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:
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
2x + 3y - z = 9
x - 3y + z = -6
3x + y - 4z = 31
Solve for X Y and Z using substitution or elimination. (Same as before)
6/x - 4/y = 1
-3/x + 3/y = 2
Solve for X and Y (6/x, 4/y , -3/x and 3/y and fractions) using substitution
EDIT: Same homework help (Once again Didn't Understand what the heck their work meant but..) They Got for X=6/11 and Y=2/5
Thank you in advance and please show your work so I can understand it.
MOVED POST TO URGENT SECTION DUE TO TEST TIME CHANGE.
I dont have the time to solve all 3 exercices, but the first one:
Originally Posted by ChrisBarrett
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