# Algebra

posted by Clyde

Solving Linear Equations and Inequalities

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)

my teacher gave me the answer she said the solution set for this should be {(7,5-2)}

pleeeeeeeeeeeeeaaaseee :(

1. Damon

http://www.jiskha.com/display.cgi?id=1461567386#1461567386.1461580442

2. Reiny

Why eliminate x when the obvious choice would be to eliminate z ??

#2 - #1 ---> 3x - 4y = 1 **
#2 - #3 ---> x - y = 2 or x = y+2
sub that into **
3(y+2) - 4y = 1
3y + 6 - 4y = 1
-y = -5
y = 5 , then
x = 7
back into #1:
7 + 5 + 5z = 2
5z = -10
z = -2

so the solution is (7,5,-2)

Depending on the coefficients in the equations, it is often easier to use a combination of methods, rather than one rigid approach

3. Clyde

xis what needs to be eliminated. . . my teacher already try eliminated the z

4. Reiny

Ok,

multiply the first by 4, then subtract the 2nd

multiply the first by 3, then subtract the third.

Now you have two equations with y's and z's

carry on

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