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C3 functions

http://www.examsolutions.net/a-level-maths-papers/Edexcel/Core-Maths/Core-Maths-C3/2010-June/paper.php#Q4

What does part b what me to do?? i don't understand
Reply 1
Avatar for B_9710
B_9710
18
∣2x−5∣=15+x |2x-5|=15+x .
Reply 2
Avatar for bigdonger
bigdonger
OP
2
Original post by B_9710
∣2x−5∣=15+x |2x-5|=15+x .


how do i solve something with modulus in it then??? it's new to me....
Reply 3
Avatar for B_9710
B_9710
18
Original post by bigdonger
how do i solve something with modulus in it then??? it's new to me....


Solve for 2 cases:
2x−5=±(15+x) 2x-5=\pm (15+x) .
Reply 4
Avatar for bigdonger
bigdonger
OP
2
Original post by B_9710
Solve for 2 cases:
2x−5=[b]±[/b](15+x) 2x-5=[b]\pm[/b] (15+x) .


where does that come from?
Reply 5
Avatar for rxns_00
rxns_00
15
Original post by bigdonger
how do i solve something with modulus in it then??? it's new to me....


Two solutions:

2x-5=15+X

And the other
-2x+5=15+X

So to remember it simply, for the first solution ignore the modulus and continue as it is. For the second solution, change the signs on the two variables included in the modulus, so instead of
+2x
You write
-2x
And instead of
-5
You write
+5
Reply 6
Avatar for B_9710
B_9710
18
mod.png
Original post by bigdonger
where does that come from?
Reply 7
Avatar for bigdonger
bigdonger
OP
2
Original post by rxns_00
Two solutions:

2x-5=15+X

And the other
-2x+5=15+X

So to remember it simply, for the first solution ignore the modulus and continue as it is. For the second solution, change the signs on the two variables included in the modulus, so instead of
+2x
You write
-2x
And instead of
-5
You write
+5


Original post by B_9710
mod.png


ah thanks both of you

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