Hi.

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