# Thread: Determine the constants A and B so that the similarity becomes an identity?

1. determine the constants A and B so that the similarity becomes an identity

12=A(x-1)+B(5+x)

The answers are
A=-2
B=2

My friend tried to explain it to me but i did not understand!!

Ax - A + 5B + Bx = 12
For this to be true for any value of x, the x's must cancel each other out, so:
A + B = 0
A = -B
That will leave us with:
-A + 5B = 12
B + 5B = 12
6B = 12
B = 12/6
B = 2
A = -2  2.

3. I didn't get the part about the x's canceling out, but if the equation has to be true for any x, then you can just choose two values of x to get two simultaneous equations, which you can solve for A and B. In your case, you used x=0 and x=1, but you could have chosen any two x's.

Let's try x=2 and x=3.
x=2: 12=A(2-1)+B(5+2)
x=3: 12=A(3-1)+B(5+3)
12=A+7B
12=2A+8B
24=2A+14B
12=2A+8B
12=6B
B=2
12=A+14
A=-2
Yep, it works.  4. BTW, in the particular case of 12=A(x-1)+B(5+x), the easiest x's to use would be 1 and -5, which gives 12 = 0 A + 6 B, and 12 = -6 A + 0 B, which gives you your A and B pretty much directly. (Zeros and ones, either as input or output, tend to simplify a lot of equations.)  5. 12=A(x-1)+B(5+x) (original equation)
To make it identity, the constant term and the x term must be separately satisfied.
12 = -A + 5B (constant)
0 = Ax + Bx (x term) leads to 0 = A + B.
Solve for A and B.  6. Let's try x=2 and x=3.
x=2: 12=A(2-1)+B(5+2)
x=3: 12=A(3-1)+B(5+3)
12=A+7B
12=2A+8B
24=2A+14B What did you do here?
12=2A+8B
12=6B
B=2
12=A+14
A=-2  7. Originally Posted by sunshine345 Let's try x=2 and x=3.
x=2: 12=A(2-1)+B(5+2)
x=3: 12=A(3-1)+B(5+3)
(1)12=A+7B
(2)12=2A+8B
(3)24=2A+14B What did you do here?
Multiplied both sides of equation (1) by 2.
(4)12=2A+8B
12=6B
Subtracted equation 4 from equation 3.
B=2
12=A+14
A=-2  8. If I had a problem like that, I would pick two random values of x, say 0 and 1. And I would plug those in and get these equations:
6B = 12
-A + 5B = 12
Then I have a system of equations I can solve for A and B with. And then I have to prove that this equals twelve with ALL x values. So -2x - (-2) + 5(2) + 2x = 2 + 10 = 12.  9. Better yet, choose the x values of 1 and 5 and see what happens.

12 = A(x1) + B(x+5)

for x = 1,

12 = A(11) + B(1+5)

12 = A(0) + B(6)

B = 12/6 = 2

for x = 5

12 = A(51) + B(5+5)

12 = A(
6) + B(0)

A = 12/6 = 2   10. Better yet, choose the x values of 1 and –5 and see what happens.

12 = A(x–1) + B(x+5)

for x = 1,

12 = A(1–1) + B(1+5)

12 = A(0) + B(6)

B = 12/6 = 2

for x = –5

12 = A(–5–1) + B(–5+5)

12 = A(–6) + B(0)

A = 12/–6 = –2
Ahh, so that's why it wasn't a good idea to distribute.  11. Thanks to everyone!!   12. I have this similar problem

A(x+5)+7=B(2x+1)

I did like this:

A(x+5)+7=B(2x+1)
x=-5
A(-5+5)+7=B(-10+1)
-5A+5A+7=-10B+B
X=-7/9

But i dont know what I should take to the other X... I think I need some help with that   13. Solve 2x + 1 = 0, which gives x = -1/2. Although any other x will also work since you know B now. (x = 0 wouldn't be particularly difficult to work out.)

Also, your last line is wrong. It's B = -7/9.  14. Oh saw that now. (that the last line was wrong) Thanks. But in my book it says that the answer should be A=-14/9, not x=-1/2  15. Originally Posted by sunshine345 Oh saw that now. (that the last line was wrong) Thanks. But in my book it says that the answer should be A=-14/9, not x=-1/2
If you set x=-1/2 and solve for A, what do you get?  16. Oh saw that now. (that the last line was wrong) Thanks. <br><br>But in my book it says that the answer should be A=-14/9, not x=-1/2  17. Thank you! Now I get it !  18. Originally Posted by mathman 12=A(x-1)+B(5+x) (original equation)
To make it identity, the constant term and the x term must be separately satisfied.
12 = -A + 5B (constant)
0 = Ax + Bx (x term) leads to 0 = A + B.
Solve for A and B.
Does the form of the solution matter? I follow what you're saying, but I can also see that what everyone else is suggesting clearly works for finding an answer, and it's arguably simpler.

Do you think it's the "identity" aspect of the problem that is important, or simply finding values for A and B?

It's been a very, very long time since I've actually used anything more complex than the quadratic equation (and that only comes up about once a year)... so I've forgotten nearly everything I ever learned. I'm curious if there are applications or processes where the methodology involved in defining this as an identity, and solving it in the way that you've outlined, with those constraints, becomes important.  Bookmarks
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