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Weird simultaneous equation!Please help!

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.

Reply 1

ValerieKR

Original post by starstudent7

Rearrange the first equation for y = (something)
Replace y in the 2nd equation with (something)
We aren't supposed to provide answers, only methods.

Reply 2

Paladin597

Yes overall you should get 2 x values and 2 y values

Reply 3

ValerieKR

Original post by Paladin597

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

Reply 4

Paladin597

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 :cool:

Reply 5

Maddie088

Original post by starstudent7

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?

Reply 6

ValerieKR

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

Reply 7

JN17

Original post by Rainbowcorn

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

Reply 8

sputum

Original post by starstudent7

you could start by subtracting the first equation from the second to get an x=...

Reply 9

JezzSenpai

Original post by starstudent7

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 :frown:

I tried, I'm sorry

Reply 10

Federerr

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

I don't think it's solvable but try?

Reply 11

K-Man_PhysCheM

It is definitely solvable:

(1)

3x+2y+5=0

(2)

4x+7y+3=0

Multiply (1) by 7 and (2) by 2, to get equal coefficients of

y

, so that they can later be eliminated:

(1)

21x+14y+35=0

(2)

8x+14y+6=0

(1) - (2), to eliminate the

y

s and leave an equation with only one

x

and a number.

13x+29=0

13x=-29

x=-\frac{29}{13} = -2\frac{3}{13}

Input that value for

x

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

0 \times 2 = 2

, which is wrong).

(edited 7 years ago)

Reply 12

MrLatinNerd

Here's a start: so, we need to solve

3x+2y+5=0,\\4x+7y+3=0.

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

x

are the same:

12x+8y+20=0,\\12x+21y+9=0.

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

Now subtract first from second:

13y - 11 = 0 \ldots

... and from here, we can work out the value of

y

, and then plug it back into the original equations to find

x

. The numbers aren't nice, but they work!

Apparently a lot of people think that

0 \times \text{something} = \text{something}

. When we multiply the original equations by a constant, the right-hand sides don't change, because

0 \times \text{something} = 0

Reply 13

ValerieKR

Original post by MrLatinNerd

0 \times \text{something} = 0

Watch that exclamation mark!

Reply 14

starstudent7

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 :smile: