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 {(7,52)}
pleeeeeeeeeeeeeaaaseee :(

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

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

Algebra 
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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