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August 18th, 2013, 01:03 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 (3/3x)=(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)=3x1, 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, 02:11 PM  #2 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,929 Thanks: 1124 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, 07:22 AM  #3  
Math Team Joined: Sep 2007 Posts: 2,409 Thanks: 6  Re: Two Questions That Have Me Stumped Quote:
Quote:
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:
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:
 
August 20th, 2013, 06:18 PM  #4 
Member Joined: Aug 2013 Posts: 59 Thanks: 0 
1. If you do it correctly you should arrive with 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, 06:25 PM  #5 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,929 Thanks: 1124 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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