7*Eq1: 35a-7b+21c = 35.
--Eq2: 2a+7b-2c = 5
Sum:---37a+0+19c = 40.
5*Eq2:10a+35b-10c = 25
7*Eq3:28a-35b-49c = -455
Sum : 38a+ 0-59c = -430.
After eliminating b from two Eqs, we have two Eqs with two unknowns:
Eq4: 37a + 19c = 40.
Eq5: 38a - 59c = -430.
-38*Eq4: -38*37a - 722c = -1520.
37*Eq5 : 37*38a - 2183c = -15,910.
Sum :--------0 - 2905c = - 17,430.
C = 6.
In Eq4, replace c with 6:
37a + 19*6 = 40.
37a = 40 - 114 = -74.
a = -2.
In Eq2, replace a, and c with -2 and 6,
respectively:
2*-2 + 7b - 2*6 = 5.
-4 + 7b - 12 = 5
7b = 5 + 16 = 21.
b = 3.
Solution: a = -2, b = 3, c = 6.
i'm trying to solve this system of equations with 3 variables
5a-b+3c=5
2a+7b-2c=5
4a-5b-7c=-65
i tried to make the first equation like
a=something
so i could substitute it in to the 2nd equation, however i get decimals so it doesnt work
how do i do this problem?
1 answer
1 answer