# Is it possible to solve this (simple?) set of equations?

• Sep 14th 2011, 06:44 AM
paulkm
Is it possible to solve this (simple?) set of equations?
Hi guys,

Is it possible to solve:

d = a - (b /2);
e = d + b;
f = p - (c /2);

g = f - (d /2);
h = g + d;
i = f - (e /2) + c;
j = i + e;

Given g,h,i,j to find a,b,c,d,e,f,p? Some are easy but then I seem to be left in a situation where its not possible for me to solve the remaining ones? :(
• Sep 14th 2011, 07:41 AM
piscoau
Re: Is it possible to solve this (simple?) set of equations?
no, you can't get a unique solution, because there are more than 7 variable exist, which is more than the number of equations given.
• Sep 14th 2011, 08:14 AM
Sudharaka
Re: Is it possible to solve this (simple?) set of equations?
Quote:

Originally Posted by paulkm
Hi guys,

Is it possible to solve:

d = a - (b /2);
e = d + b;
f = p - (c /2);

g = f - (d /2);
h = g + d;
i = f - (e /2) + c;
j = i + e;

Given g,h,i,j to find a,b,c,d,e,f,p? Some are easy but then I seem to be left in a situation where its not possible for me to solve the remaining ones? :(

Hi paulkm,

Can you show us what you have done. Since you have seven unknowns (a,b,c,d,e,f,p) and seven linearly independent equations it is possible to find a solution to this system.
• Sep 14th 2011, 08:28 AM
piscoau
Re: Is it possible to solve this (simple?) set of equations?
sorry for that I give the wrong answer since I didn't notice the values of g,h,i,j are given.
this set of equation can be solved by a lots times of substitution.
• Sep 16th 2011, 04:18 AM
paulkm
Re: Is it possible to solve this (simple?) set of equations?
Thanks for the responses, I will post my working in a few days as I've taken quite ill at the moment :(
• Sep 18th 2011, 09:35 AM
paulkm
Re: Is it possible to solve this (simple?) set of equations?
Here is what I've done which appears to be wrong:

Let
g = 1
h = 2
i = 3
j = 4

Then the equations look like this:
1) d = a - (b /2)
2) e = d + b
3) f = p - (c /2)
4) 1 = f - (d /2)
5) 2 = 1 + d
6) 3 = f - (e /2) + c
7) 4 = 3 + e

To work out e using equation 7:

e = 4 - 3, which means that e = j-i and the value of e 1

To work out d using equation 5:

d = 2 -1, which means that d= h-g and the value of d is also 1

To work out f using equation 4:

f = 1 + (1/2), so f = g + (d/2) and the value of f is 1.5

To work out c I used equation 6 and came to this:

c=(e/2)+f which means the value of c is 2 and only leaves p, b and a to find.

For p using equation 3:

p = f + ( c /2 ), which gives p = 2.5

For b using equation 2:

b = d-e, which gives b = 0

For a using equation 1:

a=d+(b/2), which gives a =1

I thought this might be correct as substituting back into the source equations appeared to yield the correct answers:

1) d = a - (b /2) 1 = 1 - (0/2)
2) e = d + b 1 = 1 + 0
3) f = p - (c /2) 1.5 = 2.5 - (2/2)
4) 1 = f - (d /2) 1 = 1.5 - (1/2)
5) 2 = 1 + d 2 = 1 + 1
6) 3 = f - (e /2) + c 3 = 1.5 - (1/2)+2
7) 4 = 3 + e 4 = 3 + 1

So then I tested again with these values:

g=50
h=100
i=70
j=15

Which is where I seem to run into an issue.

Using what I think I've figured out:
e=j-i
d=h-g
f=g+(d/2)
c=(e/2)+f
p=f+(c/2)
b=d-e
a=d+(b/2)

1) d = a - (b /2)
2) e = d + b
3) f = p - (c /2)
4) 50 = f - (d /2)
5) 100 = 50 + d
6) 70 = f - (e /2) + c
7) 15 = 70 + e

e=15-70=-55
d=100-50=50
f=50+(50/2)=75
c=(-55/2)+75=47.5
p=75+(47.5/2)=98.75
b=50--55=105
a=50+(105/2)=102.5

Sub these back into make sure it works:

1) d = a - (b /2) 50= 102.5 - (105/2) ok so far!
2) e = d + b -55 = 50 + 105, wrong this is 155!

