MHB Simplify thisIs My Expression Simplification Correct?

  • Thread starter Thread starter tmt1
  • Start date Start date
  • Tags Tags
    Expression
AI Thread Summary
The expression given is simplified incorrectly by the user, who arrives at a different result for C. The textbook simplifies the expression to C = -2, while the user’s calculations lead to a more complex form. A key error is identified in the user's initial step, where they should have multiplied by 9 instead of 3 to eliminate the denominators properly. The correct approach leads to the conclusion that C = -2 is indeed accurate. Proper simplification methods are crucial for obtaining the correct result in such expressions.
tmt1
Messages
230
Reaction score
0
My textbook has the expression

$$\frac{4}{9} + 2 - 44 = C(\frac{11}{3})(\frac{17}{3})$$

and it simplifies it to $C = -2$

But I am getting

$$(\frac{4}{9} + 2 - 44) * 3 = C187$$

which simplifies to

$$(\frac{12}{9} + 6 - 132) = C187$$

and

$$(\frac{12}{9} - 126) = C187$$

then,

$$(\frac{4}{3} - \frac {378} { 3}) = C187$$$$( - \frac {374} { 3}) = C187$$

then

$$( - \frac {374} { 561}) = C$$

Am I wrong?
 
Last edited:
Mathematics news on Phys.org
In your first step, where you multiplied through by 3...you need to multiply through by 9 in order to clear both of those denominators on the right side. :)
 
tmt said:
$$\frac{4}{9} + 2 - 44 = C(\frac{11}{3})(\frac{17}{3})$$
Hints (more like suggestions):
Start with obvious:
4/9 - 42 = 187C / 9 ; multiply by 9:
4 - 378 = 187C
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Is it possible to arrange six pencils such that each one touches the other five? If so, how? This is an adaption of a Martin Gardner puzzle only I changed it from cigarettes to pencils and left out the clues because PF folks don’t need clues. From the book “My Best Mathematical and Logic Puzzles”. Dover, 1994.

Similar threads

Replies
7
Views
2K
Replies
4
Views
11K
Replies
4
Views
1K
Replies
8
Views
2K
Replies
12
Views
2K
Back
Top