What am I doing wrong?

Z

zhuuraan

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Aug 15, 2013
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Here's my word problem and what I've done so far. I have to assume I messed something up in the formulas or something because the results are negative and that doesn't make sense in the context of the problem.A family is considering child care options for their child. They have found a home based option which charges $7 per hour, with no limitations on the number of hours. A center charges $210 per week for a maximum of 40 hours, and an additional $14 per hour after that. They want to ensure their child gets the best care at the most reasonable price. After extensive research, they have determined that these two options are the best ones.
Option 1
y = 7x where x is the number of hours and y is the price.
Option 2
y = 14x + 210 where x is the number of hours over the included 40 and y is the price.
Solving by Elimination. Since these two equations have numbers which are multiples of one another, I will multiply equation 1 by -2 in order to make the x values opposite of one another so that I can eliminate them, then solve.
-2y = -14x
y = 14x + 210
So the resulting equation is...
-y = 210
or y = -210 This doesn't make sense to me as y represents the price at which the two options are the same, but a price cannot be a negative value. Still, I will continue to find x given this value.
-210 = 7x
x = -30 This can't be right either. This is the number of hours at which the values are the same, but negative hours doesn't make sense either.
In addition, upon placing the values into the system, they don't add up.
y = 7x
-210 = 7(-30) This one is equivalent on both sides, but if I put it into the second equation, I get the following.
-210 = 14(-30) + 210 Actually, these values are equivalent as well, so I suppose they are the real solution to the system, but it doesn't make sense for the purpose of the problem. All I can figure is that I represented the problem incorrectly with the system I created.
 
zhuuraan said:
Here's my word problem and what I've done so far. I have to assume I messed something up in the formulas or something because the results are negative and that doesn't make sense in the context of the problem.A family is considering child care options for their child. They have found a home based option which charges $7 per hour, with no limitations on the number of hours. A center charges $210 per week for a maximum of 40 hours, and an additional $14 per hour after that. They want to ensure their child gets the best care at the most reasonable price. After extensive research, they have determined that these two options are the best ones.
Option 1
y = 7x where x is the number of hours and y is the price.
Option 2
y = 14x + 210 where x is the number of hours over the included 40 and y is the price.
Solving by Elimination. Since these two equations have numbers which are multiples of one another, I will multiply equation 1 by -2 in order to make the x values opposite of one another so that I can eliminate them, then solve.
-2y = -14x
y = 14x + 210
So the resulting equation is...
-y = 210
or y = -210 This doesn't make sense to me as y represents the price at which the two options are the same, but a price cannot be a negative value. Still, I will continue to find x given this value.
-210 = 7x
x = -30 This can't be right either. This is the number of hours at which the values are the same, but negative hours doesn't make sense either.
In addition, upon placing the values into the system, they don't add up.
y = 7x
-210 = 7(-30) This one is equivalent on both sides, but if I put it into the second equation, I get the following.
-210 = 14(-30) + 210 Actually, these values are equivalent as well, so I suppose they are the real solution to the system, but it doesn't make sense for the purpose of the problem. All I can figure is that I represented the problem incorrectly with the system I created.
Click to expand...

Actually, you have done a great job of recognizing that negative answers are unreasonable solutions to this type of "real world" problem. That should cause us to question the equations we have constructed. Sometimes it is also helpful to figure out actual costs for some "obvious" time period, just to get a feel for the problem. In this case, we know that one option costs $210 for a 40 hour week. That was a "given" piece of information. The other option costs ($7/hr)(40hr) = $280 for one week.

Now let's test our equations to see if they give us the same answers:

y = 7x
y = 7(40) = 280 Yes, that works.

y = 14x + 210
y = 14(40) + 210 = 560 + 210 = 770 Hmmm. That doesn't seem right.

Upon inspecting our second equation, we realize that we are DOUBLE-CHARGING. We are charging both the weekly charge of $210 AND the hourly charge of $14/hr at the same time. However, the $14/hr should only be applied to hours after the first 40hrs in a week. Therefore, we actually need two separate equations to define the second option, depending on how many hours of child care are desired. For x less than or equal to 40 hrs, the cost is simply y = $210. For x>40, your equation is y = $210 + (14)(x - 40). Make sense?

You will find it helpful to graph all of this.
 
Thank you so much! That makes a lot of sense, and it's not half as confusing now. I didn't even think of the concept of making a 3 equation system. I really appreciate the explanation though, because this really had me stumped!
 
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