My Math Forum Two Questions That Have Me Stumped

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 August 18th, 2013, 02:03 PM #1 Newbie   Joined: Aug 2013 Posts: 3 Thanks: 0 Two Questions That Have Me Stumped Hey everyone, Could really use help on these two problems so I can finish studying for my math placement test tomorrow. 1) I need to find the set of real solutions if (3/3-x)=(1/3)-(1/x). Answer Choices are: A. empty set, B. {3}, C. {-3}, D. {3, -3}, E. {3i, -3i} 2) For which of the following functions is f(a+b)=f(a)+f(b) for all values of a and b? Answer Choices are: A. f(x)=3x-1, B. f(x)=x+3, C. f(x)=2x, D. f(x)=x^2, E. x^2+1 Just completely blanked on how to solve these two. Thanks a lot in advance!!
 August 18th, 2013, 03:11 PM #2 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,912 Thanks: 1110 Math Focus: Elementary mathematics and beyond Re: Two Questions That Have Me Stumped 1) Solve 3/(3 - x) = 1/3 - 1/x for x: Multiply through by 3x(3 - x): 9x = x(3 - x) - 3(3 - x) Expand: 9x = 3x - x² - 9 + 3x Rearrange and simplify: x² + 3x + 9 = 0. 3² - 4(1)(9) = 9 - 36 < 0 hence this quadratic has no real roots, so answer choice A is correct. 2) For which of the following functions is f(a + b) = f(a) + f(b)? A: f(x) = 3x - 1. f(a + b) = 3(a + b) - 1 = 3a + 3b - 1. f(a) + f(b) = 3a - 1 + 3b - 1 = 3a + 3b - 2. B: f(x) = x + 3. f(a + b) = a + b + 3. f(a) + f(b) = a + 3 + b + 3 = a + b + 6. C: f(x) = 2x. f(a + b) = 2(a + b) = 2a + 2b. f(a) + f(b) = 2a + 2b. D: f(x) = x². f(a + b) = (a + b)² = a² + 2ab + b². f(a) + f(b) = a² + b². E: f(x) = x² + 1. f(a + b) = (a + b)² + 1 = a² + 2ab + b² + 1. f(a) + f(b) = a² + b² + 2. Only choice C has f(a + b) = f(a) + f(b).
August 20th, 2013, 08:22 AM   #3
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Re: Two Questions That Have Me Stumped

Quote:
 Originally Posted by Tyrean Hey everyone, Could really use help on these two problems so I can finish studying for my math placement test tomorrow. 1) I need to find the set of real solutions if (3/3-x)=(1/3)-(1/x).
I assume that on the left you mean 3/(3- x), not (3/3)- x= 1- x.

Quote:
 Answer Choices are: A. empty set, B. {3}, C. {-3}, D. {3, -3}, E. {3i, -3i}
Since you are given choices, one way to answer this, and perhaps the easiest, is to try each possible answer.
If x= 3 the left side has denominator is 0 which is impossible.
If x= -3, the left side is 3/(3+ 3)= 3/6= 1/2 while the left is (1/3)- (-1/3)= 2/3, not equal to 1/2.
Since neither 3 nor -3 is a solution, {3, -3} is not a solution set.
If x= 3i, 3/(3- 3i)= 1/(1- i). "Rationalize the denominator" by multiplying both numerator and denominator by 1+ i: (1+ i)/2. On the right, (1/3)- (1/3i)= (1/3)(1- 1/i)= (1/3)(1+ i). 1/3 is not equal to 1/2 so this is not a solution set.
The only option left is "A. empty set".

Quote:
 2) For which of the following functions is f(a+b)=f(a)+f(b) for all values of a and b? Answer Choices are: A. f(x)=3x-1, B. f(x)=x+3, C. f(x)=2x, D. f(x)=x^2, E. x^2+1
Again: if f(x)= 3x- 1 then f(a+ b)= 3(a+ b)- 1= 3a+ 3b- 1 while f(a)+ f(b)= (3a+ 1)+ (3b+ 1)= 3a+ 3b+ 2. Those are not the same.
If f(x)= x+ 3, then f(a+ b)= a+ b+ 3 while f(a)+ f(b)= (a+ 3)+ (b+ 3)= a+ b+ 6. Those are not the same.
If f(x)= 2x, then f(a+ b)= 2(a+ b)= 2a+ 2b while f(a)+ f(b)= (2a)+ (2b). Yes, those are the same!
If f(x)= x^2, then f(a+ b)= (a+ b)^2= a^2+ 2ab+ b^2 while f(a)+ f(b)= a^2+ b^2. Those are not the same.
If f(x)= x^2+ 1, then f(a+ b)= (a+ b)^2+ 1= a^2+ 2ab+ b^2+ 1 while f(a)+ f(b)= a^2+ 1+ b^2+ 1= a^2+ b^2+ 2. Those are not the same.

Quote:
 Just completely blanked on how to solve these two. Thanks a lot in advance!!

 August 20th, 2013, 07:18 PM #4 Member   Joined: Aug 2013 Posts: 59 Thanks: 0 1. If you do it correctly you should arrive with $x\,=\,\frac{-3\,\pm\,3\sqrt{3}i}{2}$ but by doing the process of elimination i.e. it can't be any other correct answer but A. empty set. 2. Seriously, come on . . . just plug in x = a + b, a or b (e.g. A. 3(a + b) - 1 = 3a - 1 + 3b - 1 but this isn't correct, so it's not A.)! Sometimes I'm just too lazy . . . with the benefit of hindsight, C. f(x) = 2x is the correct answer (because B. and E. are similar situations to A. but again D. would include 2ab which is additionally incorrect.
 August 20th, 2013, 07:25 PM #5 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,912 Thanks: 1110 Math Focus: Elementary mathematics and beyond Re: Two Questions That Have Me Stumped Question 1 asks for real solutions, so why would choice E even be considered?

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