# Effects of replacing a value in (a)^2=(b)^2 + 25 with some other value

#### navneet9431

##### New member
suppose we change the right-hand side of the equation (a)^2=(b)^2 + 25 with (b-1)^2 + 25 then, would the Left-hand side still be equal to the Right-hand side?

If not, then what is the reason?

Also, what changes do we need to make to the Left-hand side in order to make Left-hand side = Right-hand side?

I will be thankful for help!
Note: English is my second language.

#### lev888

##### Junior Member
suppose we change the right-hand side of the equation (a)^2=(b)^2 + 25 with (b-1)^2 + 25 then, would the Left-hand side still be equal to the Right-hand side?

If not, then what is the reason?

Calculate the right side for some values of b, then b-1. Does result stay the same for b-1? Why?

#### Dr.Peterson

##### Elite Member
suppose we change the right-hand side of the equation (a)^2=(b)^2 + 25 with (b-1)^2 + 25 then, would the Left-hand side still be equal to the Right-hand side?

If not, then what is the reason?

Also, what changes do we need to make to the Left-hand side in order to make Left-hand side = Right-hand side?
The equation is only true for certain pairs of values (a,b), not for all pairs. If you replace b with (b-1), then it will be true for a DIFFERENT set of pairs. So the answer to your question, given a specific pair (a,b), is NO: for that pair, the equation will no longer be true.

The reason is that you have changed the role b plays in the equation, which can be interpreted as changing the definition of b. If the original equation is true when b has a specific value, the new equation will be true when b-1 has that value -- that is, when the "new b" is 1 more than the "old b".

There is nothing you can do to the left side so that the equation will still be true for the same set of pairs (that is, will have the same solution set) as the original, unless it involves b as well as a.

But you can gain an understanding of this by "playing" with it. Find some pairs (a,b) that work, such as (13,12) and (13,-12), or various non-integer pairs; then compare these to pairs for which the new equation works, such as (13,13) and (13,-11). See for yourself that (13,12) does not satisfy the new equation.

#### navneet9431

##### New member
Is the new b really greater than the old b?

the "new b" is 1 more than the "old b"
I think I have decreased the value of the new 'b' by subtracting 1 from(b-1) it.But, here you have mentioned that the "new b" is 1 more than the "old b".
I think I must be missing some point.
I will be thankful for this.

#### Dr.Peterson

##### Elite Member
I think I have decreased the value of the new 'b' by subtracting 1 from(b-1) it.But, here you have mentioned that the "new b" is 1 more than the "old b".
I think I must be missing some point.
I will be thankful for this.
The original equation was true for, for example, a=13 and b=12. The new equation is true when a=13 and b=13. This "new b" is 1 more than the "old b". Did you try doing something like this to see what I meant? Often the only way to understand an abstract idea (such as this with unknown variables) is to experiment with specific numbers so you can see what is happening.

It's important (and will be even more important when you get to transformations of equations) to distinguish between "subtracting 1 from b" and "replacing b in an equation with b-1". This is the reason students get confused by horizontal shifts (translation) in a graph.

But we also have to ask a question back to you: Why do you think you COULD replace b with b-1 and make the equation still true (for the same values)? What lies behind your question? Is there some situation where you think this should be done?

#### Jomo

##### Elite Member
I think I have decreased the value of the new 'b' by subtracting 1 from(b-1) it.But, here you have mentioned that the "new b" is 1 more than the "old b".
I think I must be missing some point.
I will be thankful for this.
Dr Peterson is right on target.
Original: a=13 and b=12 works
In new equation, a=13 and b-1=12 also works. But b-1=12 implies b=13, which is one more than 12.

Consider this game. You want me to say a certain number and what ever number you tell me, I'll say one less (as in b-1). If you want me to say 23, will you say to me 22, 23 or 24? Keep in mind that whatever number you say, I'll say one less!

#### navneet9431

##### New member
The original equation was true for, for example, a=13 and b=12. The new equation is true when a=13 and b=13. This "new b" is 1 more than the "old b". Did you try doing something like this to see what I meant? Often the only way to understand an abstract idea (such as this with unknown variables) is to experiment with specific numbers so you can see what is happening.

It's important (and will be even more important when you get to transformations of equations) to distinguish between "subtracting 1 from b" and "replacing b in an equation with b-1". This is the reason students get confused by horizontal shifts (translation) in a graph.

But we also have to ask a question back to you: Why do you think you COULD replace b with b-1 and make the equation still true (for the same values)? What lies behind your question? Is there some situation where you think this should be done?
So, I am going to clarify everything regarding the question.

Actually, I was trying to solve this equation (a)^2=(b)^2 + 25, with restriction a+b=25

I was able to solve it and got the values, b=12 and a=25-12=13.

Now, what I wanted to do next was to check whether any more, such pair existed or not,

For this, I tried to play around with the equation by subtracting '-1' from b for reducing b by 1 and increasing 'a' by 1
,i.e., (26-b)^2=(b-1)^2 + 25 with a restriction a+b=25

Here I expected that the value of 'b' would decrease by '1' from the previous value as ** I have subtracted 1 from 'b'** and hence the value of the new 'a' would be '1' more than the previous value of a.

But, I got b=13, which is '1' more than the previous value of 'b'.

I feel that I am wrong in saying that *I have subtracted 1 from 'b'*.But, I do not understand why?

As you can see I have really reduced the value of 'b' by 1 by replacing 'b' with 'b-1'.I do not understand why doesn't the value gets decreased?

Please, tell me where and why I am wrong!

I will be thankful for help!

#### Dr.Peterson

##### Elite Member
So, I am going to clarify everything regarding the question.

Actually, I was trying to solve this equation (a)^2=(b)^2 + 25, with restriction a+b=25

I was able to solve it and got the values, b=12 and a=25-12=13.

Now, what I wanted to do next was to check whether any more, such pair existed or not,

For this, I tried to play around with the equation by subtracting '-1' from b for reducing b by 1 and increasing 'a' by 1
,i.e., (26-b)^2=(b-1)^2 + 25 with a restriction a+b=25

Here I expected that the value of 'b' would decrease by '1' from the previous value as ** I have subtracted 1 from 'b'** and hence the value of the new 'a' would be '1' more than the previous value of a.

But, I got b=13, which is '1' more than the previous value of 'b'.

I feel that I am wrong in saying that *I have subtracted 1 from 'b'*.But, I do not understand why?

As you can see I have really reduced the value of 'b' by 1 by replacing 'b' with 'b-1'.I do not understand why doesn't the value gets decreased?

Please, tell me where and why I am wrong!

I will be thankful for help!
When you replace b with b-1 in the equation, b has to be INCREASED by 1 in order to satisfy the equation! For a given value of a, what is now b-1 must be equal to what b was previously; and that means that the new b must be the old b PLUS 1.

Read what Jomo said, and DO it to see what is happening.

But also, your idea that increasing a by 1 and reducing b by 1 would create another point on the graph has no basis. I think you are too familiar with graphs of lines, and not with curves. Have you tried graphing a^2 = b^2 + 25 and a + b = 25 to see what they look like?

I'd also like to see how you solved this pair of equations, because the method of solution should make it clear whether there are any other solutions to look for.