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

1. ## 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? 2. ## 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.

3. ## Re: Is it possible to solve this (simple?) set of equations? 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.

4. ## 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.

5. ## 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 6. ## 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?

7. ## Re: Is it possible to solve this (simple?) set of equations? 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.

8. ## Re: Is it possible to solve this (simple?) set of equations? 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 9. ## Re: Is it possible to solve this (simple?) set of equations? 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 .

10. ## 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?

11. ## Re: Is it possible to solve this (simple?) set of equations? 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?
.

12. ## 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
(add 25 to both sides)
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)
(Add 16.25 to both sides)
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?

13. ## Re: Is it possible to solve this (simple?) set of equations? 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
(add 25 to both sides)
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)
(Add 16.25 to both sides)
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?
.

14. ## Re: Is it possible to solve this (simple?) set of equations? 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

15. ## 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