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maxonik [38]
4 years ago
12

Solve the following system of equations. Write each of your answers as a fraction reduced to lowest terms. In other words, write

the numbers in the exact form that the row operation tool gives them to you when you use the tool in fraction mode. No decimal answers are permitted.
15x + 15y + 10z = 106
5x + 15y + 25z = 135
15x + 10y - 5z = 42
x = _____
y = _____
z = _____
Mathematics
1 answer:
Genrish500 [490]4 years ago
4 0

Answer:

x\ =\ \dfrac{209}{30}

y\ =\ \dfrac{29}{18}

z\ =\ \dfrac{64}{15}

Step-by-step explanation:

Given equations are

15x + 15y + 10z = 106

5x + 15y + 25z = 135

15x + 10y - 5z = 42

The augmented matrix by using above equations can be written as

\left[\begin{array}{ccc}15&15&10\ \ |106\\5&15&25\ \ |135\\15&10&-5|42\end{array}\right]

R_1\ \rightarrow\ \dfrac{R_1}{15}

=\ \left[\begin{array}{ccc}1&1&\dfrac{10}{15}|\dfrac{106}{15}\\5&15&25|135\\15&15&-5|42\end{array}\right]

R_1\rightarrowR_2-5R1\ and\ R_3\rightarrow\ R_3-15R_1

=\ \left[\begin{array}{ccc}1&1&\dfrac{10}{15}|\dfrac{106}{15}\\\\0&10&\dfrac{65}{3}|\dfrac{299}{3}\\\\0&0&-15|-64\end{array}\right]

R_2\rightarrow\ \dfrac{R_2}{10}

=\ \left[\begin{array}{ccc}1&1&\dfrac{10}{15}|\dfrac{106}{15}\\\\0&1&\dfrac{65}{30}|\dfrac{299}{30}\\\\0&0&-15|-64\end{array}\right]

R_3\rightarrow\ \dfrac{R_3}{-15}

=\ \left[\begin{array}{ccc}1&1&\dfrac{10}{15}|\dfrac{106}{15}\\\\0&1&\dfrac{65}{30}|\dfrac{299}{30}\\\\0&0&1|\dfrac{64}{15}\end{array}\right]

R_1\rightarrow\ R_1-R_2

=\ \left[\begin{array}{ccc}1&0&\dfrac{-3}{2}|\dfrac{17}{30}\\\\0&1&\dfrac{65}{30}|\dfrac{299}{30}\\\\0&0&1|\dfrac{64}{15}\end{array}\right]

R_1\rightarrow\ R_1+\dfrac{3}{2}R_3

=\ \left[\begin{array}{ccc}1&0&0|\dfrac{209}{30}\\\\0&1&\dfrac{65}{30}|\dfrac{299}{30}\\\\0&0&1|\dfrac{64}{15}\end{array}\right]

R_2\rightarrow\ R_2-\dfrac{65}{30}R_3

=\ \left[\begin{array}{ccc}1&0&\0|\dfrac{209}{30}\\\\0&1&0|\dfrac{29}{18}\\\\0&0&1|\dfrac{64}{15}\end{array}\right]

Hence, we can write from augmented matrix,

x\ =\ \dfrac{209}{30}

y\ =\ \dfrac{29}{18}

z\ =\ \dfrac{64}{15}

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