System of four equations

Question

4a+6b+2c+4d=-12. -2a+3b+c+d=-3.                          6a-3b+c-d=9.                             -10a-6b-c-2d=1

Answer 1

Technology is great, but some teachers/professors/math departments are meanies that don't allow calculators. So here's an old fashioned algebraic solution:  

 

First, try to reduce the system to a system with 3 equations in 3 unknowns.

 

For example, if we add equation 1 to -2(equation 2) we get                 8a + 2d = -6.  So, 4a + d = -3.  Thus, d = -3 - 4a.

 

Replace d by -3-4a in each of the last three equations.  This gives us the system:

 

-6a + 3b + c = 0

10a - 3b + c = 6

-2a -6b - c = -5

 

Now, reduce this system to a system involving two equations in two unknowns.  I will eliminate c, but you could also eliminate either a or b:

 

Subtract the second equation from the first to get:  -16a + 6b = -6

Add the first equation to the third to get:                      -8a - 3b  = -5

 

If we then add the first equation to twice the second, we have:

 

-32a = -16   So, a = 1/2

 

Substitute for a in the equation -16a + 6b = -6 to get -8 + 6b = -6

                                                                                                      6b = 2

                                                                                                        b = 1/3

 

-6a + 3b + c = 0 gives us -3 + 1 + c = 0   so, c = 2

 

Finally, since 4a + 6b + 2c + 4d = -12, we have 2 + 2 + 4 + 4d = -12

 

Therefore, 8 + 4d = -12

 

                           4d = -20

 

                             d = -5

 

Solution:  a = 1/2    b = 1/3     c = 2     d = -5 

Answer 2

This system can be solved by using substitution multiple times, by Cramer's rule or matrix inversion.   However, the rref (reduced row echelon formation) function can be found in the catalog of of the TI-84 calculator.   

  Make  matrix A a 5x4 and enter the coefficients into the matrix

       4 6 2 4 -12

       -2 3 1 1 -3

       6 -3 1 -1  9

      -10 -6 -1 -2  1

 

  Then compute  rref(A).   The answers are in the last column of the returned matrix.   They

   are a  = 1/2

         b = 1/3

         c = 2

        d = -5