Solve {l}{frac{30-v_{0}}{60}+frac{v_{1}-3v_{0}}{30}-frac{v_{0}}{29}=0}{-frac{30-v_{0}}{60}+3-frac{v_{1}}{60}=0}

$$\left. \begin{array} { l } { \frac { 30 - v _ { 0 } } { 60 } + \frac { v _ { 1 } - 3 v _ { 0 } } { 30 } - \frac { v _ { 0 } } { 29 } = 0 } \\ { - \frac { 30 - v _ { 0 } } { 60 } + 3 - \frac { v _ { 1 } } { 60 } = 0 } \end{array} \right.$$

Solve for v_0, v_1

$$v_{0} = \frac{1914}{41} = 46\frac{28}{41} \approx 46.682926829$$

$$v_{1} = \frac{8064}{41} = 196\frac{28}{41} \approx 196.682926829$$

Steps Using Substitution

Consider the first equation. Multiply both sides of the equation by 1740, the least common multiple of 60,30,29.

$$29\left(30-v_{0}\right)+58\left(v_{1}-3v_{0}\right)-60v_{0}=0$$

Use the distributive property to multiply 29 by 30-v_{0}.

$$870-29v_{0}+58\left(v_{1}-3v_{0}\right)-60v_{0}=0$$

Use the distributive property to multiply 58 by v_{1}-3v_{0}.

$$870-29v_{0}+58v_{1}-174v_{0}-60v_{0}=0$$

Combine -29v_{0} and -174v_{0} to get -203v_{0}.

$$870-203v_{0}+58v_{1}-60v_{0}=0$$

Combine -203v_{0} and -60v_{0} to get -263v_{0}.

$$870-263v_{0}+58v_{1}=0$$

Subtract 870 from both sides. Anything subtracted from zero gives its negation.

$$-263v_{0}+58v_{1}=-870$$

Consider the second equation. Multiply both sides of the equation by 60.

$$-\left(30-v_{0}\right)+180-v_{1}=0$$

To find the opposite of 30-v_{0}, find the opposite of each term.

$$-30+v_{0}+180-v_{1}=0$$

Add -30 and 180 to get 150.

$$150+v_{0}-v_{1}=0$$

Subtract 150 from both sides. Anything subtracted from zero gives its negation.

$$v_{0}-v_{1}=-150$$

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.

$$-263v_{0}+58v_{1}=-870,v_{0}-v_{1}=-150$$

Choose one of the equations and solve it for v_{0} by isolating v_{0} on the left hand side of the equal sign.

$$-263v_{0}+58v_{1}=-870$$

Subtract 58v_{1} from both sides of the equation.

$$-263v_{0}=-58v_{1}-870$$

Divide both sides by -263.

$$v_{0}=-\frac{1}{263}\left(-58v_{1}-870\right)$$

Multiply -\frac{1}{263} times -58v_{1}-870.

$$v_{0}=\frac{58}{263}v_{1}+\frac{870}{263}$$

Substitute \frac{870+58v_{1}}{263} for v_{0} in the other equation, v_{0}-v_{1}=-150.

$$\frac{58}{263}v_{1}+\frac{870}{263}-v_{1}=-150$$

Add \frac{58v_{1}}{263} to -v_{1}.

$$-\frac{205}{263}v_{1}+\frac{870}{263}=-150$$

Subtract \frac{870}{263} from both sides of the equation.

$$-\frac{205}{263}v_{1}=-\frac{40320}{263}$$

Divide both sides of the equation by -\frac{205}{263}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$v_{1}=\frac{8064}{41}$$

Substitute \frac{8064}{41} for v_{1} in v_{0}=\frac{58}{263}v_{1}+\frac{870}{263}. Because the resulting equation contains only one variable, you can solve for v_{0} directly.

$$v_{0}=\frac{58}{263}\times \frac{8064}{41}+\frac{870}{263}$$

Multiply \frac{58}{263} times \frac{8064}{41} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$v_{0}=\frac{467712}{10783}+\frac{870}{263}$$

Add \frac{870}{263} to \frac{467712}{10783} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$v_{0}=\frac{1914}{41}$$

The system is now solved.

$$v_{0}=\frac{1914}{41},v_{1}=\frac{8064}{41}$$

Steps Using Matrices

Consider the first equation. Multiply both sides of the equation by 1740, the least common multiple of 60,30,29.

$$29\left(30-v_{0}\right)+58\left(v_{1}-3v_{0}\right)-60v_{0}=0$$

Use the distributive property to multiply 29 by 30-v_{0}.

$$870-29v_{0}+58\left(v_{1}-3v_{0}\right)-60v_{0}=0$$

Use the distributive property to multiply 58 by v_{1}-3v_{0}.

$$870-29v_{0}+58v_{1}-174v_{0}-60v_{0}=0$$

Combine -29v_{0} and -174v_{0} to get -203v_{0}.

$$870-203v_{0}+58v_{1}-60v_{0}=0$$

Combine -203v_{0} and -60v_{0} to get -263v_{0}.

$$870-263v_{0}+58v_{1}=0$$

Subtract 870 from both sides. Anything subtracted from zero gives its negation.

$$-263v_{0}+58v_{1}=-870$$

Consider the second equation. Multiply both sides of the equation by 60.

$$-\left(30-v_{0}\right)+180-v_{1}=0$$

To find the opposite of 30-v_{0}, find the opposite of each term.

$$-30+v_{0}+180-v_{1}=0$$

Add -30 and 180 to get 150.

