Absolutely! Let's solve the equation step-by-step:
Given equation:
[tex]\[
-\frac{2x}{2x - 10} + 3 = \frac{10}{2x - 10}
\][/tex]
First, we want to simplify and solve this rational equation. To make it simpler, let's rewrite it with a common denominator. Notice that both fractions share the denominator [tex]\(2x - 10\)[/tex].
Rewrite the left-hand side:
[tex]\[
3 = \frac{3(2x - 10)}{2x - 10}
\][/tex]
So the equation becomes:
[tex]\[
-\frac{2x}{2x - 10} + \frac{3(2x - 10)}{2x - 10} = \frac{10}{2x - 10}
\][/tex]
Combine the fractions on the left-hand side:
[tex]\[
\frac{-2x + 3(2x - 10)}{2x - 10} = \frac{10}{2x - 10}
\][/tex]
Simplify the numerator on the left-hand side:
[tex]\[
-2x + 3(2x - 10) = -2x + 6x - 30 = 4x - 30
\][/tex]
Thus, the equation becomes:
[tex]\[
\frac{4x - 30}{2x - 10} = \frac{10}{2x - 10}
\][/tex]
Since the denominators are equal, we can set the numerators equal to each other:
[tex]\[
4x - 30 = 10
\][/tex]
Solve for [tex]\(x\)[/tex]:
[tex]\[
4x - 30 = 10 \\
4x = 40 \\
x = 10
\][/tex]
Before we conclude that [tex]\(x = 10\)[/tex] is indeed a solution, we must check if it makes any denominators zero, because division by zero is undefined.
Check the denominator [tex]\(2x - 10\)[/tex]:
If [tex]\(x = 10\)[/tex],
[tex]\[
2(10) - 10 = 20 - 10 = 10 \neq 0
\][/tex]
Since the denominator does not become zero, [tex]\(x = 10\)[/tex] is a valid solution.
Therefore, the solution to the equation is:
[tex]\[
x = 10
\][/tex]