I can't see how this has gone wrong unless e=j-i should be e=i-j? But that doesn't make sense because 15 = 70 + -55 works for equation 7?
• Sep 18th 2011, 06:44 PM
Sudharaka
Re: Is it possible to solve this (simple?) set of equations?
Quote:

Originally Posted by paulkm
Here is what I've done which appears to be wrong:

Let
g = 1
h = 2
i = 3
j = 4

Then the equations look like this:
1) d = a - (b /2)
2) e = d + b
3) f = p - (c /2)
4) 1 = f - (d /2)
5) 2 = 1 + d
6) 3 = f - (e /2) + c
7) 4 = 3 + e

To work out e using equation 7:

e = 4 - 3, which means that e = j-i and the value of e 1

To work out d using equation 5:

d = 2 -1, which means that d= h-g and the value of d is also 1

To work out f using equation 4:

f = 1 + (1/2), so f = g + (d/2) and the value of f is 1.5

To work out c I used equation 6 and came to this:

c=(e/2)+f which means the value of c is 2 and only leaves p, b and a to find.

For p using equation 3:

p = f + ( c /2 ), which gives p = 2.5

For b using equation 2:

b = d-e, which gives b = 0

For a using equation 1:

a=d+(b/2), which gives a =1

I thought this might be correct as substituting back into the source equations appeared to yield the correct answers:

1) d = a - (b /2) 1 = 1 - (0/2)
2) e = d + b 1 = 1 + 0
3) f = p - (c /2) 1.5 = 2.5 - (2/2)
4) 1 = f - (d /2) 1 = 1.5 - (1/2)
5) 2 = 1 + d 2 = 1 + 1
6) 3 = f - (e /2) + c 3 = 1.5 - (1/2)+2
7) 4 = 3 + e 4 = 3 + 1

So then I tested again with these values:

g=50
h=100
i=70
j=15

Which is where I seem to run into an issue.

Using what I think I've figured out:
e=j-i
d=h-g
f=g+(d/2)
c=(e/2)+f
p=f+(c/2)
b=d-e
a=d+(b/2)

1) d = a - (b /2)
2) e = d + b
3) f = p - (c /2)
4) 50 = f - (d /2)
5) 100 = 50 + d
6) 70 = f - (e /2) + c
7) 15 = 70 + e

e=15-70=-55
d=100-50=50
f=50+(50/2)=75
c=(-55/2)+75=47.5
p=75+(47.5/2)=98.75
b=50--55=105
a=50+(105/2)=102.5

Sub these back into make sure it works:

1) d = a - (b /2) 50= 102.5 - (105/2) ok so far!
2) e = d + b -55 = 50 + 105, wrong this is 155!

I can't see how this has gone wrong unless e=j-i should be e=i-j? But that doesn't make sense because 15 = 70 + -55 works for equation 7?

Hi paulkm,

Your answers for a,b,c,d,e,f and p when g=1, h=2, i=3 and j=4 are correct. However when, g=50, h=100, i=70, j=15 the answers are,

a=-2.5000
b=-105.0000
c=-32.5000
d=50.0000
e=-55.0000
f=75.0000
p=58.7500

Substitute them back into the equations and see if they work.
• Sep 19th 2011, 07:50 AM
paulkm
Re: Is it possible to solve this (simple?) set of equations?
Quote:

Originally Posted by Sudharaka
Hi paulkm,

Your answers for a,b,c,d,e,f and p when g=1, h=2, i=3 and j=4 are correct. However when, g=50, h=100, i=70, j=15 the answers are,

a=-2.5000
b=-105.0000
c=-32.5000
d=50.0000
e=-55.0000
f=75.0000
p=58.7500

Substitute them back into the equations and see if they work.

