Solving 2-Unknown Algebra Problems: A Challenge!

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The discussion centers on solving algebra problems with two unknowns, particularly in the context of physics applications. A user seeks challenging algebra problems to enhance their skills, specifically mentioning difficulties with equations involving multiple variables. It is noted that to solve for multiple unknowns, more equations are required, as illustrated by the example of needing to express one variable in terms of another. The conversation also touches on the relevance of systems of linear equations and matrices for solving such problems. Ultimately, the focus remains on improving algebra proficiency to tackle complex physics problems effectively.
Warwick
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Well, my physics teacher loves to make his problems hard, so he makes the physics portion of the problem med difficulty but he makes it so you have to solve with more than 1 unknown present. With big problems on forces I get lost in the algebra so.. I'm trying to improve my algebra skills. If anyone could give a couple algebra problems with 2 unknowns and solve for both, I would be grateful. Give me some nasty ones :)

I was making one up my self and couldn't figure it out.
8=2(2^2)
8=x(k^2)
I tired solving that but I couldn't get the answer,

thanks
 
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If you have more than 1 unknown, you need more than 1 equation, unless you want one unknown in terms of the others.

8 = xk^2 => k = (8/x)^1/2

It is impossible to get k (or x) with such little info.
 
http://www.ping.be/~ping1339/index.html#Main-Purpose-=-MATH-

Just look at systems of linear equations and look also under matrices and determinants...

regards
marlon
 
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Hmm, yes I remember doing these(not how, heh), never understood the use of them then. So this will work when I have 3 sum of the forces equations in physics when there is more than one unknown?

Can you show me how you would go about solving this one?

u=.065,m=2.5,a=0.12,g=9.8

Tcos20-ukn-mgsin25=ma
n-mgcos25+Tsin20=0

Well this one has only 1 unknown but can you use that matrice technique on it?
 
not necessary, you only need the first equation.

ukn = -ma -mdsin25 + Tcos20

that's all

regards
marlon
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
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