Question

Help with algebra question?

Answer 1

The system is

i)
`\displaystyle{ (-12+4y)/(2x)-(5x-6)/(x)=-1`

ii)
`\displaystyle{ 8^(5y-4)=8^(10x-4)`


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:


`\displaystyle{ (-12+8x)/(2x)-(5x-6)/(x)=-1`.

Multiplying the second fraction by 2/2 and -1 by 2x/2x, we have:


`\displaystyle{ (-12+8x)/(2x)-(2(5x-6))/(2x)=-(2x)/(2x)`.

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)