I get this:

1) d = a - (b /2) 50 = -2.5 -(-105/2)
2) e = d + b -55 = 50 + -105
3) f = p - (c /2) 75 = 58.7 - (-32/2)
4) 50 = f - (d /2) 50 = 75 - (50/2)
5) 100 = 50 + d 100 = 50 + 50
6) 70 = f - (e /2) + c 70 = 75 - (-55/2)+-32
7) 15 = 70 + e 15 = 70 + -55

All seem ok, but for equation 3) 75 = 58.7 - (-32/2) is actually 74.7, should it not be exactly 75? I.e 59 - (-32/2) ?

The same goes for equation 6) where the result is 70.5 not 70?

My other question is how did you get these values:

a=-2.5000
b=-105.0000
c=-32.5000
d=50.0000
e=-55.0000
f=75.0000
p=58.7500

Since using the other equations I worked out:

e=15-70=-55
d=100-50=50
f=50+(50/2)=75
c=(-55/2)+75=47.5
p=75+(47.5/2)=98.75
b=50--55=105
a=50+(105/2)=102.5

Or in order:

a=50+(105/2)=102.5
b=50--55=105
c=(-55/2)+75=47.5
d=100-50=50
e=15-70=-55
f=50+(50/2)=75
p=75+(47.5/2)=98.75

I don't understand how I ended up with the wrong values? Thanks for your help :)
• Sep 19th 2011, 08:49 AM
Sudharaka
Re: Is it possible to solve this (simple?) set of equations?
Quote:

Originally Posted by paulkm
I get this:

1) d = a - (b /2) 50 = -2.5 -(-105/2)
2) e = d + b -55 = 50 + -105
3) f = p - (c /2) 75 = 58.7 - (-32/2)
4) 50 = f - (d /2) 50 = 75 - (50/2)
5) 100 = 50 + d 100 = 50 + 50
6) 70 = f - (e /2) + c 70 = 75 - (-55/2)+-32
7) 15 = 70 + e 15 = 70 + -55

All seem ok, but for equation 3) 75 = 58.7 - (-32/2) is actually 74.7, should it not be exactly 75? I.e 59 - (-32/2) ?

I have written p=58.75 and c=-32.5

The same goes for equation 6) where the result is 70.5 not 70?

You have made a similar mistake. The result is 70. Note that, c=-32.5. You have substituted c=-32. Please substitute the exact values that I have given.

My other question is how did you get these values:

a=-2.5000
b=-105.0000
c=-32.5000
d=50.0000
e=-55.0000
f=75.0000
p=58.7500

Since using the other equations I worked out:

e=15-70=-55
d=100-50=50
f=50+(50/2)=75
c=(-55/2)+75=47.5
p=75+(47.5/2)=98.75
b=50--55=105
a=50+(105/2)=102.5

Or in order:

a=50+(105/2)=102.5
b=50--55=105
c=(-55/2)+75=47.5
d=100-50=50
e=15-70=-55
f=50+(50/2)=75
p=75+(47.5/2)=98.75

I don't understand how I ended up with the wrong values? Thanks for your help :)

.
• Sep 19th 2011, 09:41 AM
paulkm
Re: Is it possible to solve this (simple?) set of equations?
Sorry that was a very stupid mistake on my part. I guess that's also why my values for a-p where wrong previously too..

Edit:

So doing it again..

e=15-70
e=-55

d=100-50
d=50

f = 50 + (50/2)
f = 75

c = (-55/2) + 75
c = 47.5

But here c is clearly wrong as you calculate it as -32.5? Can you explain how its gone wrong for me?
• Sep 19th 2011, 06:36 PM
Sudharaka
Re: Is it possible to solve this (simple?) set of equations?
Quote:

Originally Posted by paulkm
Sorry that was a very stupid mistake on my part. I guess that's also why my values for a-p where wrong previously too..

Edit:

So doing it again..

e=15-70
e=-55

d=100-50
d=50

f = 50 + (50/2)
f = 75

c = (-55/2) + 75
c = 47.5

You have taken c=e/2+f; but there is no equation as such. Please try to do the problem from the beginning. If you want to check the values I have given, substitute them in the original equations;

d = a - (b /2);
e = d + b;
f = p - (c /2);

g = f - (d /2);
h = g + d;
i = f - (e /2) + c;
j = i + e;

with g=50, h=100, i=70, j=15

But here c is clearly wrong as you calculate it as -32.5? Can you explain how its gone wrong for me?

