(5y)/(6y^2-7y-3)+(4y-3)/(2y^2-15y+18)consider nonpermissible values for the addition of this equation, the product of the nonpermissible values is.
5y/(2y-3)(3y+1)+(4y-3)(2y-3)(y-6)
i not get what it asking for.
5y/( (2y-3)(3y+1) )+(4y-3)/((2y-3)(y-6) )
look at the denominators .
We know we cannot divide by zero, so the "nonpermissible" values of x would be all those that make either one of the denominators zero.
But you factored it, so that is good, all we have to find out which values make any of the factors equal to zero
so.....
2y-3 ≠ 0
2y ≠ 3
y ≠ 3/2
3y+1 ≠ 0
3y ≠ -1
y ≠ -1/3
and
y ≠ 6
so the nonpermissible values of x are -1/3, 2/3 and 6
thanks very much reiny :)
The question is asking for the nonpermissible values of the expression (5y)/(6y^2-7y-3) + (4y-3)/(2y^2-15y+18), and then finding the product of those nonpermissible values.
To find the nonpermissible values of the expression, we need to identify the values of 'y' that would make the denominators equal to zero. This is because division by zero is undefined.
First, we identify the factors of the denominators:
6y^2 - 7y - 3 can be factored as (2y - 3)(3y + 1)
2y^2 - 15y + 18 can be factored as (2y - 3)(y - 6)
Now, we set each factor equal to zero and solve for 'y':
For (2y - 3) = 0:
2y = 3
y = 3/2
For (3y + 1) = 0:
3y = -1
y = -1/3
For (2y - 3) = 0:
2y = 3
y = 3/2
For (y - 6) = 0:
y = 6
So, the nonpermissible values of the expression are y = 3/2, y = -1/3, and y = 6.
Finally, to find the product of the nonpermissible values, we multiply them together:
(3/2) * (-1/3) * 6 = -3
Therefore, the product of the nonpermissible values is -3.