Given:
The equation is:
To find:
The error in the given equation and correct it.
Solution:
We have,
Taking left-hand side, we get
![[\because a^2-b^2=(a-b)(a+b)]](https://tex.z-dn.net/?f=%5B%5Cbecause%20a%5E2-b%5E2%3D%28a-b%29%28a%2Bb%29%5D)
![[\because (ab)^x=a^xb^x]](https://tex.z-dn.net/?f=%5B%5Cbecause%20%28ab%29%5Ex%3Da%5Exb%5Ex%5D)
It is not equal to right-hand side 
. In the right hand side, there must be a negative sign instead of positive sign.
Therefore, 
.
ABC ~ A’B’C’
So, AB=A’B’ , BC=B’C’ , AC = A’C’
Given, AB=15 and A’C’=4
So, AC = 4 , A’B’ = 15
It's <span>Diophantine equation.
First, we need to found gcd(6,(-2)):</span>
6=2*3
(-2)=(-1)*2
So, gcd(6,-2)=2
Now, the question is. Can we dived c=24, by gcd(6,-2) and in the end get integer?
Yes we can.
So, we can solve it.
Now is the formula:
Second, we need the first pair (x0,y0)
if
x=0
then
Third, we gonna use that formula:
Congratulations! We solve it.
The answer is c. Because there inside and opposite
Step 1: Set the two equations equal to each other and solve for x.
3x -5 = 6x - 8
3x + (-5+5) = 6x -8 + 5
(3x - 6x) = (6x - 6x) - 3
-3x/-3 = -3/-3
x = 1
Step 2: To solve for y take one of the given equation of your choice (for the purpose of this explanation I will only do y = 3x - 5) and replace x with 1, then solve for y
y = 3(1) - 5
y = 3 - 5
y = -2
(1,-2)
Check:
-2 = 3(1) - 5 ---> - 2 = -2
-2 = 6(1) - 8 ---> -2 = -2
Hope this helped!
