Posted by john on Sunday, October 7, 2012 at 1:29pm.
i'm trying to solve this system of equations with 3 variables
5ab+3c=5
2a+7b2c=5
4a5b7c=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: 35a7b+21c = 35.
Eq2: 2a+7b2c = 5
Sum:37a+0+19c = 40.
5*Eq2:10a+35b10c = 25
7*Eq3:28a35b49c = 455
Sum : 38a+ 059c = 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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