Question

if x and y are integers then which of the following could be the value of x where x^2-y^2 =121.
a.15
b.33
c.61
d.91​

Answer 1

Answer:

i wont say

Step-by-step explanation:

15 is the answer

Answer 2

So last time I answered your latest question, but I found some missing steps, so I'm posting again.

Not all the steps are required. I hope this helps.

 

Unit: Equations, Multiples

Given equation:

$x^2-y^2=121\implies (x+y)(x-y)=11^2$

 

First, we expect $x+y$ and $x-y$ as integers.

We can see that $11^2$ is a multiple of $x+y$ and $x-y$.

So, $x+y$ and $x-y$ are some factors of $11^2$.

 

Let's list the factors.

The factors of $11^2$:-

• $1,11,11^2$ and $-1,-11,-11^2$.

Now, we find pairs of which product is $11^2$.

We write the combinations like $\underset{\begin{matrix}1\\ 11\\ 11^2\\ -1\\ -11\\ -11^2\end{matrix}}{(x+y)}\underset{\begin{matrix}11^2\\ 11\\ 1\\ -11^2\\ -11\\ -1\end{matrix}}{(x-y)}=11^2$.

 

This gives a list of linear equations in two variables.

$\displaystyle{\left \{ {{x + y = 1} \atop {x - y = 11^2}} \right. }$, $\displaystyle{\left \{ {{x + y = 11} \atop {x - y = 11}} \right. }$, $\displaystyle{\left \{ {{x + y = 11^2} \atop {x - y = 1}} \right. }$

and $\displaystyle{\left \{ {{x + y = -1} \atop {x - y = -11^2}} \right. }$, $\displaystyle{\left \{ {{x + y = -11} \atop {x - y = -11}} \right. }$, $\displaystyle{\left \{ {{x + y = -11^2} \atop {x - y = -1}} \right. }$.

 

Now here's the solution list.

$\displaystyle{\left \{ {{x=61} \atop {y=-60}} \right. }$, $\displaystyle{\left \{ {{x=11} \atop {y=0}} \right. }$, $\displaystyle{\left \{ {{x=61} \atop {y=60}} \right. }$ and $\displaystyle{\left \{ {{x=-61} \atop {y=60}} \right. }$, $\displaystyle{\left \{ {{x=-11} \atop {y=0}} \right. }$, $\displaystyle{\left \{ {{x=-61} \atop {y=-60}} \right. }$.

 

Interesting facts:

If A is a multiple of B, B is the factor of A.

 

These points lie on the graph of $y^2=x^2-121$.

The graph is symmetrical against the x, y axes, and origin.

Here are the non-negative solutions:

• $\displaystyle{\left \{ {{x=11} \atop {y=0}} \right. }$(x-axis) and $\displaystyle{\left \{ {{x=61} \atop {y=60}} \right. }$(1st Quadrant)

Points are symmetrical as well as the graph. (*For more information, refer to the attachment.)