Sure! Let's solve the given equation step-by-step.
Given equation:
[tex]\[
\frac{3}{x-1} + 5 = \frac{2x}{x-1}
\][/tex]
Step 1: Get a common denominator
Both terms on the left and the right side of the equation have the denominator [tex]\(x-1\)[/tex]. We can combine the terms on the left-hand side to match the denominator on the right-hand side:
[tex]\[
\frac{3 + 5(x-1)}{x-1} = \frac{2x}{x-1}
\][/tex]
Step 2: Simplify the numerator on the left side
Simplify the numerator [tex]\(3 + 5(x-1)\)[/tex]:
[tex]\[
3 + 5(x-1) = 3 + 5x - 5 = 5x - 2
\][/tex]
So the equation becomes:
[tex]\[
\frac{5x - 2}{x-1} = \frac{2x}{x-1}
\][/tex]
Step 3: Since the denominators are equal, set the numerators equal to each other
Given that the denominators are the same, we can set the numerators equal to each other:
[tex]\[
5x - 2 = 2x
\][/tex]
Step 4: Solve for [tex]\(x\)[/tex]
Subtract [tex]\(2x\)[/tex] from both sides to isolate the terms involving [tex]\(x\)[/tex]:
[tex]\[
5x - 2x - 2 = 0
\][/tex]
[tex]\[
3x - 2 = 0
\][/tex]
Add 2 to both sides:
[tex]\[
3x = 2
\][/tex]
Divide both sides by 3 to solve for [tex]\(x\)[/tex]:
[tex]\[
x = \frac{2}{3}
\][/tex]
Step 5: Verify the solution
Substitute [tex]\(x = \frac{2}{3}\)[/tex] back into the original equation to verify if the solution is correct.
When [tex]\(x = \frac{2}{3}\)[/tex]:
1. Left-hand side (LHS):
[tex]\[
\frac{3}{\frac{2}{3}-1} + 5 = \frac{3}{\frac{2}{3}-\frac{3}{3}} + 5 = \frac{3}{-\frac{1}{3}} + 5 = -9 + 5 = -4
\][/tex]
2. Right-hand side (RHS):
[tex]\[
\frac{2 \cdot \frac{2}{3}}{\frac{2}{3}-1} = \frac{\frac{4}{3}}{-\frac{1}{3}} = -4
\][/tex]
Since LHS equals RHS, the solution is correct.
Conclusion:
The solution to the equation [tex]\(\frac{3}{x-1} + 5 = \frac{2x}{x-1}\)[/tex] is:
[tex]\[
x = \frac{2}{3}
\][/tex]
