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1. How do u solve the simultaneous equations: 3x+2y+5=0 and 4x+7y+3=0?
I am really stuck on this problem as I tried to do it but is kept on getting weird answers.It would be great if someone could answer it and show a method if possible please. Thanks in advance for ur help.
2. (Original post by starstudent7)
How do u solve the simultaneous equations: 3x+2y+5=0 and 4x+7y+3=0?
I am really stuck on this problem as I tried to do it but is kept on getting weird answers.It would be great if someone could answer it and show a method if possible please. Thanks in advance for ur help.
Rearrange the first equation for y = (something)
Replace y in the 2nd equation with (something)
We aren't supposed to provide answers, only methods.
3. Yes overall you should get 2 x values and 2 y values
Yes overall you should get 2 x values and 2 y values
Nope - they should only get one of each - it would be quadratic or higher simultaneous equations for 2
5. (Original post by ValerieKR)
Nope - they should only get one of each - it would be quadratic or higher simultaneous equations for 2
rip, just realised
6. (Original post by starstudent7)
How do u solve the simultaneous equations: 3x+2y+5=0 and 4x+7y+3=0?
I am really stuck on this problem as I tried to do it but is kept on getting weird answers.It would be great if someone could answer it and show a method if possible please. Thanks in advance for ur help.
times 1st by 4 and 2nd by 3 to get the X's the same (12x)
eg
3x+2y+5=0
4x+7y+3=0

12x+8y+20=4
12x+21y+9=3

cancel out thee X's

8y+20=4
21y+9=3

i got up to here but i keep getting weird answers too. Are you sure its solve-able and you copied it correctly?
7. (Original post by Rainbowcorn)
i got up to here but i keep getting weird answers too. Are you sure its solve-able and you copied it correctly?
It is solvable because it's two straight non-parallel lines
8. (Original post by Rainbowcorn)
times 1st by 4 and 2nd by 3 to get the X's the same (12x)
eg
3x+2y+5=0
4x+7y+3=0

12x+8y+20=4
12x+21y+9=3

cancel out thee X's

8y+20=4
21y+9=3

i got up to here but i keep getting weird answers too. Are you sure its solve-able and you copied it correctly?
In your first step you multiplied by four on one side, but added four to the other side (same for three), and you can't just cancel out the x's, find an equivalence by making 12x = ... for both, then setting the ... equal to eachother for one equation in terms of y
9. (Original post by starstudent7)
How do u solve the simultaneous equations: 3x+2y+5=0 and 4x+7y+3=0?
I am really stuck on this problem as I tried to do it but is kept on getting weird answers.It would be great if someone could answer it and show a method if possible please. Thanks in advance for ur help.
you could start by subtracting the first equation from the second to get an x=...
10. (Original post by starstudent7)
How do u solve the simultaneous equations: 3x+2y+5=0 and 4x+7y+3=0?
I am really stuck on this problem as I tried to do it but is kept on getting weird answers.It would be great if someone could answer it and show a method if possible please. Thanks in advance for ur help.
3x+2y+5=0 (1)
4x+7y+3=0 (2) < you have to make the coefficient of 'y' the same. To do that you multiply 2 by 7 to give you 14y and times 7 by 2 which will also give you 14y. Therefore the coefficient of y is now the same. Also you must multiply 3x by 7 for the first equation and 4x by 2. So now you not the pattern

It'll look like this

21x+14y+35=7
4x+14y+6=2

Because the symbols are the same ( + ) you must subtract

21x+14y+35=7
4x +14y+6=2
At this part the 14y cancels each other out and you subtract the rest
17x + 29 = 9

Subtract 29 from the equation and subtract it from 9

17x = -20 Now you must get x on its own so

I GIVE UP I tried, I'm sorry
11. Tasnia'x

I've been trying this for 5 mins. I'm so baffed

I don't think it's solvable but try?
12. It is definitely solvable:

(1)
(2)

Multiply (1) by 7 and (2) by 2, to get equal coefficients of , so that they can later be eliminated:

(1)
(2)

(1) - (2), to eliminate the s and leave an equation with only one and a number.

Input that value for in any of the other 2 equations and you get the value for y. Overall: do not be daunted by fractions, and remember that multiplying by 0 gives just 0 (some others of you did , which is wrong).
13. Here's a start: so, we need to solve

We start by doing what a lot of people have suggested: multiplying each equation by an appropriate number so that the coefficients of, say, are the same:

(Multiply the first equation by 4 and the second by 3.)

Now subtract first from second:

... and from here, we can work out the value of , and then plug it back into the original equations to find . The numbers aren't nice, but they work!

Apparently a lot of people think that . When we multiply the original equations by a constant, the right-hand sides don't change, because !
14. (Original post by MrLatinNerd)
!
Watch that exclamation mark!
15. (Original post by Federerr)
Tasnia'x

I've been trying this for 5 mins. I'm so baffed

I don't think it's solvable but try?
Finally Maths in such longg time

Haha. I've been like that when I saw the question.

Its solvable at the beginning but towards the Middle/end- I'm just clueless. Show me what you got, I'll help ya out
16. Guys I've managed to get an answer of 0.85 for y and -2.24 for x. It works when you put it back into the equation. Thanks for trying to work it out for me, it took me a bit of time to figure it out
17. (Original post by JN17)
In your first step you multiplied by four on one side, but added four to the other side (same for three), and you can't just cancel out the x's, find an equivalence by making 12x = ... for both, then setting the ... equal to eachother for one equation in terms of y
oh yeah thank you! havent done maths for a year at school haha

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