Algebra SOLVING LINEAR EQUATIONS AND INEQUALITIES
posted by Clyde .
Please help Im am Grade 5 and my teacher is letting me do this.
1. Solve the solution by using elimination and substitution
3/x  2/y = 14
6/x + 3/y = 7
2. Solve by eliminating x (this is solving 3 linear equations) then substitute to the other two equation
x + y + 5z =2 (1)
4x  3y + 5z =3 (2)
3x  2y + 5z=1 (3)
the solution set for this should be {(7,52)}
Please I really need your help
teach me how do this

Algebra SOLVING LINEAR EQUATIONS AND INEQUALITIES 
Steve
#1 you can solve for 1/x and 1/y in the usual ways:
elimination: double the 1st and subtract
6/x  4/y = 28
6/x + 3/y = 7

7/y = 21
1/y = 3
then, 1/x = 8/3
or, y = 1/3 and x = 3/8
using substitution,
3/x = 14+2/y
2(14+2/y) + 3/y = 7
28 + 4/y + 3/y = 7
7/y = 21
1/y = 3
as above
For elimination, you can enter your coefficients at
http://www.gregthatcher.com/Mathematics/GaussJordan.aspx
and see all the details.
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