child 1 = a
child 2 = b
a + b = 23
a x b = 132
a = 23 - b
(23 - b) x b = 132
-b^2 + 23b - 132 = 0
Now it has become a quadratic, you can use the quadratic rule, but i'll bore you no longer
child b = 11 or 12
So The first born child would be 12 years old and the second born would be 11 years old
X + Y = 23 XY = 132 Y = 23-x xy = x (23-x) = 23x -x2= 132 x2-23x+132 = 0 (X-11)(x-12) = 0 X = 11 Y = 12
11 and 12
There is an infinite number of multiples of 132. Choose any number and multiply, this will give you a product. Which is the name given to the result of multiplying numbers together. If you choose a number less than 1 then the result will be less than 132 but is also a multiple of it.
11, 12 -11, -12
12 x 11 = 132.
X + Y = 23 XY = 132 Y = 23-x xy = x (23-x) = 23x -x2= 132 x2-23x+132 = 0 (X-11)(x-12) = 0 X = 11 Y = 12
11 & 12 are the integers whose product is 132.
1+131=132
The product of two consecutive integers is 132. Find the two integers. They are: 11*12 = 132
The product is 132
17,556.
They are: 11 times 12 = 132
As a product of its prime factors: 2*2*3*11 = 132
2 x 2 x 3 x 11 = 132
169 = 132
The least common multiple of 3, 11 and 12 is 132.
11 and 12