Help urgent mathematics

Help I don’t get which solution is the answer for part b…
The graph intersect (meet) each other at one point when x equals -3/2a which I understand.
But for the other solutions I don’t know whichside to make negative or positive

(edited 11 months ago)

Hello, can I do this one for you?

Reply 2

mqb2766

21

Just be explicit and write down the intervals and the functions on each interval so
x < -5a/3, so |x| = -x, |3x+5a| = -(3x+5a)
-5a/3 < x < 0, so ...
x > 0, so ...

Its pretty much what the sketch shows and what theyve done.

Reply 3

Alevelhelp.1

OP

15

Original post by johnny_kyalo

Hello, can I do this one for you?

Hi, yes plzzz

Reply 4

Alevelhelp.1

OP

15

Original post by mqb2766

Just be explicit and write down the intervals and the functions on each interval so
x < -5a/3, so |x| = -x, |3x+5a| = -(3x+5a)
-5a/3 < x < 0, so ...
x > 0, so ...

Its pretty much what the sketch shows and what theyve done.

So are there 2 solutions

Reply 5

mqb2766

21

The sketch pretty much shows that. You have 4 half lines (each function is composed of 2 half lines) and if you understand the sketch (the x and 3x terms are important) then you understand that there are 2 solutions.

Reply 6

Alevelhelp.1

OP

15

Original post by mqb2766

The sketch pretty much shows that. You have 4 half lines (each function is composed of 2 half lines) and if you understand the sketch (the x and 3x terms are important) then you understand that there are 2 solutions.

So do I make either both sides of the equation positive (or negative both sides as it will give same ans) and then the other eq one side should be negative and other positive… then solve?

Reply 7

mqb2766

21

Youve got the working and #3 describes which halflines you use on each interval. You have to be systematic about it. Then when you solve, you have to check the solution lies in that interval.

If youre "guessing" signs for the |.| arguments, then youre unlikely to check the solution lies in the appropriate interval which is important as youd get one extraneous solution in this case. Do you undestand the sketch? If not, try and draw it accurately and think about how the intervals, half lines and hence the points of intersection.

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