Question

solve the following system for a.
 a+3c=1
-5b+c-a=-3
2a+3b+c=-3

Answer 1

From the first equation a=1-3c. Substitute this value for a in the other equations:

-5b+c-(1-3c)=-5b+c-1+3c=-5b+4c-1=-3, -5b+4c=-2;

2(1-3c)+3b+c=2-6c+3b+c=2+3b-5c=-3, 3b-5c=-5.

Now we have two equations and two unknowns:

-5b+4c=-2,

3b-5c=-5.

Multiply the first equation by 3 and second by 5:

-15b+12c=-6,

15b-25c=-25.

Now add: -13c=-31, 13c=31, c=31/13.

Since a=1-3c, a=1-3(31/13)=1-93/13=-80/13.

(3b=5c-5=5(31/13)-5=155/13-5=90/13, b=30/13.)

Solution is a=-80/13.