2x – 5y = 8
4x + ky = 17
For which of the following values of k will the system of equations above have no solution?
A)-10
B) -5
C) 0
D) 5
E) 10
Is it -10?
Yup. It’s -10. Plug in A, B, C, D, E and you’ll get your answer.
Is there a shorter way to do it than plugging in? I just thought I should make the two equations have the same slope
How about this one?
Jane has a set of marbles. There are X marbles total, Y red marbles, and Z marbles will be removed. If the red marble/total marble ratio remains the same after Z marbles are removed, what answer represents the amount of red marbles removed?
A) xy/z
B) xz/y
C) z/xy
D) yz/x
E) y/xz
@marsha.ss Yep. Because the system is given by two lines in the xy-plane, they will have no solution iff the lines are parallel and not equivalent (e.g. 2x-5y = 8 and 4x-10y = 16 are the same line → infinitely many solutions).
For the first problem, I strongly suggest not plugging in answer choices for this one - a little know-how about linear systems will save you lots of time.
The red marble/total marble ratio originally is y/x. You are removing z marbles and the ratio afterwards is preserved, so out of the z marbles, you want the same ratio of them to be red. So z*(y/x) = D) yz/x red marbles should be removed.
Your first problem above can be done very quickly. Since we multiply 2 by 2 to get to 4, we multiply -5 by 2 to get -10, choice (A).
Equivalently, we can just set up the ratio 2/4 = -5/k. Cross multiplying gives 2k = -20, so that k = -20/2 = -10.
The general form of an equation of a line is ax + by = c where a, b and c are real numbers. If b ≠ 0, then the slope of this line is m = -a/b. If b = 0, then the line is vertical and has no slope.
Let us consider 2 such equations.
ax + by = c
dx + ey = f
(1) If there is a number r such that ra = d, rb = e, and rc = f, then the two equations represent the same line. Equivalently, the two equations represent the same line if a/d = b/e = c/f. In this case the system of equations has infinitely many solutions.
(2) If there is a number r such that ra = d, rb = e, but rc ≠ f, then the two equations represent parallel but distinct lines. Equivalently, the two equations represent parallel but distinct lines if a/d = b/e ≠ c/f. In this case the system of equations has no solution.
(3) Otherwise the two lines intersect in a single point. In this case a/d ≠ b/e, and the system of equations has a unique solution.
Additional examples: The following two equations represent the same line.
2x + 8y = 6
3x + 12y = 9
To see this note that 2/3 = 8/12 = 6/9 .(or equivalently, let r = 3/2 and note that (3/2)(2) = 3, (3/2)(8) = 12, and (3/2)(6) = 9.
The following two equations represent parallel but distinct lines.
2x + 8y = 6
3x + 12y = 10
This time 2/3 = 8/12 ≠ 6/10 .
The following two equations represent a pair of intersecting lines.
2x + 8y = 6
3x + 10y = 9
This time 2/3 ≠ 8/10 .
thank you! what about the second one Dr Steve? How would you explain that?
The second one is an example where making up numbers will suddenly help you to see the algebra. For example:
Say there were originally x = 100 marbles, y=20 of which were red. And now you remove z=10 of them. To keep the ratios the same, 2 of the 10 that you remove will have to be red. So plug in x = 100, y = 20 and z = 10 into each answer choice, ruling out any answer that doesn’t come out to 2.
But before you even get as far as plugging the numbers back in, think about how you got 2. To keep the ratio the same, of the 10 marbles you removed, (20/100) of them were red. So the red ones were 10 x (20/100). That’s also what you get when you do z(y/x) or yz/x.
@Marsha Picking numbers as pckeller did is the way to go for most students for this problem. It’s just a bit too confusing algebraically. Pckeller’s solution is fine, but let me also tell you how I thought about it.
I started with a simple unreduced fraction that reduces easily: 6/8 = 3/4. Using these fractions as a guide, I’m going to start with x = 8 total, y = 6 red, and remove z = 4 total (to get from 8 to 4). So we see that 3 red marbles must have been removed (we went from 6 to 3).
Put a nice, dark circle around the number 3 so it’s easy to find later, and plug x = 10, y = 6, z = 4 into each answer choice. Make sure you try all 5 choices. You will see that only choice (D) gives the correct answer of 3.
Also, and this may be obvious, but in case it isn’t…
The key idea you need to notice when you make up numbers is that to remove a sample without changing the ratio, the sample you remove has to have the same contents ratio as the overall collection. I think both DrSteve and I may have zipped past that step without explaining it.
It helps if you spend a lot of time watching baseball and thinking about math (as many nerdy people do). Say a player is batting .200. On any given day, if their batting average does not change at all, they must also have batted .200 for that day alone, perhaps going 1 for 5. If what you add (or remove) from a population has the same ratio of contents as the population, then the combined ratio won’t change.
I don’t know if that helps, but it came to mind as I was checking on a ball game.