Solve Algebra: (0.97)(1+ Beta gas (-Ti)) = (1 + Beta steel (-Ti))

In summary, the student is asking for help with an algebraic equation for their physics homework. They are struggling with distributing and solving for the variable Ti, despite attempting to work it out on paper. The expert suggests distributing and manipulating the equation to free the variable.
  • #1
chantalprince
54
0

Homework Statement



Hi- this is actually part of my physics homework, but all I need help with deals strictly with algebra.

This is my equation:

(0.97)(1 + Beta gas (-Ti)) = (1 + Beta steel (-Ti))



Homework Equations





The Attempt at a Solution




How do I solve for Ti? To me it seems like they cancel out, but from what I can remember, they don't and I can solve for it. I think it has to do with distributing before I start to solve, but I'm having issues. I am sorry I don't have more of an attempt, but I've tried to work it out on paper and they just cancel out.

Thanks so much.
 
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  • #2
First, distribute the 0.97 over the left-hand side. Then add -0.97*(Beta gas)*Ti to both sides. If the variable you're interested in is trapped inside a distribution, distribute it out so you can free the variable for independent algebraic manipulation.
 
  • #3
Thanks slider :smile:
 

FAQ: Solve Algebra: (0.97)(1+ Beta gas (-Ti)) = (1 + Beta steel (-Ti))

How do I solve for Beta gas and Beta steel in this equation?

In order to solve for Beta gas and Beta steel, you will need to first isolate them on one side of the equation. Then, you can use algebraic techniques such as factoring, combining like terms, and applying the distributive property to simplify and solve for the variables.

What is the significance of the numbers 0.97 and 1 in this equation?

The numbers 0.97 and 1 represent coefficients that are used to scale the variables in the equation. They are typically chosen based on the units and context of the problem.

How does temperature (Ti) affect the solution of this equation?

The temperature (Ti) is a variable in this equation, which means that it can vary and affect the overall solution. It is important to pay attention to the units and context of the problem when solving for Ti.

Can I use any value for Beta gas and Beta steel in this equation?

Yes, as long as the values are consistent and correspond to the units and context of the problem. However, some values may lead to more complex or difficult solutions.

Are there any practical applications of this equation?

Yes, this equation can be used in various scientific and engineering fields, such as thermodynamics and materials science, to analyze and compare properties of different materials under different conditions.

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