$$150+v_{0}-v_{1}=0$$

Subtract 150 from both sides. Anything subtracted from zero gives its negation.

$$v_{0}-v_{1}=-150$$

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

$$-263v_{0}+58v_{1}=-870,v_{0}-v_{1}=-150$$

Write the equations in matrix form.

$$\left(\begin{matrix}-263&58\\1&-1\end{matrix}\right)\left(\begin{matrix}v_{0}\\v_{1}\end{matrix}\right)=\left(\begin{matrix}-870\\-150\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}-263&58\\1&-1\end{matrix}\right).

$$inverse(\left(\begin{matrix}-263&58\\1&-1\end{matrix}\right))\left(\begin{matrix}-263&58\\1&-1\end{matrix}\right)\left(\begin{matrix}v_{0}\\v_{1}\end{matrix}\right)=inverse(\left(\begin{matrix}-263&58\\1&-1\end{matrix}\right))\left(\begin{matrix}-870\\-150\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}v_{0}\\v_{1}\end{matrix}\right)=inverse(\left(\begin{matrix}-263&58\\1&-1\end{matrix}\right))\left(\begin{matrix}-870\\-150\end{matrix}\right)$$

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

$$\left(\begin{matrix}v_{0}\\v_{1}\end{matrix}\right)=inverse(\left(\begin{matrix}-263&58\\1&-1\end{matrix}\right))\left(\begin{matrix}-870\\-150\end{matrix}\right)$$

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.

$$\left(\begin{matrix}v_{0}\\v_{1}\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{-263\left(-1\right)-58}&-\frac{58}{-263\left(-1\right)-58}\\-\frac{1}{-263\left(-1\right)-58}&-\frac{263}{-263\left(-1\right)-58}\end{matrix}\right)\left(\begin{matrix}-870\\-150\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}v_{0}\\v_{1}\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{205}&-\frac{58}{205}\\-\frac{1}{205}&-\frac{263}{205}\end{matrix}\right)\left(\begin{matrix}-870\\-150\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}v_{0}\\v_{1}\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{205}\left(-870\right)-\frac{58}{205}\left(-150\right)\\-\frac{1}{205}\left(-870\right)-\frac{263}{205}\left(-150\right)\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}v_{0}\\v_{1}\end{matrix}\right)=\left(\begin{matrix}\frac{1914}{41}\\\frac{8064}{41}\end{matrix}\right)$$

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

$$v_{0}=\frac{1914}{41},v_{1}=\frac{8064}{41}$$

Steps Using Elimination

Consider the first equation. Multiply both sides of the equation by 1740, the least common multiple of 60,30,29.

$$29\left(30-v_{0}\right)+58\left(v_{1}-3v_{0}\right)-60v_{0}=0$$

Use the distributive property to multiply 29 by 30-v_{0}.

$$870-29v_{0}+58\left(v_{1}-3v_{0}\right)-60v_{0}=0$$

Use the distributive property to multiply 58 by v_{1}-3v_{0}.

$$870-29v_{0}+58v_{1}-174v_{0}-60v_{0}=0$$

Combine -29v_{0} and -174v_{0} to get -203v_{0}.

$$870-203v_{0}+58v_{1}-60v_{0}=0$$

Combine -203v_{0} and -60v_{0} to get -263v_{0}.

$$870-263v_{0}+58v_{1}=0$$

Subtract 870 from both sides. Anything subtracted from zero gives its negation.

$$-263v_{0}+58v_{1}=-870$$

Consider the second equation. Multiply both sides of the equation by 60.

$$-\left(30-v_{0}\right)+180-v_{1}=0$$

To find the opposite of 30-v_{0}, find the opposite of each term.

$$-30+v_{0}+180-v_{1}=0$$

Add -30 and 180 to get 150.

$$150+v_{0}-v_{1}=0$$

Subtract 150 from both sides. Anything subtracted from zero gives its negation.

$$v_{0}-v_{1}=-150$$

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.

$$-263v_{0}+58v_{1}=-870,v_{0}-v_{1}=-150$$

To make -263v_{0} and v_{0} equal, multiply all terms on each side of the first equation by 1 and all terms on each side of the second by -263.

$$-263v_{0}+58v_{1}=-870,-263v_{0}-263\left(-1\right)v_{1}=-263\left(-150\right)$$

Simplify.

$$-263v_{0}+58v_{1}=-870,-263v_{0}+263v_{1}=39450$$

Subtract -263v_{0}+263v_{1}=39450 from -263v_{0}+58v_{1}=-870 by subtracting like terms on each side of the equal sign.

$$-263v_{0}+263v_{0}+58v_{1}-263v_{1}=-870-39450$$

Add -263v_{0} to 263v_{0}. Terms -263v_{0} and 263v_{0} cancel out, leaving an equation with only one variable that can be solved.

$$58v_{1}-263v_{1}=-870-39450$$

Add 58v_{1} to -263v_{1}.

$$-205v_{1}=-870-39450$$

Add -870 to -39450.

$$-205v_{1}=-40320$$

Divide both sides by -205.

$$v_{1}=\frac{8064}{41}$$

Substitute \frac{8064}{41} for v_{1} in v_{0}-v_{1}=-150. Because the resulting equation contains only one variable, you can solve for v_{0} directly.

$$v_{0}-\frac{8064}{41}=-150$$

Add \frac{8064}{41} to both sides of the equation.

$$v_{0}=\frac{1914}{41}$$

The system is now solved.

$$v_{0}=\frac{1914}{41},v_{1}=\frac{8064}{41}$$