Three Variables Please help

Question

Can someone show me the steps on how to solve this I have been trying for more the 4 hours and cant seem to get the answer correctly

  5a + 3b = 22..
12c - 85b = -244
33c + 13a = 290

 

answer in (2,4,8)

Answer 1

5a + 3b            = 22

     -85b + 12c = -244

13a        + 33c = 290

 

Multiply the first equation by 13 and the third equation by -5:

 

65a + 39b         = 286

        -85b + 12c = -244

-65a          -165c = -1450

 

Add the first and third equations and replace the third equation by the result:

 

65a + 39b         = 286

       -85b + 12c = -244     (*)    

        39b - 165c = -1164   (**)

 

Use the last two equations to eliminate b:

 

Multiply equation (*) by 39 and equation (**) by 85:

         -3315b + 468c = -9516

          3315b -14025c = -98940

 

Add these equations to get -13557c = -108456

                                                       c = 8

 

Substitute 8 for c in equation (*):  

                                   -85b + 12(8) = -244

                                   -85b + 96 = -244

                                              -85b = -340

                                                  b = 4

 

Replace b by 4 in the equation 5a + 3b = 22:

 

                              5a + 3(4) = 22

                              5a + 12 = 22

                                       5a = 10

                                         a = 2

 

SOLUTION:  a = 2, b = 4, c = 8

 

Answer 2

Ace:

 

This is not a difficult problem but it is full of places to make mistakes.  Be careful of your multiplication and your plus and minus signs.

 

First, I am going to add the first two equations together and eliminate the b variable.  Once I have done that I will add that equation to the original equation #3 and lose variable c.

 

To lose the b variable, I will multiply the first equation by 85 and the second equation by 3.

 

So 85(5a + 3b) = 85(22)       425a + 255b = 1870

     3(12c - 85b = 3(-244)        36c -255b =-732        425a + 36c = 1138

 

425a + 36c +1138

13a + 33c = 290

 

Multiply first equation by -33   so -33(425a + 36c) = -33(1138)

Multiply second equation by 36 so     36(13a + 33c) = 36(290)

 

First equation becomes         -14,025a -1188c = -37,554

Second equation becomes       468a +1188c = 10,440

 

Add the two equations together and c variables cancel  -13,557a = -27,114

 

Divide both sides by -13,557 so a becomes 2.

 

Go back and substitute 2 for a and find c and b.

 

This is a complicated problem with big numbers so be careful with multiplying and signs.

 

Hope this helps.

 

Denny Rose

Galion, OH