# Algebra SOLVING LINEAR EQUATIONS AND INEQUALITIES

posted by Clyde

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 {(7,5-2)}

teach me how do this

1. 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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