Question

50 POINTS! false answers will be reported and u wont get points

Can a solution to this equation be 3, because then both sides would have divide by 0?

Answer 1

Answer: You may divide by a variable. But only if you know the variable is non-zero. So, if setting the variable to zero makes both sides of the equation zero, we have a solution that that variable is 0, or the equation with the variable divided out holds. E.g. solve x(x-1) = 0.

Answer 2

Explanation:

If we multiply x - 7 by x - 3, we get x^2 - 10x + 21. This shows that the LCD is x^2 - 10x + 21, or (x - 7)(x - 3). Let's convert each of the two fractions on the right to their equivalents, with the LCD in the denominator.

Multiplying both the numerator and the denominator of 6/(x - 7) by (x - 3), we get 6(x - 3) over x^2 - 10x + 21, which is the LCD.

Similarly, we get 5(x - 7) over x^2 - 10x + 21.

Now all three of the original fractions have the same denominator, x^2 - 10x + 21. We can now "drop" the LCD, which leaves us with:

x^2 - 29 = 6(x - 3) + 5(x - 7)

Performing the indicated multiplications, we get:

x^2 - 29 = 6x - 18 + 5x - 35. Combining like terms:

x^2 - 29 = 11x - 53, or x^2 - 11x + 24 = 0

This factors as follows: x^2 - 11x + 24 = 0 = (x - 3)(X - 8) = 0

Then the preliminary solutions are x = 3 and x = 8. Unfortunately, we cannot substitute x = 3 into the original equation, because that would lead to division by zero. The only valid solution is x = 8, since every term of the original equation is defined at x = 8, and the equation is true for this value.

Repeat: 3 cannot be a solution because the 3rd fraction is undefined at

x = 3. So we have only one valid solution (8) with no other solution to multiply by 8.