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Simultaneos Equations - help.

Hello,

I need some help with the following equations. I missed the lesson and am being forced to complete some equations.

1.

3x - 2y = 11
7x - 2y = 27

2.

y - x = -1
3x - y = 5

Thanks.

Reply 1

KAISER_MOLE

hello

1.

with this, minus the first equation from the second equation, as -2y is in both of them, and we can get rid of it. this basically means;

(7x-2y) - (3x-2y) = (27) - (11)

which becomes

4x = 16 , so x=4 . You can work out y by putting this value of x into either of the original equations

2. Knowing a particular method, we would like the same amount of y in each equation so we can eliminate y. However, at the moment we do not have that. If we multiply the first equation by -1 though, giving x-y = 1 , we do have (-y) in each, and so we can again subtract one equation from the other and find x

too late

Another method for 2:
3x-y=5
-y=5-3x
y=3x-5
Substitute this into the other equation:
y-x=-1
So 3x-5-x=-1
2x-5=-1
2x=4
x=2

Choose the mothod you find easier.

Reply 4

Esquire

you can also solve simultaneous equations using matrices, but i dunno how to draw one for u here. They make the process much easier, I always use them.

Reply 5

Dekota

Esquire

you can also solve simultaneous equations using matrices, but i dunno how to draw one for u here. They make the process much easier, I always use them.

Reply 6

Christophicus

2.
y - x = -1
3x - y = 5

Thanks.

Add the two equations to eliminate y. :wink:

3x-x = 5-1
2x = 4
x = 2

y = -1+x = -1+2 = 1

Reply 7

ElWilson

Could anyone sort out some of these Simultaneos Equations?

a) 3x - 2y = 21
4x + 3y = 11

b) 3x + y = 11
3x + 2y = 13

Cheers.

Reply 8

screenager

ElWilson

Could anyone sort out some of these Simultaneos Equations?

a) 3x - 2y = 21
4x + 3y = 11

b) 3x + y = 11
3x + 2y = 13

Cheers.

A) (5,-3)
B) (3,2)

Reply 9

Zerforax

a) 3x - 2y = 21 (x4)
4x + 3y = 11 (x3)
giving;
12x - 8y = 84
12x + 9y = 33 (subtract them to get rid of x)
so -17 = 51 hence y = -3
sub y into 3x - 2y = 21
3x - (-3) = 21
hence 3x = 15 therefore x = 5
and just to check...
4x + 3y = 11
(4 x 5) + (3 x -3) = 11 (tick :biggrin:

)

b) 3x + 2y = 13
3x + y = 11 (subtract)
y = 2
3x + 4 = 13 therefore x = 3
(3 x 3) + 2 = 11 (tick :biggrin:

)

i wish my maths was this easy again...

Reply 10

Excalibur

a) You multiply the top one by 3, and the bottom by 2, so that you have a common coefficient (i.e. 6y). You get:

9x - 6y = 63
8x + 6y = 22

Add them together, as 6y have different signs. You get: 17x = 85, so x = 5. Substitute x=5 into the equation, and you eventually get y = -3

b) 3x is already common, so just minus one from the other.

3x + 2y = 13
3x + y = 11

makes y=2. Substitute that into the equation, to get x=3.