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 :(

4 answers

http://www.jiskha.com/display.cgi?id=1461567386#1461567386.1461580442
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
xis what needs to be eliminated. . . my teacher already try eliminated the z
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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