The system is
i) [tex]\displaystyle{ \frac{-12+4y}{2x}-\frac{5x-6}{x}=-1[/tex]
ii)[tex]\displaystyle{ 8^{5y-4}=8^{10x-4}[/tex]
In the second equation, the bases are equal, so we make the exponents equal to set the equation:
5y-4=10x-4. This means that 5y=10x, that is y=2x.
So, we can substitute y with 2x in the first equation:
[tex]\displaystyle{ \frac{-12+8x}{2x}-\frac{5x-6}{x}=-1[/tex].
Multiplying the second fraction by 2/2 and -1 by 2x/2x, we have:
[tex]\displaystyle{ \frac{-12+8x}{2x}-\frac{2(5x-6)}{2x}=-\frac{2x}{2x}[/tex].
Since the denominators are all equal, we can only work with the numerators, setting the equation:
-12+8x-10x+12=-2x,
simplifying we get 0-2x=-2x, that is -2x=-2x, which is always true. (Here remember that x cannot be 0 since x was in the denominator).
This means that for any (x, y) such that y=2x, the system is satisfied.
We can check that the correct statements are:
3) The point (1, 2) is a solution. (Since we have y=2x).
6) The system has infinitely many solutions. (Since any pair (x, 2x) is a solution, except when x=0)
