SOLUTION: Please help my solve this. It was in the form of a word problem and this is what I got it to. Now I am clueless on what to do. {{{d+e-f=5}}} {{{e=f+d+7}}} {{{f=2e-11}}} Th

Algebra ->  Coordinate Systems and Linear Equations  -> Lessons -> SOLUTION: Please help my solve this. It was in the form of a word problem and this is what I got it to. Now I am clueless on what to do. {{{d+e-f=5}}} {{{e=f+d+7}}} {{{f=2e-11}}} Th      Log On


   

Question 26915: Please help my solve this. It was in the form of a word problem and this is what I got it to. Now I am clueless on what to do.
d%2Be-f=5
e=f%2Bd%2B7
f=2e-11
Thank you!
Answer by bmauger(101) About Me  (Show Source):
You can put this solution on YOUR website!
To make it a bit easier, I'd rewrite it so every line is in the same form like:
Eq1:d%2Be-f=5
Eq2:-d%2Be-f=7
Eq3:-2e%2Bf=-11
Now in this form we can see that we can use combination and add Eq1 & Eq2 to write an equation only in terms of E&F:
2e-2f=12
Now that we have two equations of only two variables (this one & Eq3) we can solve for e & f. Adding this new equation to Eq 3 we get:
-f=1
f=-1
Putting this back into Eq 3 we get:
-2e%2Bf=-11
-2e%2B-1=-11
-2e=-10
e=5
Now we know e & f, so we can use any equation (in this example Eq 1.) to find d:
d%2Be-f=5
d%2B5-%28-1%29=5
d=-1
Double checks:
Eq1:d%2Be-f=5=-1%2B5-%28-1%29=5check.
Eq2:-d%2Be-f=7=-%28-1%29%2B5-%28-1%29=7check.
Eq3:-2e%2Bf=-11=-2%2A5%2B%28-1%29=-11check.