Posted by **john** on Sunday, October 7, 2012 at 1:29pm.

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?

- algebra 2 -
**Henry**, Monday, October 8, 2012 at 9:47pm
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.

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