.
• Sep 19th 2011, 07:12 PM
paulkm
Re: Is it possible to solve this (simple?) set of equations?
Sorry I'm getting really confused here!

Here is the full detail of how I've attempted to solve it:

g=50
h=100
i=70
j=15

Which gives:

1) d = a - (b /2)
2) e = d + b
3) f = p - (c /2)
4) 50 = f - (d /2)
5) 100 = 50 + d
6) 70 = f - (e /2) + c
7) 15 = 70 + e

First I solve e by using 7):

15 = 70 + e
70 + e = 15
(-70 both sides)
e = -55

Which means that e = j-i

Now solve d using equation 5):

100 = 50 + d
50 + d = 100
(-50 from both sides)
d = 50

Which means that d = h-g

Now I solve f using equation 4):

50 = f - (d /2)
f - (d /2) = 50
Sub d since we know it
f - (50 /2) = 50
f - 25 = 50
f = 75
which means that f = g + (d/2)

Now solve c using equation 6):

70 = f - (e /2) + c
f - (e /2) + c = 70
Sub f as we know its value
75 - (e /2) + c = 70
Sub e as we know its value
75 - (-55 /2) + c = 70
102.5 + c = 70
(-102.5 from both sides)
c = -32.5

Which means c = f -( e/2)

Now solve P using equation 3):
f = p - (c /2)
Sub f
75 = p - (c /2)
Sub c
75 = p - (-32.5 /2)
75 = p - (-16.25)
91.25 = p or
p = 91.25

Which means p = f + (c/2)

Now solve b using equation 2):

e = d + b
Sub e
-55 = d + b
Sub d
-55 = 50 + b
(-50 from both sides)
-105 = b or
b = -105

Which means b = e-d

Finally solve a using equation 1):

d = a - (b /2)
Sub d
50 = a - (b /2)
Sub b
50 = a - (-105 /2)
50 = a - (-52.5)
(+52.5 to both sides)
102.5 = a or
a = 102.5
Which means a = d + (b/2)

So:
a=102.5 (wrong somehow..)
b=-105
c=-32.5
d=50
e=-55
f=75
p=91.25 (wrong somehow..)

Testing gives:

1) 50 = 102.5 - (-105 /2) (wrong because 155)
2) -55 = 50 + -105 (ok)
3) 75 = 91.25 - (-32.5 /2) (wrong because 107.5)
4) 50 = 75 - (50 /2) (ok)
5) 100 = 50 + 50 (ok)
6) 70 = 75 - (-55 /2) + -32.5 (ok)
7) 15 = 70 + -55 (ok)

So I can see I've still managed to get the values of a and p wrong?
• Sep 20th 2011, 02:33 AM
Sudharaka
Re: Is it possible to solve this (simple?) set of equations?
Quote:

Originally Posted by paulkm
Sorry I'm getting really confused here!

Here is the full detail of how I've attempted to solve it:

g=50
h=100
i=70
j=15

Which gives:

1) d = a - (b /2)
2) e = d + b
3) f = p - (c /2)
4) 50 = f - (d /2)
5) 100 = 50 + d
6) 70 = f - (e /2) + c
7) 15 = 70 + e

First I solve e by using 7):

15 = 70 + e
70 + e = 15
(-70 both sides)
e = -55

Which means that e = j-i

Now solve d using equation 5):

100 = 50 + d
50 + d = 100
(-50 from both sides)
d = 50

Which means that d = h-g

Now I solve f using equation 4):

50 = f - (d /2)
f - (d /2) = 50
Sub d since we know it
f - (50 /2) = 50
f - 25 = 50
f = 75
which means that f = g + (d/2)

Now solve c using equation 6):

70 = f - (e /2) + c
f - (e /2) + c = 70
Sub f as we know its value
75 - (e /2) + c = 70
Sub e as we know its value
75 - (-55 /2) + c = 70
102.5 + c = 70
(-102.5 from both sides)
c = -32.5

Which means c = f -( e/2)

Now solve P using equation 3):
f = p - (c /2)
Sub f
75 = p - (c /2)
Sub c
75 = p - (-32.5 /2)
75 = p - (-16.25)
91.25 = p or
p = 91.25

You haven't considered the order of operations. Multiplication should be done before addition. So -16.25 should be multiplied by -1 before adding whatever necessary to subject p.

