Solving for variables using 3 different equations (simultaneous equations)

In summary, the problem involves solving for A, T1, and T2 in three equations. The first two equations involve isolating T1 and T2 respectively, while the third equation involves substituting for T1 and T2 and solving for a.
  • #1
ReMa
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Homework Statement



Solve for A, T1 and T2.

Equa 1) T1 + 25.0a = 245
Equa 2) T2 − 10.0a = 98.0
Equa 3) T1 − T2 − 80.0a = 78.4

Homework Equations



As above.

The Attempt at a Solution



I isolated a in the first equation:

a = (245-T1)/25

and T2 in the second equation:

T2 = 98 + 10(245-T1)/25

And when subbing these values in I end up getting an answer WAY off for T1. Any help would be highly appreciated as I feel like I'm just doing improper calculations with my fractions...
 
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  • #2
Solve for T1 in equation 1 and for T2 in equation 2. Both T1 and T2 will be in terms of a. Substitute for T1 and T2 in equation 3, and solve for a. After you get a, you can get T1 and T2.
 
  • #3


I can provide a methodical approach to solving these equations. The most effective way to solve for A, T1, and T2 would be to use the method of substitution. This involves isolating one variable in one equation and substituting it into the other equations.

Let's start by isolating A in the first equation:

T1 + 25A = 245
25A = 245 - T1
A = (245 - T1)/25

Now we can substitute this value for A into the second equation:

T2 - 10((245 - T1)/25) = 98
T2 = 98 + 10((245 - T1)/25)

Next, we can substitute this value for T2 into the third equation:

T1 - (98 + 10((245 - T1)/25)) - 80A = 78.4

Simplifying this equation, we get:

T1 - 9.8(245 - T1) - 80A = 78.4
T1 - 2450 + 9.8T1 - 80A = 78.4
10.8T1 - 80A = 2528.4

Lastly, we can substitute the values for A and T1 into this equation and solve for T1:

10.8T1 - 80((245 - T1)/25) = 2528.4
10.8T1 - 80(9.8 - 0.8T1) = 2528.4
10.8T1 - 784 + 64T1 = 2528.4
74.8T1 = 3312.4
T1 = 44.3

Now that we have solved for T1, we can use this value to find T2:

T2 = 98 + 10((245 - 44.3)/25)
T2 = 98 + 81.9
T2 = 179.9

And finally, we can use both T1 and T2 to find A:

A = (245 - 44.3)/25
A = 8.2

Therefore, the solutions to the three equations are A = 8.2, T1 = 44.3, and T2 = 179.9. It is important to check these
 

Related to Solving for variables using 3 different equations (simultaneous equations)

What are simultaneous equations?

Simultaneous equations are a set of two or more equations that have to be solved together in order to find the values of the variables.

What is the purpose of solving for variables using 3 different equations?

The purpose of solving for variables using 3 different equations is to eliminate any possible errors and to find a more precise solution to the problem.

What methods can be used to solve for variables using 3 different equations?

The most commonly used methods are substitution, elimination, and graphical representation.

How do you know if a system of equations has a unique solution?

A system of equations has a unique solution if the number of variables is equal to the number of equations and the equations are independent of each other.

Can simultaneous equations be used in real-life situations?

Yes, simultaneous equations can be used in various real-life situations such as in business, economics, engineering, and physics to find the values of multiple variables simultaneously.

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