Solving for variables using 3 different equations (simultaneous equations)

[ReMa]

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...

 

[Mark44]

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.

 

[Architeuthis Dux]

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