Question

What is the sum of the two consecutive numbers, the difference of whose squares is 19?

Answer 1

Hey, the answer is very simple
Let the numbers be x and x+1
(x+1)²-x²=19

x²+2x+1-x²=19

2x+1=19

x=9
So x+1=10

Sum= 10+9
=19
hope this helped u :)

Answer 2

Answer:

The sum of the two consecutive numbers 9 and 10 be 19.

Solution:

Let the consecutive numbers be taken as x and x + 1

Given that, the difference of their squares will be 19.

Hence,

$\begin{array} { c } { ( x + 1 ) ^ { 2 } - x ^ { 2 } = 19 } \\\\ { x ^ { 2 } + 1 + 2 x - x ^ { 2 } = 19 }\\ \\ { 2 x + 1 = 19 } \\\\ { 2 x = 19 - 1 } \\\\ { 2 x = 18 } \\\\ { x = 9 } \end{array}$

Thus, the two consecutive numbers be x = 9 and (x+1) = 10

The sum of the numbers will be,

$x+(x+1)=9+10\\\\x+(x+1)=19$

Thus, their sum will also be 19.