You can
put this solution on YOUR website!PLease help ASAP: Solve the system with the addition method:
3/7x + 5/9y = 27 ----(1)
1/9x + 2/7y = 7 ----(2)
lcm of 7 and 9 is 63. So multiply (1) and (2) through out by 63. We get
27x +35y = 27X63 ----(3)X 7
7x +18y = 7X63 ----(4)X 27
(to equalise the coefficients of x we need to get the lcm (27X7)as the common coefficient in the resulting equations
(27X7)x +(35)X(7)y = 27X63X7 ----(5)
(7X27)x +(18)X(27)y = 7X63X27 ----(6)
(5) - (6) gives
0 + [35X7 - 18X27]y =0
[()]y = 0 which implies y=0
(product of two things =0 with one of them not zero implies the other is zero)
y=0 in (1)
(3/7)x= 27
3x = 27X7
x = (27X7)/3 = (9X3X7)/3 = 63
Answer: x= 63 and y=0
Note: If you retain the fractions and try to solve you might get boggled in a different way. Once you make up your mind to look at these huge products which let them be as they are(their actual value need not be got) and follow the steps one by one I think you should be fine.
The eyes see what the mind wishes to see.So focus your mind first
and then the eyes.
For instance (5) - (6) gives RHS = 0 why?
because the product is the same of the same three numbers
(whatever the huge product be)
If still you do not understand this method.Please write. Shall give you an alternate method.
(