Moved - im having a prob with a word prob also, not in the book

boogie

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stan is two thirds as old as adam. in 7 years stan will be 3/4 as old as adam how old are they now?
worked out other probs, but not with a fraction wont fit the equation the answer is 14 and 21 but how did they get that?
 
equation


Hello, boogie!


Stan is two thirds as old as Adam.
In 7 years Stan will be 3/4 as old as Adam.
How old are they now?

\(\displaystyle \begin{Bmatrix} \text{Let} & A &=& \text{Adam's age now} \\ \text{Then} & \frac{2}{3}\!A &=& \text{Stan's age now}\end{Bmatrix}\)

\(\displaystyle \text{In 7 years, they both will be 7 years older: }\:\begin{Bmatrix} \text{Adam will be }A+7\text{ years old.} \\ \text{Stan will be }\frac{2}{3}\!A + 7\text{ years old.}\end{Bmatrix}\)

\(\displaystyle \text{At that time, Stan's age will be }\frac{2}{3}\text{ of Adam's age.}\)

\(\displaystyle \text{Our equation: }\ \frac{2}{3}\ A + 7 \ = \ \frac{3}{4}(A + 7)\)

\(\displaystyle \text{Multiply by 12: }\:8A + 84 \:=\:9(A + 7)\)

. . . . . . . . . . . . .\(\displaystyle 8A + 84 \:=\:9A + 63\)

. . . . . . . . . . . . . . . . . \(\displaystyle A \:=\:21\)

. . . . . . . . . \(\displaystyle \text{Stan}\:=\:\frac{2}{3}\!A \:=\:14\)
 
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wow, good answer that helped!!

this is the first time ive found useful info online. The name is dan (cat is boogie) the prentice hall book is not the best. im 44 and back to do something i should have done 25 years ago. all i have left is algebra 1&2. i hate algebra, I passed US govt and history in about 2 months, but now this man oh man, anyway,, i have one more question and i think i have this section worked out


four pencils and two pens cost $.74. six pencils and five pens cost $1.53. find the cost of a pencil and a pen



i can do this in my head ,but thats not going to help on the quiz.. thank you stranger

DAN
 
this is the first time ive found useful info online. The name is dan (cat is boogie) the prentice hall book is not the best. im 44 and back to do something i should have done 25 years ago. all i have left is algebra 1&2. i hate algebra, I passed US govt and history in about 2 months, but now this man oh man, anyway,, i have one more question and i think i have this section worked out


four pencils and two pens cost $.74. six pencils and five pens cost $1.53. find the cost of a pencil and a pen



i can do this in my head ,but thats not going to help on the quiz.. thank you stranger

DAN
Let x = the price of a pen.

Let y = the price of a pencil.

So you have TWO unknowns. To solve such a problem you must have TWO equations.

Solving for n unknowns calls for n equations (which must be independent and consistent).

So what looks like two equalities in that problem?

How do you express them mathematically using x and y?

Do you know how to solve the resulting problem?
 
word prob (not on quiz)

im using x=.74+y

x+6=1.53(y+5) uummmmmm ex-1 { x+y= } prentice hall classic algebra 1

how about 4x +2y=.74 { x-y= } try this (E) pg 375

6x +5y= 1.53 ex-2 x+y=
x=n+y

2x+3y=.79 hhhmmm ex-3 x=n+y
x+n=n(y+n)
ok 6+5=1.53 (x+y=1.53)

x= no i cant do it
 
post got mangled when sent

didnt post the way i wrote it. in short , no i cant do the prob.... but thank you anyway
 
didnt post the way i wrote it. in short , no i cant do the prob.... but thank you anyway
2x + 4y = 74. The problem says that 2 pens and 4 pencils cost 74 cents, and we identified the price of pen and pencil as x and y respectively.

5x + 6y = 1.53. Just took that from the problem as well.

This is the general way to do word problems.

Briefly define each unknown in writing and assign a letter to each.

Translate the conditions of the problem into equations using the letters assigned above.

Then solve the resulting math problem.

Do you know how to solve equations in more than one unknown?
 
im using x=.74+y
Sigh. You were told that "four pencils and two pens cost $.74". Do you realy think that a single pen will cost $.74 plus the cost of a pencil?


four pencils and two pens cost $.74. six pencils and five pens cost $1.53. find the cost of a pencil and a pen


If a pencil cost "x", how much will 4 pencils cost? If a pen cost "y", how much much do two pens cost? Add those and set them equal to $.74.

If a pencil cost "x", how much will 6 pencils cost? If a pen cost "y", how much will 6 pencils and one pen cost? Set that equal to $1.53. That gives you two equations to solve for x and y.

 
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one pen does not

one pen isnt 74 cents. thanks i know. my post was jumbled up.
i came up with 4x+2y
and 6x+5y . didnt know what to do next. im off to school. got the side note thanks

all i need is a few examples of something before i get it,,i think im getting it
 
im using x=.74+y
Why? Jeff told you that 2x + 4y = 74. (His x and y are in cents, yours apparently in dollars which owould give you 2x+ 4y= .74.) That is not the same as "x= .74+ y".

x+6=1.53(y+5) uummmmmm ex-1 { x+y= } prentice hall classic algebra 1
I have no idea what this means.

how about 4x +2y=.74
Okay, now this is what Jeff suggested.

{ x-y= } try this (E) pg 375
You can't just take equations from other problems and expect them to work for this problem!

6x +5y= 1.53
Okay, this is "6 pens and 5 pencils cost $1.53"

ex-2 x+y=
x=n+y
Again, I don't know what "x+ y= " has to do with this. And where did "n" come from??

2x+3y=.79
I think you are subtracting each side of 4x+ 2y= .74 from 6x+ 5y= 1.53. Yes, (6x+ 5y)- (4x+ 2y)= 2x+ 3y. But 1.53- .74= .79, not .78. But what was your purpose in doing that? You ought to be trying to eliminate one of "x" or "y"?

hhhmmm ex-3 x=n+y
x+n=n(y+n)
ok 6+5=1.53 (x+y=1.53)
again, where did this "n" come from? What does it mean?

Of course, "6+ 5" is NOT equal to "1.53"! You meant 6x+ 5y= 1.53. That is NOT the same as "x+ y= 1.53". I don't know where you got that.

x= no i cant do it
You seem to be just putting numbers and letters together pretty much at random! Use some logic. You did, finally get the two equations:
"four pencils and two pens cost $.74": 4x+ 2y= .74

:six pencils and five pens cost $1.53. find the cost of a pencil and a pen": 6x+ 5y= 1.53.

Just subtracting one from the other doesn't help. You need to have a reason for doing things like that. If you multiply the first equation by 3, you get 12x+ 6y= 2.22. If you multiply the second equation by 2 you get 12x+ 10y= 3.06.
My "reason" for doing that was to get the same coefficient for x in both equations. Now you can subtract one equation from the other- because each equation has "12x", those will cancel: (12x+ 10y)- (12x+ 10y)= 2y= 3.06- 2.22= 0.84.
 
This thread seems to be another example of a person singing the praises of camera-ready copy as the most-helpful thing ever, only to immediately demonstrate that they learned practically nothing from receiving the most-helpful thing ever.

Is it a good idea to include some tutoring (i.e., explanation) with completely-work solutions? Then, is it a better idea to do it using a similar example, instead? I'm not a teacher, so you tell me.

"Spoonfeeding, in the long run, teaches us nothing but the shape of the spoon" ~ EM Forster
 
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