What is the solution to the equation y y-4 - 4 y+4 = 32 y^ 2 -16 ? y = - 4 and y = 4 y = 0 all real numbers no solution

[tex]$\frac{y}{y-4}+\frac{y}{y-4}=\frac{32}{y^{2}-16}$The least common multiple is$y^{2}-16=(y-4)(y+4)$We multiply through by the LCM to obtain,$(y-4)(y+4) \times \frac{y}{y-4}+(y-4)(y+4) \times \frac{y}{y-4}=\frac{32}{y^{2}-16} \times(y-4)(y+4)$[/tex]

[tex]$y^{2}+y^{2}+4 y-4 y=32$We simplify to get,$2 y^{2}+0=32$[/tex]

we divide through by 2

y²= 16

Taking the square root of both sides gives

y=±√16

y = ± 4

This implies that,

y = 4 or y= -4

but the above solution is not within the domain of the function, which is

y≠ 4 or y ≠ -4

Therefore the equation has no solution.

What does an equation having no solution mean?

The solution x = 0 means that the value 0 satisfies the equation, so there is a solution.
“No solution” means that there is no value, not even 0, which would satisfy the equation.
Do not make the mistake of thinking that the equation 4 = 5 means that 4 and 5 are values for x that are solutions.

To learn more about the solution of the equation, refer

https://brainly.com/question/4344292

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