Solve a system

Question

6r-5s+2t=-14

5r+s-4t=27

-6r+3s-3t=15

Answer 1

We need to find the values of r, s, and t, so that when we plug in those values, all the equations are satisfied.  Since the second equation has a coefficient of 1 for the s term, we can substitute that equation into the other equations.

 

Note: We must be careful in our algebra as we tackle this problem.

 

 

From the second equation,

 

s = -5r + 4t + 27

 

 

New form of the first equation:

 

6r - 5(-5r + 4t + 27) + 2t = -14

 

6r + 25r - 20t - 135 + 2t = -14

 

31r - 18t = 121

 

 

New form of the third equation:

 

-6r + 3(-5r + 4t + 27) - 3t = 15

 

-6r - 15r + 12t + 81 - 3t = 15

 

-21r + 9t = -66

 

 

The bolded equations are the equations we will work with.  Notice that we have one less equation and one less variable to deal with.

 

 

31r - 18t = 121             new eq1

 

-21r + 9t = -66            new eq3

 

 

Multiply new eq3 by 2.  Keep new eq1

 

31r - 18t = 121             new eq1

-42r + 18t = -132          new eq3

 

 

Add the two equations to eliminate the t terms.

 

-11r = -11

 

     r = 1

 

Next, you will substitute this value of r into new eq1 to solve for t.  After you do that, substitute both values of r and t into the original first equation to solve for s.