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GCSE simultaneous equation help please

3x+5y=19
4x-2y=-18

so basiccally i am stuck and got this far:

12x+20y=76
12x-6y=54
20y+6y=76--54?
26y=?

Any help could be good.

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Reply 1

the bear

it is going well...

if 26y = 130

then y = ...

Reply 2

Kangie

Original post by Green22

3x+5y=19
4x-2y=-18

so basiccally i am stuck and got this far:

12x+20y=76
12x-6y=54
20y+6y=76--54?
26y=?

Any help could be good.

3x+5y=19
4x-2y=-18

(x4) 12x+20y=76
(x3) 12x-6y=-54

-12x-20y=-76
12x-6y=-54

-26y=-130
-y=-5
y=5

Then substitute y into the equation or do the same for x

(Hope I did this right)

Reply 3

Green22

Original post by the bear

it is going well...

if 26y = 130

then y = ...

I thought too it was something to do with 130. Anyway 130/26=?

Reply 4

Green22

Original post by Kangie

3x+5y=19
4x-2y=-18

(x4) 12x+20y=76
(x3) 12x-6y=-54

-12x-20y=-76
12x-6y=-54

-26y=-130
-y=-5
y=5

Then substitute y into the equation or do the same for x

(Hope I did this right)

May i ask where did you get the -26y from :confused:

Reply 5

Placeboo123

Original post by Green22

I thought too it was something to do with 130. Anyway 130/26=?

Y = 130/26 yah
What you aiming to get in your maths gcse? xd

Reply 6

Kangie

Original post by Green22

May i ask where did you get the -26y from :confused:

When you change the top equation so that it's negative the y changes to -20, then when you minus -6 from -20 it goes to -26

Reply 7

Green22

Original post by Kangie

When you change the top equation so that it's negative the y changes to -20, then when you minus -6 from -20 it goes to -26

ohh ok.

Reply 8

Green22

Original post by Placeboo123

Y = 130/26 yah
What you aiming to get in your maths gcse? xd

God knows tbh. I have tried a question before posting this one and i got the answers rights but this one is just messed up.

Reply 9

Green22

Original post by Kangie

3x+5y=19
4x-2y=-18

(x4) 12x+20y=76
(x3) 12x-6y=-54

-12x-20y=-76
12x-6y=-54

-26y=-130
-y=-5
y=5

Then substitute y into the equation or do the same for x

(Hope I did this right)

I am doubting this for some reason.

Reply 10

Green22

Original post by the bear

it is going well...

if 26y = 130

then y = ...

I am starting to think one of the answers will not be a whole number? I will keep trying.

Reply 11

Kangie

Original post by Green22

I am doubting this for some reason.

Why? I even did the whole simultaneous equation and the answer is correct.

Reply 12

Green22

Original post by Kangie

3x+5y=19
4x-2y=-18

(x4) 12x+20y=76
(x3) 12x-6y=-54

-12x-20y=-76
12x-6y=-54

-26y=-130
-y=-5
y=5

Then substitute y into the equation or do the same for x

(Hope I did this right)

I thought it would be 20y+6y=26y? thats what i got.?

Reply 13

Placeboo123

Original post by Green22

God knows tbh. I have tried a question before posting this one and i got the answers rights but this one is just messed up.

Haha private message me or quote if you need any help, should be able to give you some tips at least :biggrin:

year 10 or 11?

Reply 14

Kangie

Original post by Green22

I thought it would be 20y+6y=26y? thats what i got.?

No the bottom equation has a negative y number unless you wrote the question wrong in your original post

Reply 15

Green22

Original post by Kangie

Why? I even did the whole simultaneous equation and the answer is correct.

Thanks for that but it is just that may working out is a bit different. Maybe your answer is right though.

Reply 16

Green22

Original post by Placeboo123

Haha private message me or quote if you need any help, should be able to give you some tips at least :biggrin:

year 10 or 11?

Cool. i have done tons of these over the past and have got them right but this question just messed me up. I am 16 btw.

Reply 17

ChaoticButterfly

Original post by the bear

it is going well...

if 26y = 130

then y = ...

Reply 18

Blaze3211

Original post by Green22

ohh ok.

Why did you make the top equation negative?

Reply 19

Green22

Original post by Kangie

No the bottom equation has a negative y number unless you wrote the question wrong in your original post

I am sure the question i wrote in my op is right.
Here is the link:
http://keshgcsemaths.files.wordpress.com/2013/11/87_simultaneous-equations.pdf
It is the second sim equation question on that paper.

Thanks for trying to help me.