GCSE Maths question (Simultaneous Equations)

[mossypne]

For my maths Homework i have to complete some simultaneous equations. I have come across some negative ones and we haven't done those.

 

Is there anyway you could explain to me how to do them. The question is: (Please note: im not asking you to do it)

 

3a + 2b = 1
15a + 3b = -51

and

5m + 6n = -33
10m + 5n = -45

 

[Shadow]

There are many ways to solve simultaneous equations; my favorite is the substitution method.

[imatfaal]

OK - there are a few pointers.

 

1. You can add and subtract simultaneous equations from each other.

Your equations are made up of three terms (using the first question) an 'a' term, a 'b' term, and a constant - ie it is 3 times 'a' plus 2 times 'b' equals 1. When you add or subtract a pair of equations you get the 'like terms' together. You cannot subtract 15a from 2b and get a simpler answer, but if you subtract 3a from 15a you get a simpler answer 12a.

 

2. An equation remains the same if you multiply all of it by a number

If you multiply each part of 3a+2b =1 by a number, say for example 4, you get the same equation but with a fixed multiple 3a+2b=1 says the same thing as 12a+8b=4

 

Play with these two rules and you will find that multiplying equation one (remember every single term) by a number and then taking it away from equation two (gather 'like terms') will give you a simpler equation where one of the terms (the 'a' term or the 'b' term disappear).

 

Try it and put up your ideas

 

 

 

If I was helpful, let me know by clicking the [+] sign ->

[mossypne]

No i know how to do them it's just im not sure about the negative ones. Look closely at the = on the equations above.

[imatfaal]

Mossy

 

Put down what you have done to the two questions and we will push you in the right direction. There is really no difference at all - ie the fact that the constant is positive or negative makes no odds

 

If I was helpful, let me know by clicking the [+] sign ->

[mossypne]

For the 1st question i have:

 

3a + 2b = 1 (x5) = 15a + 10b = 5
15a + 3 = -51

(15a + 10b = 5) - (15a + 3 = -51)
= 13b = -46

 

 

Not sure where to go from there and for the second i have:

 

5m + 6n = -33 (x2) = 10m + 12n = -66
10m + 5n = -4

(10m + 2n = -66) - (10m + 5n = -45)
= 7n = -21

 

 

Same im not sure where to go from here.

[Cap'n Refsmmat]

For the 1st question i have:

 

3a + 2b = 1 (x5) = 15a + 10b = 5
15a + 3b = -51

(15a + 10b = 5) - (15a + 3b = -51)
= 13b = -46

If [math]13b = -46[/math], you just divide by 13 to get [imath]b=\frac{-46}{13}[/imath].

 

However, you have made an error in this step:

 

[math] (15a + 10b = 5) - (15a + 3b = -51)[/math]

 

10b - 3b = 7b, not 13b, and similarly for the others. You subtracted 15a but added the rest, which is incorrect -- every term must be subtracted. You'll get:

 

[math]7b=56[/math]

 

And you can divide by 7 to get b.

 

5m + 6n = -33 (x2) = 10m + 12n = -66
10m + 5n = -4

(10m + 2n = -66) - (10m + 5n = -45)
= 7n = -21

You've made the same error here. 2n - 5n = -3n, not 7n.

 

Once you have solved for b and n, you can use that value to substitute into the other equation and solve for the other variable.

[imatfaal]

Ok Mossy

 

Now we are cooking with gas. First equation

firstly put stuff in nice neat columns if I was you - its helps keep things together

 

secondly do your basic arithmetics more carefully

 

15a-15a= 0

10b-3b =??

5 - (-51) =??

 

do these sums and it will work out a bit nicer

 

 

edit

the captain is damn quick with his [maths] markup

Edited June 9, 2011 by imatfaal

[mossypne]

Oh the mistake is me just rushing it slightly. Im not that bad at maths (honest ).

[imatfaal]

Cool - glad you have got your head round it