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Q: i need help with solving equations for intermediate algebra.
A: This conversation has been flagged as incorrect. New answers have been added below ....
Original conversation
User: i need help with solving equations for intermediate algebra.

Weegy: so how can I help u ??

User: (3x+1)(5x-4)=0

Weegy: (3x+1)(5x-4)=0, 15x^2-12x+5x-4=0, 15x^2-7x-4=0

User: so is that the final answer? or can it be simplified more?

Weegy: ya that is the final answer and it cannot be simplified more.....Is there anything that I can help u with

User: yes....

Weegy: so how can I help u

User: solve using square root property: x^2=121

Weegy: x^2=121 then when the square moves to the other side it becomes root . so now x=sqrt(121) then x=11.......Is that understandable

User: Yes, for some reason i thought it was both -11 and+11

Weegy: no it cant be -11 , its just +11 b'coz as 121 is a positive value so obviously sqrt(121) is sqrt(11*11) , so it is +11 but not -11.

User: ok, i understand

Weegy: do you have any other questions?

User: what about solving for: x(3x+4)=4

Weegy: x(3x+4)=4, 3x^2+4x=4, 3x^2+4x-4=0.....

User: After it is set to 0 is where im lost, i got that far but im not sure how to finish it

Question
Asked 7/7/2009 8:03:28 PM
Updated 5/30/2014 8:56:46 PM
This conversation has been flagged as incorrect.
Flagged by andrewpallarca [5/30/2014 8:56:46 PM]
Rating
3
(3x+1)(5x-4)=0;

3x + 1 = 0; 3x = 0 - 1; 3x = -1; x = -1/3;
5x - 4 = 0; 5x = 0 + 4; 5x = 4; x = 4/5

The solution for (3x+1)(5x-4)=0 is x = -1/3 or x = 4/5
Added 5/30/2014 8:56:41 PM
This answer has been confirmed as correct and helpful.

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