+0

Simultaneous equations with 5 unknowns

0
407
4

Here are simultaneous equations with 5 unknowns:

b=d+f

e=c+f

12=10c+20e

0=10f+20b-10c

0=10d-20e-10f

Guest Dec 3, 2015

#1
+80928
+10

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

CPhill  Dec 3, 2015
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#1
+80928
+10

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

CPhill  Dec 3, 2015
#2
+26399
+5

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

.

Alan  Dec 4, 2015
#3
+91436
0

Thanks Chris and Alan,

I had not even seen that question :/

Melody  Dec 4, 2015
#4
+80928
+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)

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
edited by CPhill  Dec 27, 2015

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