Algebra help

[Crimson Dynamo]

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:

[Smithy]

Yeah

[Crimson Dynamo]

you dont know :oh:

i need smarts

[The Slim Reaper]

:smug: Jake is the maths genius on here.

[Niamh.]

I hate Maths sorry

[James]

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:

-3x +4y=0 I would have thought?

If you imagine, say, y=3 and x=4.

[Beso]

https://youtu.be/jvMhQmSZIU8

[Cherie]

Alegebra :cloud:

[DouglasS]

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:



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

[Crimson Dynamo]

-3x +4y=0 I would have thought?

If you imagine, say, y=3 and x=4.

so to draw a line on the x and y axis

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?

[Redway]

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:

You move the 3x from the last one to the left-hand side so it’s - 3x + 4y = 0.

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.

[Redway]

You said these equations were supposed to be solved via line graphs but here’s the working out:

[Josy]

Wow now I have a sore head

[Toy Soldier]

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.

[Smithy]

Yeah