Solve each equation. Watch for extraneous solutions

5/y-3 = y+7/2y-6 +1

You need to use parentheses to clarify what the numerators and denominators are.

For example, do you mean
(y+7)/(2y-6) +1, or
y + 7/(2y-6) +1 ?

5 is numerator/ y-3 is denominator

y + 7/(2y-6) +1
y +7 is numerator, 2-y is denominator the whole thing +1

If Sandra's comments are correcta about "Joe's" question,

5/(y-3) = (y+7)/[2(y-3)] +1
10/[2(y-3)]- (y+7)/[2(y-3)] = 1
(10 -y -7)/(y-3) = 2
3-y = 2(y-3) = 2y -6
3y = 9
y = 3
However, the original equation is undefined when y = 3

To solve the given equation, we need to simplify both sides of the equation and then bring the terms involving y to one side.

Let's start by simplifying the right side of the equation:

y + 7 / 2y - 6 + 1

To simplify this expression, we need to get a common denominator for the fractions. The denominator of the second fraction is 2y - 6, so we need to multiply the first fraction by (2y - 6) / (2y - 6):

= (y * (2y - 6) + (y + 7)) / (2y - 6)

= (2y^2 - 6y + y + 7) / (2y - 6)

Now, we can simplify further:

= (2y^2 - 5y + 7) / (2y - 6)

Now, let's rewrite the original equation with the simplified right side:

5 / (y - 3) = (2y^2 - 5y + 7) / (2y - 6)

To eliminate the fractions, we can cross-multiply:

5(2y - 6) = (y - 3)(2y^2 - 5y + 7)

10y - 30 = 2y^3 - 5y^2 + 7y - 6y^2 + 15y - 21

Combining like terms:

10y - 30 = 2y^3 - 11y^2 + 22y - 21

Now, let's arrange this equation in standard form (with all terms on one side and set equal to zero):

2y^3 - 11y^2 + 22y - 21 - 10y + 30 = 0

2y^3 - 11y^2 + 12y + 9 = 0

Now, to solve this cubic equation, we can use factoring, the rational roots theorem, or numerical methods. However, finding the exact solutions for this equation is quite challenging.

Considering the difficulty, it might be more appropriate to use numerical methods or a calculator/graphing tool to find the solutions.