help in math 3!! (1 Viewer)

[thuynguyen239]

could anyone please help me with these tricky questions

|3y-1|+ |2y+3|>5
|3y-1|+|2y+3|+|5y-2|>5
THANK YOU!!!

 

[thongetsu]

1. 3y-1+2y+3>5
5y+2>5
y>3/5
that's only half the answer i think. im too lazy to do the rest.

 

[Drongoski]

could anyone please help me with these tricky questions

|3y-1|+ |2y+3|>5
|3y-1|+|2y+3|+|5y-2|>5
THANK YOU!!!

wrong forum!

Don't know if my analysis is right.

For 1) y < -7/5 and y > 3/5

For 2) y < 1/6 and y > 1/2

 

[muzeikchun852]

could anyone please help me with these tricky questions

|3y-1|+ |2y+3|>5
|3y-1|+|2y+3|+|5y-2|>5
THANK YOU!!!

here are the answer.
|3y-1|+ |2y+3|>5

y = 1/3 y= -3/2
on a number line:

---------I----- -3/2-----II-----1/3------III--------

I -(3y-1) - (2y+3) >5
-3y + 1 - 2y - 3 > 5
-5y - 2 > 5
5y < -7
y < -7/5

II -(3y-1) + (2y+3) >5
-3y + 1 + 2y + 3 > 5
-y > 1
y < 1
rejected!

III (3y-1) + (2y+3) > 5
5y > 3
y > 3/5

therefore the ansmwer is y < -7/5 or y > 3/5.


|3y-1|+|2y+3|+|5y-2|>5
y = 1/3, -3/2, 2/5

on a number plane:
___I___-3/2____II____1/3___III____2/5____IV___

I -(3y-1) - (2y+3) - (5y-2) > 5
-3y +1 - 2y - 3 - 5y + 2 > 5
-10y > 5
y < -1/2

II -(3y-1) + (2y+3) - (5y-2) > 5
-3y +1 + 2y + 3 - 5y + 2 > 5
-6y + 6 > 5
-6y > -1
y < 1/6

III (3y-1) + (2y+3) - (5y-2) > 5
3y - 1 + 2y + 3 - 5y + 2 > 5
5y - 5y + 4 > 5
no solution

IV 3y-1 + 2y+3 + 5y-2 > 5
10y > 5
y > 1/2

therefore the answer is y < -1/2
y is not equal to 1/6 and 1/2.

 

[Mr Incredbile]

I think you should consult your teacher. I obtained a different answer from muzeikchun852.

My answer finally is Y < 2/7 or Y>5 /11

Simplify the equation, then just work it out from there.

E.g.

|3y| + |2y+3| > 5

Simplified: 5y+3 = 5

Subtract 3 from both sides:

5y = 2

Divide by 5

y = 2/5

Because you divided the arrow switches direction so:

Y < 2/5

Do this for the second equation, simplify then solve. When you divide change the sign.

Then you get your number line from your final answer ;
Y < 2/7 or Y>5 /11:

# represents the circle over the number)
<------# #---------->
====2/7============5/11==== (both have open circles on them)


Then substitute something into your answer to test it out, try something like 2..

[2x2 = 4] + [2x2 + 3 = 7] = 11 > 5

Don't take my word for it though, I only just learnt it the other day..

 

[muzeikchun852]

E.g.

|3y| + |2y+3| > 5

Simplified: 5y+3 = 5

Subtract 3 from both sides:

5y = 2

Divide by 5

y = 2/5

Because you divided the arrow switches direction so:

Y < 2/5

Do this for the second equation, simplify then solve. When you divide change the sign.

u got the equation wrong. :uhhuh::uhhuh:
is |3y-1|+ |2y+3| > 5

i think wat u did is a 2U maths method.
and this question is 3U maths question so wat u did wont work with two absolute value.

 

[namesAsh]

E.g.

|3y| + |2y+3| > 5

Simplified: 5y+3 = 5

Subtract 3 from both sides:

5y = 2

Divide by 5

y = 2/5

Because you divided the arrow switches direction so:

Y < 2/5

Don't really want to be a jerk or anything but the inequality sign only switches when you divide or multiply by a negative number.

Eg.

-3y > 4
y < -4/3

Check:

-3 x -4/3 = 4
therefore anything lower than -4/3 should turn out to be greater than 4
-3 x -5/3 = 5

So yea just for future reference