- #1

Ry122

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**[SOLVED] simultaneous equations**

Why is it that when solving some simultaneous equations two different ways you can get two different answers? One example is this:

0=-tsin(x)+85.7sin(30) eq. 1

0=-tcos(x)+85.7cos(30)-t eq. 2

I solved these equations in mathematica and got t=49.4789 and x=1.0472. These were the values I submitted for an engineering assignment I recently had and I only got t correct, the correct answer for x was 60 degrees.

So knowing which value was correct i subbed it into eq1. and i still got x=1.04723

Then I tried using a different method. I equated both equations so eq1=eq2 and i used x=60 to solve for t and i got 49.476. Why did these two different methods yield two different values for x and how am I supposed to know which one is correct?