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3 years ago

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How do you know or what indicates to you that the solution set of an equation is all real numbers?

1 answer:

Harrizon [31]3 years ago

3 0

Answer:

the equation is a tautology, or the system of equations is under-specified

Step-by-step explanation:

By definition, an equation that is a tautology is true for all values of the variable(s). Hence, its solution set is "all real numbers."

One such equation is ...

x = x

__

An under-specified system of equations* may also have "all real numbers" as a solution set. For example, two equations in 3 unknowns:

x + y = 3

x + z = 5

The solution set is ...

(x, y, z) = (x, 3-x, 5-x)

Every value of x will give a solution.

__

Similarly, a dependent system of equations may have "all real numbers" as a solution. A system is dependent when one or more of the equations can be derived from the others.

For example, adding the equation 2x+y+z=8 to the above set will give 3 equations in 3 unknowns. However, this added equation can be obtained by adding together the two above. It adds nothing that would restrict the solution set.

A system of two equations in two unknowns will be <em>dependent</em> if one of the equations is a constant multiple of the other. For example, x+y=1, 2x+2y=2 has a second equation that is 2 times the first. This set of equations has "all real numbers" as its solution set. (Of course, y = 1-x.)

_____

* Non-linear equations may impose limits on the variables, so that "all real numbers" cannot be a solution.

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3 years ago

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Solve the 3 × 3 system shown below. Enter the values of x, y, and z. X + 2y – z = –3 (1) 2x – y + z = 5 (2) x – y + z = 4 (3)

Svet_ta [14]

<h2>Answer:</h2>

$\boxed{x=1, \y=-1, \ z=2}$

<h2>Step-by-step explanation:</h2>

We will use the Gaussian elimination method to solve this problem. To do so, let's follow the following steps:

Step 1: Let's multiply first equation by −2. Next, add the result to the second equation. So:

$\begin{array}{ cccc }~~ x&+~~2~ y&-~~~~~ z&~=~-3\\&-~~~5~ y&+~~3~ z&~=~11\\~~ x&-~~~~~ y&+~~~~ z&~=~4\end{array}$

Step 2: Let's multiply first equation by −1. Next, add the result to the third equation. Thus:

$\begin{array}{ cccc }~~ x&+~~2~ y&-~~~~~ z&~=~-3\\&-~~~5~ y&+~~3~ z&~=~11\\&-~~~3~ y&+~~2~ z&~=~7\end{array}$

Step 3: Let's multiply second equation by −35, Next, add the result to the third equation. Therefore:

$\begin{array}{ cccc }~~ x&+~~2~ y&-~~~~~ z&~=~-3\\&-~~~5~ y&+~~3~ z&~=~11\\&&+~~\frac{ 1 }{ 5 }~ z&~=~\frac{ 2 }{ 5 }\end{array}$

Step 4: solve for z, then for y, then for x:

$\frac{ 1 }{ 5 } ~ z & = \frac{ 2 }{ 5 } \\ \\ \boxed{z & = 2}$

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kupik [55]

Answer:

P= 28.28 m

diagonal (d) 10m

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