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# GCSE maths question help - Functions watch

1. "f(x) = 2x + c
g(x) = cx + 5
fg(x) = 6x + d
c and d are constants.
Work out the value of d. "

I've done the first two steps to get to 2cx+10+c... but I'm not quite sure what to do next. I'm guessing you make that equal to 6x+d but I don't know what to do after that.

Help would be appreciated, cheers.
2. I recognise this as a MyMaths question.

Think of it more simply...

6 = 2 * ?

Then, once you have worked out the value of c - work out fully the function fg(x) or f(g(x)).

Then compare your equation to the equation given for fg(x) as equal and work out for the missing constant.
3. Or, to put it another way, you have already worked out 2cx + 10 + c...

Thus,

2cx + 10 + c = 6x + d

What must c be if x is not manipulated in any other term?
4. (Original post by ns_2)
Or, to put it another way, you have already worked out 2cx + 10 + c...

Thus,

2cx + 10 + c = 6x + d

What must c be if x is not manipulated in any other term?
To be very honest with you, I do not know. I can't think of any method to work that out. Could you cancel out the x's?

I'm not sure what you mean by x not being manipulated by any other term; I didn't know you could use the word "manipulated" in a mathematical context!

Thank you for the help thus far.
5. (Original post by anonymous_1947)
To be very honest with you, I do not know. I can't think of any method to work that out. Could you cancel out the x's?

I'm not sure what you mean by x not being manipulated by any other term; I didn't know you could use the word "manipulated" in a mathematical context!

Thank you for the help thus far.
The only thing/term that has x in your equation is 2cx

No other bit has x.

On the provided equation, the only term of x is 6x.

Hence,

2cx must equal 6x

And 10 + c must equal d
6. (Original post by ns_2)
The only thing/term that has x in your equation is 2cx

No other bit has x.

On the provided equation, the only term of x is 6x.

Hence,

2cx must equal 6x

And 10 + c must equal d
Thank you so much!
7. (Original post by anonymous_1947)
Thank you so much!
No problem and good luck

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