Linear Equations

Hi.

Quote:

Originally Posted by ChrisBarrett

Hello, I have been having trouble with these 3 questions that are on my upcoming test. I have tried it many times without getting an answer and I could really use some help.
Here are the questions:
1)
2x - 3y = 5
4y + 2z = -6
5x + 7z = -15
Solve for X Y and Z using substitution or elimination.
EDIT :I Recently tried to use a homework help online service who gave me the answer ( I didnt understand their work though) For this they got x=4 y=1 and z=-5

I dont have the time to solve all 3 exercices, but the first one:
I think the easiest way is solving by substitution, that means

2x - 3y = 5 => 2x = 5 +3y => x = (5+3y)/2

You have to other equation
(2) 4y + 2z = -6 here is no x in it -> nothing to substitute

5x + 7z = -15 but here is a x, you know x, x = (5+3y)/2

Therefor 5((5+3y)/2) + 7z = -15

(3) (25+15y)/2+7z = -15

Using (2) 4y + 2z = -6 => z = (-6-4y)/2

(3) (25+15y)/2+7z = -15 you know z, thus

(25+15y)/2+7*((-6-4y)/2) = -15 (solve it!) => y = 1

then solve z by z = (-6-4y)/2 (you already know y ...), same for x

Rapha