Number Problem 3

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mathdad

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Separate 90 into two parts do that one part is four times the other part. DO NOT SOLVE. Set up the equation(s).

Set up:

Let 90 = x and (90 - x)

Equation:

x = 4(90 - x)

Is this right?

Since one part is four times the other part, I can also say 4x = 90 - x.

Yes?
 
Yes, both are right, but your method is not, in my opinion, the way to learn how to do word problems. For one thing, you do not really mean that x = 90 although that is what you say.

Identify unknowns and assign a variable to each. (Yes, I know that it is not necessary in this problem, but it can be essential in harder problems. Practice what you will need in the hard problems, not the easy ones.)

[MATH]s = \text {smaller number.}[/MATH]
[MATH]b = \text {bigger number.}[/MATH]
Now, if you have n unknowns, you need to find n (independent) numeric relationships among the n unknowns and express them as equations.

[MATH]b = 4s.[/MATH]
[MATH]b + s = 90.[/MATH]
You have a purely mechanical problem to solve.

[MATH]b = 4s \implies 90 = b + s = 4s + s = 5s.[/MATH]
 
Yes, both are right, but your method is not, in my opinion, the way to learn how to do word problems. For one thing, you do not really mean that x = 90 although that is what you say.

Identify unknowns and assign a variable to each. (Yes, I know that it is not necessary in this problem, but it can be essential in harder problems. Practice what you will need in the hard problems, not the easy ones.)

[MATH]s = \text {smaller number.}[/MATH]
[MATH]b = \text {bigger number.}[/MATH]
Now, if you have n unknowns, you need to find n (independent) numeric relationships among the n unknowns and express them as equations.

[MATH]b = 4s.[/MATH]
[MATH]b + s = 90.[/MATH]
You have a purely mechanical problem to solve.

[MATH]b = 4s \implies 90 = b + s = 4s + s = 5s.[/MATH]

Another one for the files.
 
As JeffM said, you said x=90 which is wrong. It is the sum of the two parts that equal 90
 
As JeffM said, you said x=90 which is wrong. It is the sum of the two parts that equal 90

No. I said 90 = x and (90 - x), the word and replacing the plus sign.

90 = x + (90 - x)
 
You can do this:

Let x = one part
Let 90 - x = the other part

Then 4x = 90 - x

or

x = 4(90 - x)
 
Separate 90 into two parts do that one part is four times the other part. DO NOT SOLVE. Set up the equation(s).
If matters not what we call the variables. Just pick two.
\(\displaystyle \left\{ \begin{array}{l}x + y = 90\\x = 4y\\4y+y=90\\5y=90\\y=18,~x=72\end{array} \right.\)
The important thing is the simplicity of design.
Look at the above. That would quite easy to grade
 
Is this not my original answer?

No, that is not your original answer. Your let-statements are incorrect.
The x and (90 - x), which you used, would add to 0, because you set x = 90.
You cannot set x equal to 90 when you are trying to solve for the unknown
value of x.

One part plus the other part equals the whole.

The whole is 90.

First, you can anchor down what the one part is by calling it "x," for example.
Then, express the other part in terms of x.

x + (the other part) = 90

Therefore, (the other part) = 90 - x
 
No, that is not your original answer. Your let-statements are incorrect.
The x and (90 - x), which you used, would add to 0, because you set x = 90.
You cannot set x equal to 90 when you are trying to solve for the unknown
value of x.

One part plus the other part equals the whole.

The whole is 90.

First, you can anchor down what the one part is by calling it "x," for example.
Then, express the other part in terms of x.

x + (the other part) = 90

Therefore, (the other part) = 90 - x

Very nice explanation.
 
... x = 4(90 - x)

... I can also say 4x = 90 - x

Yes?
As you're showing two versions, I would use a different symbol in the second version, because x does not represent the same quantity in both versions.

x = 4(90 - x)

Symbol x represents the bigger part (72)

4z = 90 - z

Symbol z represents the smaller part (18)

?
 
Good but directions specifically say DO NOT SOLVE. I could have found the numbers, too. In fact, for most of the word problems that will be posted, I simply will show my work in terms of setting up the needed equation(s). This is what matters most to me.
Sorry I ever replied to any of your posts. Please forgive me.
I am so naive as to not see you for what you are: A TROLL.
 
Sorry I ever replied to any of your posts. Please forgive me.
I am so naive as to not see you for what you are: A TROLL.

Since I am a troll to you, please ignore my questions as I will ignore you from now on.
 
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