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August 17th, 2017, 12:26 PM  #1 
Newbie Joined: Aug 2017 From: glasgow Posts: 2 Thanks: 0  Straight Line workings
Hi, Need help with the following please: equation is Y=3x 1 I have two rows X and Y and trying to work the answers for the Y coordinate, struggling when I get to the minus figures. X 3 2 1 0 1 2 3 Y ? ? ? 1 2 5 8 What are the (Y) answers for 1, 2 and 3 To get the first answer 8, I done 3x31 =8 and 3x21 = 5 Many thanks Last edited by jl1974; August 17th, 2017 at 12:28 PM. 
August 17th, 2017, 12:34 PM  #2  
Senior Member Joined: Sep 2015 From: Southern California, USA Posts: 1,413 Thanks: 717  Quote:
for $x=1$ $y = 3(1)  1 = 3 1 = 4$ I'm sure you can continue for $x=2,~3$  
August 17th, 2017, 12:48 PM  #3 
Newbie Joined: Aug 2017 From: glasgow Posts: 2 Thanks: 0 
Thanks for response. This is for my daughter, can you explain like my example if possible please. like 3 times 3 = 9 minus 1 = 8 Can you tell me what the answer is for 1? Thanks. Last edited by jl1974; August 17th, 2017 at 01:08 PM. 
August 17th, 2017, 02:47 PM  #4 
Member Joined: Aug 2017 From: United Kingdom Posts: 93 Thanks: 26  
August 17th, 2017, 04:00 PM  #5 
Senior Member Joined: Sep 2015 From: Southern California, USA Posts: 1,413 Thanks: 717  
September 13th, 2017, 09:53 AM  #6 
Newbie Joined: Sep 2017 From: malaysia Posts: 1 Thanks: 0 
Hi everyone, please help me to solve this equation, my brain stuck already... given f(x) = 2x  6, determine value of k if inverse function f(2k) = f(k+1). Please help me with the detailed solution. Last edited by skipjack; September 13th, 2017 at 12:30 PM. 
September 13th, 2017, 12:35 PM  #7 
Global Moderator Joined: Dec 2006 Posts: 17,919 Thanks: 1386 
The function f(x) = 2x  6 has inverse g(x) = x/2 + 3. As g(2k) = k + 3 and f(k+1) = 2k + 2  6 = 2k  4, these are equal for k = 7. 
September 13th, 2017, 11:55 PM  #8  
Senior Member Joined: Jun 2017 From: India Posts: 227 Thanks: 6  Quote:
Quote:
$\displaystyle y=3x1$ So, If, $\displaystyle x=3$ Then, $\displaystyle y=3\times31=91=8$ $\displaystyle x=2$ Then, $\displaystyle y=3\times21=61=5$ $\displaystyle x=1$ Then, $\displaystyle y=3\times11=31=2$ $\displaystyle x=0$ Then, $\displaystyle y=3\times01=01=1$ $\displaystyle x=1$ Then, $\displaystyle y=3\times{1}1=31=4$ $\displaystyle x=2$ Then, $\displaystyle y=3\times{2}1=61=7$ $\displaystyle x=3$ Then, $\displaystyle y=3\times{3}1=91=10$  

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