Matt - Copy this out and when you've read through this, possibly several times, turn your book over and try to do it yourself without looking at the copy. label the equations 1, 2 & 3. Notice (1) & (3) both have a + 3z, so SUBTRACT (1) from (3) to get 4x + y = 5 ...(4) ; multiply (2) by 3 to get - 6x - 3y + 3z = - 9, so again SUBTRACT this from (3) to get 8x + 6y = 14, or 4x + 3y = 7 (1.e. divide by 2) and this is equation (5), so we take (4) & (5) together; subtract (4) from (5) and we have 2y = 2 so y = 1; substitute this into (4) to get 4x + 1 = 5, sso 4x = 4, x = 1; put those two values into (2) (in fact, any of them, just choose the easiest) and we get - 2 - 1 + z = -3, so -3 + z = -3, so z = 0. Now heck by using all three values in a different equation to make sure it checks out (if you use (1) and put in those 3 values, check to see if the answer is 0)
(2) For substitution, rewrite (1) as x + y + z = 8 or z = 8 - (x + y) and use that expression in (2) to get - 4x + 4y + 5[8 - (x + y)]= 7, or 4x + 4y + 40 - 5x - 5y = 7 or 40 - x - y = 7 or x + y = 33; this we can write as y = 33 - x, so use z = 8 - x - y = 8 - x - (33 - x) = - 25 ; use that in (3) and x = 2 - z = 2 + 25 = 27; put both those values into (1) and this will giuve you the value of y. You can, of course,take a totally different approach and find e.g. x & y hence z or y & z hence x. Substitution is always a bit more difficult for 3 variables than elimination, and given the choice most people would automatically go for elimination instead. Good Luck in your test.
Retired Maths Teacher