SOLUTION: I am so lost on this lesson. Help please! :) 5r - 4s - 3t = 3 r = 3s + 1 t = s + r

Algebra ->  Coordinate Systems and Linear Equations -> SOLUTION: I am so lost on this lesson. Help please! :) 5r - 4s - 3t = 3 r = 3s + 1 t = s + r      Log On


   

Question 180219This question is from textbook Algebra 2
: I am so lost on this lesson. Help please! :)
5r - 4s - 3t = 3
r = 3s + 1
t = s + r This question is from textbook Algebra 2

Found 2 solutions by Fombitz, stanbon:
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
Substitute and get to an equation with one variable.
5r-4s-3t=3
5%283s%2B1%29-4s-3%28s%2Br%29=3
5%283s%2B1%29-4s-3%28s%2B%283s%2B1%29%29=3
You now have an equation only in s.
Now solve for s and go back to find r and t.
Repost if you get stuck or need more help.

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
5r - 4s - 3t = 3
r = 3s + 1
t = s + r
----------------
You want to have one equation with one variable.
Find a variable that is associated with two others, like "s"
r = 3s+1
t =s + r = s+ 3s+1 = 4s+1
-------------------------------
Now substitute for "r" and for "t" in the 1st equation to get:
5(3s+1) - 4s - 3(4s+1) = 3
Solve for "s".
15s + 5 - 4s - 12s - 3 = 3
-s +2 = 3
-s = 1
s = -1
----------------
Substitute to solve for r and for t:
r = 3s+1 = 3(-1) + 1 = -2
---
t = 4s+1 = 4(-1) + 1 = -3
============================
Final Answer:
s = -1
r = -2
t = -3
==============
Cheers,
Stan H.