math problem POW

Question

Suppose a and b are positive real numbers such that ab=1/9 a(b+1)=7/9 b(a+1)=18, what is (a+1) (b+1)?

Answer 1

You could solve any two of the equations for a and b. But you aren't asked to do this, only to figure (a+1)*(b+1). So solve the 2nd and 3rd equations for those 2 respective factors, then combine, and use the 1st equation to plug in the value. This may seem unclear to you, so I'll show:

a(b+1)=7/9 -> (b+1) = (7/9)*(1/a)

b(a+1)=18 -> (a+1) = 18 * (1/b)

Therefore (a+1)*(b+1) = (7/9)*18*(1/(ab))

OK, since (ab) = 1/9, sub in:

(a+1)*(b+1) = (7/9)*18*(1/(1/9)) = 7*2*9 = 126, if my mental math is still working.

Do you understand WHY you shouldn't solve for a and b? It's b/c it's wasting some of your time, when a shorter way to the required answer is readily at hand!

-- Cheers, -- Mr. d.

Answer 2

Solve for (b+1)

a(b + 1) = 7/9

(1/a)[a(b+1)]=(7/9)(1/a)

(b+1)=7/9a

Solve for (a+1)

b(a+1)=18

(1/b)[b(a+1)]=(18)(1/b)

(a+1)=18/b

Multiply (a+1)(b+1) and substitute the solved values of each:

(a + 1)(b + 1) = (18/b)(7/9a)

(a + 1)(b + 1) = 126/9ab

Substitute the value of ab to simplify

(a + 1)(b + 1) = 126/[9(1/9)]

(a + 1)(b + 1) = 126

Therefore the product of (a + 1)(b + 1) is 126.