75 = p - (-16.25)
75=p+16.25
75-16.25=p
p=58.75

Which means p = f + (c/2)

Now solve b using equation 2):

e = d + b
Sub e
-55 = d + b
Sub d
-55 = 50 + b
(-50 from both sides)
-105 = b or
b = -105

Which means b = e-d

Finally solve a using equation 1):

d = a - (b /2)
Sub d
50 = a - (b /2)
Sub b
50 = a - (-105 /2)
50 = a - (-52.5)
(+52.5 to both sides)
102.5 = a or
a = 102.5

Again the same mistake as earlier. I hope you can correct it now.

Which means a = d + (b/2)

So:
a=102.5 (wrong somehow..)
b=-105
c=-32.5
d=50
e=-55
f=75
p=91.25 (wrong somehow..)

Testing gives:

1) 50 = 102.5 - (-105 /2) (wrong because 155)
2) -55 = 50 + -105 (ok)
3) 75 = 91.25 - (-32.5 /2) (wrong because 107.5)
4) 50 = 75 - (50 /2) (ok)
5) 100 = 50 + 50 (ok)
6) 70 = 75 - (-55 /2) + -32.5 (ok)
7) 15 = 70 + -55 (ok)

So I can see I've still managed to get the values of a and p wrong?

.
• Sep 20th 2011, 05:39 AM
CaptainBlack
Re: Is it possible to solve this (simple?) set of equations?
Quote:

Originally Posted by paulkm
Hi guys,

Is it possible to solve:

d = a - (b /2);
e = d + b;
f = p - (c /2);

g = f - (d /2);
h = g + d;
i = f - (e /2) + c;
j = i + e;

Given g,h,i,j to find a,b,c,d,e,f,p? Some are easy but then I seem to be left in a situation where its not possible for me to solve the remaining ones? :(

Yes, you just have to systematically solve for unknowns in terms of knowns, starting from the bottom and working up seems easiest.

CB
• Sep 20th 2011, 05:31 PM
paulkm
Re: Is it possible to solve this (simple?) set of equations?
Thanks a lot I can see what I've done wrong now :)

Also should it be ok to always solve it using this method?

Derived equations which should always yeild the correct values for e,d,f,c,p,b,a:
e = j-i
d = h-g
f = g + (d/2)
c = i - (f -( e/2))
p = f + (c/2)
b = e-d
a = d + (b/2)
These come from the "manual" solution of each unknown.

Selected constants:
g=50
h=100
i=70
j=15

Constants subed into orignal equations:

1) d = a - (b /2)
2) e = d + b
3) f = p - (c /2)
4) 50 = f - (d /2)
5) 100 = 50 + d
6) 70 = f - (e /2) + c
7) 15 = 70 + e

Solve derived equations by subbing:
e = j-i
e = 15-70
e = -55

d = h-g
d = 100 - 50
d = 50

f = g + (d/2)
f = 50 + (50/2)
f = 75

c = i - (f -( e/2))
c = 70 - (75 - (-55/2))
c = -32.5

p = f + (c/2)
p = 75 + (-32.5/2)
p = 58.75

b = e-d
b = -55-50
b = -105

a = d + (b/2)
a = 50 + (-105/2)
a = -2.5

Sub solved equation answers back into orignal equations with subbed constants:

1) 50 = -2.5 - (-105 /2)
2) -55 = 50 + -105
3) 75 = 58.75 - (-32.5 /2)
4) 50 = 75 - (50 /2)
5) 100 = 50 + 50
6) 70 = 75 - (-55 /2) + -32.5
7) 15 = 70 + -55