Here are simultaneous equations with 5 unknowns:

b=d+f

e=c+f

12=10c+20e

0=10f+20b-10c

0=10d-20e-10f

I need to find out "f". Please help

Guest Dec 3, 2015

#1**+10 **

Best Answer

According to WolframAlpha......no solutions exist for this system......

CPhill
Dec 3, 2015

#2**+5 **

I get the following solutions (one of us has probably typed in something incorrectly!):

.

Alan
Dec 4, 2015

#4**+5 **

I wanted to come back to this one and solve it with substitutions......

(1) b=d+f → f = b - d

(2) e=c+f → f = e - c

(3) 12=10c+20e → 6 = 5c + 10e

(4) 0=10f+20b-10c → 0 = f + 2b - c

(5) 0=10d-20e-10f → 0 = d - 2e - f

Subtract (2) from (1)

b - e = d - c (6)

Add (4) and (5)

0 = 2b - 2e + d - c → d - c = 2e - 2b

Substitute the previous result into (6) for d - c

b - e = 2e - 2b → 3b = 3e → b = e and from(6) this implies that → d = c

Substitute (2) and the fact that b = e into the rearrangement of (4)

0 = e - c + 2e - c → 2c = 3e → e = [2/3]c

Substitute the previous result into the rearrangement of (3)

6 = 5c + 10(2/3)c → 18 = 15c + 20c → 18 = 35c → c = 18/35 → d = 18/35

And e = [2/3]c = [2/3](18/35) = 12/35 = b

And f = b - d .....so f = 12/35 - 18/35 = -6/35

Alan's answers are correct......I may have mis-typed something when I put the system into WolframAlpha.....!!!!

[When I put this system back into WolframAlpha, it still says no solutions exist.....odd!! ]

CPhill
Dec 27, 2015