elimination word problems

jhawk555

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Sep 26, 2006
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I know I have the equations right, I have reread the question and I can't write the equations any other way.

Beth has $218 in $10, $5 and $1 bills. There are forty-six bills in all and four more fives than tens. How many bill of each kind are there?

For my first equation: 10x + 5y + 1z = 218
Second equation: x + y + z = 46
The third equation: y=4x


I used y=4x and entered it into the first and second equation and came up with.

First equation: 10x + 5(4x) + z = 218 and got 30x + z =218

Second equation: x + 4x + z = 46 resulting in 5x + z =46.

I thought I would solve for x by eliminating the variable z. So I multiplied the second equation by -1. Giving me -5x-z=-46. Thus setting up to solve for x.

30x+z=218
-5x -z = 46
--------------
25x=172

Divided both sides by 25 and got 6.88. I didn't think that I could get this kind of answer and be right. I tried rounding up to 7 and continuing to solve for y and z, but when I checked my work it didn't add up right. Where did I go wrong?
 
jhawk555 wrote:
Beth has $218 in $10, $5 and $1 bills. There are forty-six bills in all and four more fives than tens. How many bill of each kind are there?

For my first equation: 10x + 5y + 1z = 218
Second equation: x + y + z = 46
The third equation: y=4x

You quoted the problem as saying that there are "four more fives than tens." What is the mathematical equivalent of the the words "four more?" If I have five toes on my left foot and five more on my right foot, does that give me a total of 25 toes (5x5) or 10 toes (5+5)? Apply this reasoning to modify your third equation, then re-solve the problem.

hreeve
 
jhawk555 said:
Beth has $218 in $10, $5 and $1 bills. There are forty-six bills in all and four more fives than tens. How many bill of each kind are there?
It is often helpful to define your terms clearly. I will take the following as your meaning:

. . . . .x: the number of ten-dollar bills
. . . . .y: the number of five-dollar bills
. . . . .z: the number of one-dollar bills

(Note: While these definitions are fine, as they stand, it can frequently be helpful to use variables which remind you of their meaning. For instance, you could use "t" for the tens, "f" for the fives, and "s" for the singles. Just FYI....)

jhawk555 said:
The third equation: y=4x
The exercise says "four more than", not "four times as many as", so the equation should probably be as follows:

. . . . .y = x + 4

Using this, you can substitute:

. . . . .x + y + z = 46
. . . . .10x + 5y + 1z = 218
. . . . .y = x + 4

Then:

. . . . .x + (x + 4) + z = 46
. . . . .10x + 5(x + 4) + 1z = 218

. . . . .2x + z = 42
. . . . .15x + z = 198

Subtract the first equation from the second to get:

. . . . .13x = 156

Then x = 12, and you should be able to back-solve from there.

Hope that helps! :D

Eliz.

P.S. Thank you for showing all your work and reasoning!!
 
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