11.2 Systems of linear equations: three variables  (Page 5/8)

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Algebraic

For the following exercises, determine whether the ordered triple given is the solution to the system of equations.

and $\text{\hspace{0.17em}}\left(0,1,-1\right)$

and $\left(3,-3,-5\right)$

No

and $\text{\hspace{0.17em}}\left(4,2,-6\right)$

and $\text{\hspace{0.17em}}\left(4,4,-1\right)$

Yes

and $\text{\hspace{0.17em}}\left(4,1,-7\right)$

For the following exercises, solve each system by substitution.

$\left(-1,4,2\right)$

$\left(-\frac{85}{107},\frac{312}{107},\frac{191}{107}\right)$

$\left(1,\frac{1}{2},0\right)$

For the following exercises, solve each system by Gaussian elimination.

$\left(4,-6,1\right)$

$\left(x,\frac{1}{27}\left(65-16x\right),\frac{x+28}{27}\right)$

$\begin{array}{l}\text{\hspace{0.17em}}2x+3y-4z=5\hfill \\ -3x+2y+z=11\hfill \\ -x+5y+3z=4\hfill \end{array}$

$\left(-\frac{45}{13},\frac{17}{13},-2\right)$

No solutions exist

$\left(0,0,0\right)$

$\begin{array}{l}3x+2y-5z=6\\ 5x-4y+3z=-12\\ 4x+5y-2z=15\end{array}$

$\left(\frac{4}{7},-\frac{1}{7},-\frac{3}{7}\right)$

$\left(7,20,16\right)$

$\begin{array}{l}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{1}{2}x-\frac{1}{5}y+\frac{2}{5}z=-\frac{13}{10}\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{1}{4}x-\frac{2}{5}y-\frac{1}{5}z=-\frac{7}{20}\hfill \\ -\frac{1}{2}x-\frac{3}{4}y-\frac{1}{2}z=-\frac{5}{4}\hfill \end{array}$

$\begin{array}{l}\begin{array}{l}\\ -\frac{1}{3}x-\frac{1}{2}y-\frac{1}{4}z=\frac{3}{4}\end{array}\hfill \\ -\frac{1}{2}x-\frac{1}{4}y-\frac{1}{2}z=2\hfill \\ -\frac{1}{4}x-\frac{3}{4}y-\frac{1}{2}z=-\frac{1}{2}\hfill \end{array}$

$\left(-6,2,1\right)$

$\begin{array}{l}\frac{1}{2}x-\frac{1}{4}y+\frac{3}{4}z=0\\ \frac{1}{4}x-\frac{1}{10}y+\frac{2}{5}z=-2\\ \frac{1}{8}x+\frac{1}{5}y-\frac{1}{8}z=2\end{array}$

$\left(5,12,15\right)$

$\begin{array}{l}\begin{array}{l}\\ -\frac{1}{3}x-\frac{1}{8}y+\frac{1}{6}z=-\frac{4}{3}\end{array}\hfill \\ -\frac{2}{3}x-\frac{7}{8}y+\frac{1}{3}z=-\frac{23}{3}\hfill \\ -\frac{1}{3}x-\frac{5}{8}y+\frac{5}{6}z=0\hfill \end{array}$

$\begin{array}{l}\begin{array}{l}\\ -\frac{1}{4}x-\frac{5}{4}y+\frac{5}{2}z=-5\end{array}\hfill \\ -\frac{1}{2}x-\frac{5}{3}y+\frac{5}{4}z=\frac{55}{12}\hfill \\ -\frac{1}{3}x-\frac{1}{3}y+\frac{1}{3}z=\frac{5}{3}\hfill \end{array}$

$\left(-5,-5,-5\right)$

$\begin{array}{l}\frac{1}{40}x+\frac{1}{60}y+\frac{1}{80}z=\frac{1}{100}\hfill \\ \text{\hspace{0.17em}}-\frac{1}{2}x-\frac{1}{3}y-\frac{1}{4}z=-\frac{1}{5}\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{3}{8}x+\frac{3}{12}y+\frac{3}{16}z=\frac{3}{20}\hfill \end{array}$

$\begin{array}{l}0.1x-0.2y+0.3z=2\\ 0.5x-0.1y+0.4z=8\\ 0.7x-0.2y+0.3z=8\end{array}$

$\left(10,10,10\right)$

$\begin{array}{l}0.2x+0.1y-0.3z=0.2\\ 0.8x+0.4y-1.2z=0.1\\ 1.6x+0.8y-2.4z=0.2\end{array}$

$\begin{array}{l}1.1x+0.7y-3.1z=-1.79\\ 2.1x+0.5y-1.6z=-0.13\\ 0.5x+0.4y-0.5z=-0.07\end{array}$

$\left(\frac{1}{2},\frac{1}{5},\frac{4}{5}\right)$

$\begin{array}{l}0.5x-0.5y+0.5z=10\\ 0.2x-0.2y+0.2z=4\\ 0.1x-0.1y+0.1z=2\end{array}$

$\begin{array}{l}0.1x+0.2y+0.3z=0.37\\ 0.1x-0.2y-0.3z=-0.27\\ 0.5x-0.1y-0.3z=-0.03\end{array}$

$\left(\frac{1}{2},\frac{2}{5},\frac{4}{5}\right)$

$\begin{array}{l}0.5x-0.5y-0.3z=0.13\\ 0.4x-0.1y-0.3z=0.11\\ 0.2x-0.8y-0.9z=-0.32\end{array}$

$\begin{array}{l}0.5x+0.2y-0.3z=1\\ 0.4x-0.6y+0.7z=0.8\\ 0.3x-0.1y-0.9z=0.6\end{array}$

$\left(2,0,0\right)$

$\begin{array}{l}0.3x+0.3y+0.5z=0.6\\ 0.4x+0.4y+0.4z=1.8\\ 0.4x+0.2y+0.1z=1.6\end{array}$

$\begin{array}{l}0.8x+0.8y+0.8z=2.4\\ 0.3x-0.5y+0.2z=0\\ 0.1x+0.2y+0.3z=0.6\end{array}$

$\left(1,1,1\right)$

Extensions

For the following exercises, solve the system for $\text{\hspace{0.17em}}x,y,$ and $\text{\hspace{0.17em}}z.$

$\left(\frac{128}{557},\frac{23}{557},\frac{28}{557}\right)$

$\begin{array}{l}\frac{x+4}{7}-\frac{y-1}{6}+\frac{z+2}{3}=1\\ \frac{x-2}{4}+\frac{y+1}{8}-\frac{z+8}{12}=0\\ \frac{x+6}{3}-\frac{y+2}{3}+\frac{z+4}{2}=3\end{array}$

$\begin{array}{l}\frac{x-3}{6}+\frac{y+2}{2}-\frac{z-3}{3}=2\\ \frac{x+2}{4}+\frac{y-5}{2}+\frac{z+4}{2}=1\\ \frac{x+6}{2}-\frac{y-3}{2}+z+1=9\end{array}$

$\left(6,-1,0\right)$

Real-world applications

Three even numbers sum up to 108. The smaller is half the larger and the middle number is $\text{\hspace{0.17em}}\frac{3}{4}\text{\hspace{0.17em}}$ the larger. What are the three numbers?

24, 36, 48

Three numbers sum up to 147. The smallest number is half the middle number, which is half the largest number. What are the three numbers?

At a family reunion, there were only blood relatives, consisting of children, parents, and grandparents, in attendance. There were 400 people total. There were twice as many parents as grandparents, and 50 more children than parents. How many children, parents, and grandparents were in attendance?

70 grandparents, 140 parents, 190 children

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