How would I do...

4p-3q=6

3p-2q=5

2x=3y-2

3y=1+4x

and

4c + 3t =515

3c + 5t = 565

Thanks a lot. Still havent got my head around Simultaneous Equations :(

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- October 28th 2006, 10:20 AMYppolitiaSimultaneous Equations
How would I do...

4p-3q=6

3p-2q=5

2x=3y-2

3y=1+4x

and

4c + 3t =515

3c + 5t = 565

Thanks a lot. Still havent got my head around Simultaneous Equations :( - October 28th 2006, 11:25 AMQuick
You have equation:

Subtract 2x from both sides:

But look at your first equation, you know that 2x equals 3y-2. So substitute:

Simplify:

Therefore:

Subtract 1 from both sides:

divide by 2:

Go back to original equation:

Substitute:

Multiply:

Add 2 to both sides:

Divide by 3:

Can you do the other ones? - October 28th 2006, 11:32 AMQuick
- October 28th 2006, 02:38 PMCaptainBlack
You want to modify the equations so that one of the variables has the same

coefficient in both the equations, then by subtracting the equations you

can eliminate that variable.

In this case if we multiply the first equation by 3 and the second by 4 the

coefficient of p will be 12 in both cases:

12p - 9q = 18

12p - 8q = 20.

Subtracting these equations gives:

(12p - 9q) - (12p - 8q) = 18-20,

or:

-q = -2,

so q=2.

Now we substitute this value of q back into any of the equations that

we have to find the value of p. So lets take the equation

4p - 3q = 6,

with q=2 this becomes:

4p - 6 = 6,

or:

4p = 12,

so p=3.

If you reorganise your equations into the standard form with all the unknowns on one side

(and in the same order in all the equations) and the constants on the other side like the

above equations, then this method will work for all your problems.

It has another advantage, in that it can be used with a few more tricks for equations in

more than two variables (of course with the same number of equations as variables).

RonL