

$$-27v_{0}=-1200$$

Divide both sides by -27.

$$v_{0}=\frac{400}{9}$$

Substitute \frac{400}{9} for v_{0} in -6v_{1}+11v_{0}=50. Because the resulting equation contains only one variable, you can solve for v_{1} directly.

$$-6v_{1}+11\times \frac{400}{9}=50$$

Multiply 11 times \frac{400}{9}.

$$-6v_{1}+\frac{4400}{9}=50$$

Subtract \frac{4400}{9} from both sides of the equation.

$$-6v_{1}=-\frac{3950}{9}$$

Divide both sides by -6.

$$v_{1}=\frac{1975}{27}$$

The system is now solved.

$$v_{1}=\frac{1975}{27},v_{0}=\frac{400}{9}$$