![]() |
Algebra help
SO we are drawing lines on xy axis using standard equations (simultaneous equation homework) and we have done all the standard 3x +5y=15 type ones but the last one is 4y=3x
wtf that is 2 variables but what is the number? do i move 3x to the right to get like -3x +4y=1? :bawling: |
Yeah
|
you dont know :oh:
i need smarts |
:smug: Jake is the maths genius on here.
|
I hate Maths sorry
|
Quote:
If you imagine, say, y=3 and x=4. |
|
Alegebra :cloud:
|
Quote:
So you could divide by 3x, meaning it’d be 4Y/3x =1 in order to get a number Subtracting would just make it equal 0 rather than 1 LT, so I guess you could do -3x +4Y= 0 also You can then use the other simultaneous equation to get either the x or y and work out the remaining one by plugging in the number |
Quote:
x would be 0 and y would be 0 so there is no line as 0 id the point in the middle of the cross? |
Quote:
It’s been a while since I did simultaneous equations so I’m a little rusty but my answer’s 2.2 for x and 1.7 for y. |
1 Attachment(s)
You said these equations were supposed to be solved via line graphs but here’s the working out:
|
Wow now I have a sore head
|
As simply as I can phrase it - you solve one of the equations for X and then sub in what you get so that the 2nd equation only has Y values.
In this case, I actually think starting with the 2nd equation is easier; 4y = 3x 3x = 4y (3x/3) = (4y/3) x = (4y/3) Thus 3x+5y=15 3(4y/3)+5y=15 [no more X, yay] multiply out the bracket, first by multiplying both sides by 3: 3(4y)+15y=45 then simplifying the left hand side 12y+15y=45 27y = 45 y=45/27 find lowest terms y = 5/3 (approx 1.67) Then solve X using what you got earlier x = 4y/3 x = 4(5/3)/3 (you can multiply all of this out on paper but just use a damn calculator :laugh: ) x = 20/9 (approx 2.22) :think: I think I made it seem more complicated than it is though. tl;dr you solve one as far as you can to eliminate a variable from the other. It doesn't matter which. |
Yeah
|
All times are GMT. The time now is 10:13 PM. |
Powered by vBulletin® Version 3.8.11
Copyright ©2000 - 2025, vBulletin Solutions Inc.
User Alert System provided by
Advanced User Tagging (Pro) -
vBulletin Mods & Addons Copyright © 2025 DragonByte Technologies Ltd.