Solving systems of linear equations?

Hello. Can anyone tell me how to go about solving two equations
simultaneously. Such as. a - b = 1, a/3 + b/5 = 1. The second equation is a fractional equation so i thought the correct process was to multiply the denoms by a number that both can divide into evenly.

(15/1)a/3 + (15/1)b/5 = 1(15/1) --> 5a + 3b = 15

Now take the first equation equation. a - b = 1 and multiply by -5 to drop one variable. -5(a - b = 1) --> -5a + 5b = -5

Now i end up with two equations --> 3b = 15, 5b = -5

Add the cofficents 8b = 10 --> 8b/8 = 10/8 --> b = 5/4

So if i plug in 5/4 into the first equation a - b = 1 --> a - 5/4 = 1
--> a = 5/4 + 1 --> a = 9/4

I guess i was on the right track. :rolleyes:

You should've added "resolved" to the title for the fastest-solved post ever. :rolleyes:

I managed to figure it out in mid post. After all that typing i didn't want to scrap a good thread. :blush:

Hi,

Check this link:
http://www.ibrtses.com/delphi/dequ.html

I contains the theory and the code to solve N equations with N unknown, not just this simple problem.

have fun

Pretty cool but a little too complex.

for two equations theres two simple methods: add the equations left side to left side and right side to right side. Alternatively subtract.

For several equations, try googling for linear algebra and gauss elimination

Dilenger4 wrote:

Quote:

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Hello. Can anyone tell me how to go about solving two equations simultaneously. Such as. a - b = 1, a/3 + b/5 = 1. The second equation is a fractional equation so i thought the correct process was to multiply the denoms by a number that both can divide into evenly.
(15/1)a/3 + (15/1)b/5 = 1(15/1) --> 5a + 3b = 15

Now take the first equation equation. a - b = 1 and multiply by -5 to drop one variable. -5(a - b = 1) --> -5a + 5b = -5

Now i end up with two equations --> 3b = 15, 5b = -5

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You method is correct, up to the last sentence!. You now really have two equations, namely:
5a + 3b = 15
-5a + 5b = -5

You now add these two, obtaining:

(5a-5a) + (3b+5b) = (15-5), or
0 + 8b = 10, from which
b = 10/8, or
b = 1.25

Then, substituing this value of b in the first equation:
5a + 3*1.25 = 15
5a + 3.75 = 15, from which
a = (15 - 3.75)/5, or
a = 11.25/5. or
a = 2.25