Answered

Consider the following system of linear equations:

[tex]\[
\begin{array}{l}
x + 4y - 7z = 6 \\
x + y + 2z = -8
\end{array}
\][/tex]

In this question, you are to find three distinct solutions to this system. To allow you to check your solutions independently, the answers are structured to allow you to first enter one solution, then two solutions, and finally all three. The previous solutions must be repeated to ensure your answers are distinct.

1. Enter one solution to this system of equations in the form [tex]\((1, 2/7, 3)\)[/tex]. Note that you should use exact values; for example, you would have to use [tex]\(1/3\)[/tex] rather than 0.3333.

2. Enter two distinct solutions to this system in the form [tex]\((1, 2/7, 3), (4, -1/2, 2)\)[/tex]. You may include your correct answer from the previous part.

3. Now enter three distinct solutions in the form [tex]\((1, 2/7, 3), (4, -1/2, 2), (4, 3, -2)\)[/tex].



Answer :

GinnyAnswer
To find three distinct solutions to the given system of equations:

[tex]\[ \begin{array}{l} x+4 y-7 z=6 \\ x+y+2 z=-8 \end{array} \][/tex]

we can solve it step-by-step as follows:

1. Express one variable in terms of the others: Let's solve the second equation for [tex]\( x \)[/tex]:

[tex]\[ x = -8 - y - 2z \][/tex]

2. Substitute [tex]\( x \)[/tex] into the first equation:

[tex]\[ (-8 - y - 2z) + 4y - 7z = 6 \][/tex]

Simplify to find the relationship between [tex]\( y \)[/tex] and [tex]\( z \)[/tex]:

[tex]\[ -8 - y - 2z + 4y - 7z = 6 \implies -8 + 3y - 9z = 6 \implies 3y - 9z = 14 \implies y = 3z + \frac{14}{3} \][/tex]

3. General solution: Using the relationships:
- [tex]\( y = 3z + \frac{14}{3} \)[/tex]
- [tex]\( x = -8 - y - 2z \)[/tex]

We can substitute [tex]\( y \)[/tex] into [tex]\( x \)[/tex]:

[tex]\[ x = -8 - (3z + \frac{14}{3}) - 2z = -8 - 3z - \frac{14}{3} - 2z = -8 - 5z - \frac{14}{3} \][/tex]

Convert [tex]\(-8\)[/tex] to a fraction:

[tex]\[ x = - \frac{24}{3} - 5z - \frac{14}{3} = - \frac{38}{3} - 5z \][/tex]

So, the general solution is:

[tex]\[ \begin{cases} x = - \frac{38}{3} - 5z \\ y = 3z + \frac{14}{3} \end{cases} \][/tex]

4. Choose distinct values for [tex]\( z \)[/tex]: Since [tex]\( z \)[/tex] is free, we can choose any distinct values for [tex]\( z \)[/tex] to get distinct solutions. Let's choose [tex]\( z = 0 \)[/tex], [tex]\( z = 1 \)[/tex], and [tex]\( z = 2 \)[/tex].

- For [tex]\( z = 0 \)[/tex]:

[tex]\[ \begin{cases} x = - \frac{38}{3} - 5(0) = - \frac{38}{3} \\ y = 3(0) + \frac{14}{3} = \frac{14}{3} \\ z = 0 \end{cases} \][/tex]

- For [tex]\( z = 1 \)[/tex]:

[tex]\[ \begin{cases} x = - \frac{38}{3} - 5(1) = - \frac{38}{3} - \frac{15}{3} = - \frac{53}{3} \\ y = 3(1) + \frac{14}{3} = 3 + \frac{14}{3} = \frac{23}{3} \\ z = 1 \end{cases} \][/tex]

- For [tex]\( z = 2 \)[/tex]:

[tex]\[ \begin{cases} x = - \frac{38}{3} - 5(2) = - \frac{38}{3} - \frac{30}{3} = - \frac{68}{3} \\ y = 3(2) + \frac{14}{3} = 6 + \frac{14}{3} = \frac{32}{3} \\ z = 2 \end{cases} \][/tex]

Therefore, our three distinct solutions are:

1. [tex]\(\left(- \frac{38}{3}, \frac{14}{3}, 0\right)\)[/tex]
2. [tex]\(\left(- \frac{53}{3}, \frac{23}{3}, 1\right)\)[/tex]
3. [tex]\(\left(- \frac{68}{3}, \frac{32}{3}, 2\right)\)[/tex]

So we can present the solutions as:

- One solution: [tex]\(- \left(\frac{38}{3}, \frac{14}{3}, 0\right)\)[/tex]
- Two solutions: [tex]\(- \left(\frac{38}{3}, \frac{14}{3}, 0\right), \left(- \frac{53}{3}, \frac{23}{3}, 1\right)\)[/tex]
- Three solutions: [tex]\(- \left(\frac{38}{3}, \frac{14}{3}, 0\right), \left(- \frac{53}{3}, \frac{23}{3}, 1\right), \left(- \frac{68}{3}, \frac{32}{3}, 2\right)\)[/tex]

Other Questions