Quadratic systems, the use of "and" or "or"

[dolf]

Hi There,

I am new to this forum and wondering if anyone can help,

I have the following question. I can solve the system, and i get the right answers, but i believe the answer is false due to the word used being "or", if it was "and" it would have been correct.

---

B. The solution of the system

. . . . .\(\displaystyle \left. \begin{align} 2x\, -\, y\, &=\, -1 \\ y^2\, -\, x^2\, &=\, \frac{7}{4} \end{align} \right\}\)

... is\(\displaystyle \, x\, =\, \frac{1}{6}\,\) and \(\displaystyle \, y\, =\, \frac{4}{3},\, \) or \(\displaystyle \, x\, =\, -\frac{3}{2}\,\) and \(\displaystyle \, y\, =\, -2.\)

---

Am i correct in making this assumption? And if so, what will the "or" mean in this answer, as opposed to "and"?

 

[mmm4444bot]

Hi There,

I am new to this forum and wondering if anyone can help,

I have the following question. I can solve the system, and i get the right answers, but i believe the answer is false due to the word used being "or", if it was "and" it would have been correct.

---

B. The solution of the system

. . . . .\(\displaystyle \left. \begin{align} 2x\, -\, y\, &=\, -1 \\ y^2\, -\, x^2\, &=\, \frac{7}{4} \end{align} \right\}\)

... is\(\displaystyle \, x\, =\, \frac{1}{6}\,\) and \(\displaystyle \, y\, =\, \frac{4}{3},\, \) or \(\displaystyle \, x\, =\, -\frac{3}{2}\,\) and \(\displaystyle \, y\, =\, -2.\)

---

Am i correct in making this assumption? And if so, what will the "or" mean in this answer, as opposed to "and"?

Welcome to the boards.

The way I think about this is: variables take on only one value at a time. x takes on values from a domain (from least to greatest), and y takes on corresponding values from a range. As we move along a graph, at some points we have a solution, and we have no solution at the other points.

Yes, there are two solutions in your exercise, but they don't occur "at the same time". We're either at one intersection point, OR we're at the other. It's kinda like you can't be in two places at once.

If you're not satisfied with this informal explanation, please feel free to say so. I'm sure other people here can do a better job of explaining it. :cool:

 

[dolf]

Welcome to the boards.

Yes, there are two solutions, but they don't occur "at the same time". We're either at one intersection point, OR we're at the other. Kinda like you can't be in two places at once.

Ahah!

thanks so much for the response. When i look at the equations, I can see that the one would be from a circle, and the other would be from a straight line. In that sense, there could be two intersection points, so would that mean that the use of the "or" terminology would then be correct? seeing that both values would then be on the intersection point?

I know that "and" and "or" refer to intersection and union. But with regards to this question, I just want to make sure that I do have a deep understanding of it, and your response has given me a different way to think of it...

I appreciate any help!

 

[JeffM]

This is a different way to express a viewpoint that is not much different from (perhaps the same as) mmm's view.

When we are asked to solve an equation we are asked to find A member of the set of numbers that make the equation true. That set may have no members, meaning that there is no number that makes the equation true (for example x + 8 = x - 7). That set may have an infinite number of members, (for example every real number makes the equation x = (2x / 2) true). That set may have a single member. Or that set may have a finite number of members greater than one.

The solution set for \(\displaystyle x^2 = 4\) has two members, namely 2 and - 2. So if I am describing the set of possible solutions, English grammar demands "and." However, 2 and - 2 are different numbers, and a solution is a single number. So English grammar requires "The solution is either 2 or - 2, but not both." Notice that the subject is singular, not plural.

 

[HallsofIvy]

However, you can say. as you do, "The solution set contains 2 and -2" or "Both 2 and -2 are solutions to this equation".