Calculation

[drmurat]

C = a x b +a + b
İf we know c how can we calculate a and b
For example
120 = 10 x 10 + 10 + 10

[LaurV]

We can not.
If we could, factoring would be a breeze.

[retina]

Originally Posted by drmurat

C = a x b +a + b
İf we know c how can we calculate a and b
For example
120 = 10 x 10 + 10 + 10

The solution is not unique.

e.g. We can always set a=0 and b=C.

[Dr Sardonicus]

Originally Posted by drmurat

C = a x b +a + b
İf we know c how can we calculate a and b
<snip>

By factoring C + 1.

C = a*b + a + b

C + 1 = a*b + a + b + 1

C + 1 = (a + 1)*(b + 1)

[drmurat]

Originally Posted by Dr Sardonicus

By factoring C + 1.

C = a*b + a + b

C + 1 = a*b + a + b + 1

C + 1 = (a + 1)*(b + 1)

İf

C = 2*a*b + a + b
Or
C = 3*a*b + a + b

MODERATOR NOTE:

Given the equation

C = k*a*b + a + b

there is an obvious algebraic factorization similar to the one I provided above.

Thread closed.

[drmurat]

Okay I Must solve a equation
Think that the equation
a>0 b>0 a, b integer
10 * a * b + a + b = c
İn this forum I learned
a x b + a + b + 1 = c + 1
( a + 1 ) x ( b + 1 ) = c + 1

Now the equation is
9 x a x b + ( a + 1) x ( b + 1 ) = c + 1

İt is still hard
Also I know
(5 x a + ... ) x ( 2 x b + ... ) =c + ...

Any book sugestion ?

MODERATOR NOTE: Thread merged with earlier "Calculation" thread.

[BudgieJane]

I wouldn't call this an equation, because if you reverse it it becomes an instruction for calculating c given a and b.

c = 10ab + a + b

so there is an infinite number of "solutions".

a=b=0 gives c=0
a=b=1 gives c=12
a=0 b=1 or a=1 b=0 gives c=1
and so on ad infinitum.

In my humble opinion, you do not need a book on algebra to work this out.

[Viliam Furik]

Originally Posted by BudgieJane

I wouldn't call this an equation, because if you reverse it it becomes an instruction for calculating c given a and b.

Why not? Equations can have an infinite number of solutions.

And for drmurat:

k * a * b + a + b = c
k^2 * a * b + k * a + k * b = k * c
k^2 * a * b + k * a + k * b + 1 = k * c + 1
(k * a + 1) * (k * b + 1) = k * c + 1