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Simultaneous equations

  • Thread starter 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?
 

Answers and Replies

  • #2
Dick
Science Advisor
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Using mathematica to solve the equations may not be helping you understand them. In the answer x=1.0472, x is expressed in radians. That IS 60 degrees.
 

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