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anonymous
 3 years ago
Can someone teach me how to solve an equation with variables on both sides? i need help :(
anonymous
 3 years ago
Can someone teach me how to solve an equation with variables on both sides? i need help :(

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yeah something like that

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0put all that have x one side and that have not , to the other

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0What is the equation you're trying to solve?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0myko is right, but when you move a number to the other side of the eual sign you need to change the sign of the number itself. 4 becomes 4 and 4 becomes 4

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0well i understand that part but i cant go on further, i always get stuck :(

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0after you got something like this: ax=b do this: x=b/a andyou are done

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so now how would you solve 4x=18?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1362524358288:dw I dont know if this is right or wrong but i dont know where to go from there :/

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0that is the problem i am working on now

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0If you're getting the "X's" on one side then make sure you include the x. You're combining like terms. So it would be +3x under the 3x and the 7x. But you got it right so far after you get there your going to combine the +2 and the 22.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0okay i can go on from there, thanks for your help :)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1362525009117:dw Is this right?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0can you help me please :(

Ghazi
 3 years ago
Best ResponseYou've already chosen the best response.13x + 2 = 7x22 now bring like terms at one side so terms with variable lets say the ones with "x" bring them together and numbers together ;) so 3x+7x=222 4x=24 now divide both sides by 4 and you will have x=6 \[\frac{ 4x }{ 4 }=\frac{ 24 }{ 4 }\] so x= 6 ;)

Ghazi
 3 years ago
Best ResponseYou've already chosen the best response.1dw:1362525310743:dw clear?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0wait.. how did you do that? :/

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i think you are missing something

Ghazi
 3 years ago
Best ResponseYou've already chosen the best response.1nope i did , just bring 4x in the left so sign will be changed and take +2 on the right so sign will be changed and it will be +4x and on the right it will be 2 because sign gets interchanged

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1362525671903:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0okay thank you :D your help is very much appriciated kinda Sir :D lol

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0wait.. kind * not kinda

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0haha sorry about that :D

Ghazi
 3 years ago
Best ResponseYou've already chosen the best response.1yea yea silly , post your questions please and lets do it ;)

Ghazi
 3 years ago
Best ResponseYou've already chosen the best response.1well i am talking about your question lol not mine i have loads of question but i cant ask on this post of yours but yea if you wanna arrange another post i can draw and ask my questions and we can solve our questions if you like :D your maths and mine is kinda tough ;) ? what say? lets do it ;)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0alright :) well thanks for your help, and everyone else too :P

Ghazi
 3 years ago
Best ResponseYou've already chosen the best response.1but i am sure you are smart enough
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