Math worksheet for solving rational equations, featuring eight problems with instructions to check for extraneous solutions.

Final Answer:
1. x = 5/6
2. r = -1
3. x = 17/8
4. No solution (extraneous)
5. p = 3
6. x = -4 → extraneous, so no solution
7. x = 6
8. x = 1 or x = -1

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Let’s go through each problem step by step to make sure we get the right answers and check for any tricky parts like extraneous solutions.

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Problem 1:
Equation:
 1/x = 6/(5x) + 1

Step 1: Multiply both sides by 5x to eliminate denominators:
  5x*(1/x) = 5x*(6/(5x)) + 5x*1
→ 5 = 6 + 5x

Step 2: Solve for x:
  5 - 6 = 5x → -1 = 5x → x = -1/5? Wait — let me double-check.

Wait — actually, I think I made a mistake here. Let me redo it carefully.

Original:
  1/x = 6/(5x) + 1

Multiply every term by 5x (the LCD):
  5x * (1/x) = 5x * (6/(5x)) + 5x * 1
→ 5 = 6 + 5x
→ 5 - 6 = 5x → -1 = 5x → x = -1/5

But wait — let’s plug back in to check:

Left side: 1/(-1/5) = -5
Right side: 6/(5*(-1/5)) + 1 = 6/(-1) + 1 = -6 + 1 = -5 → OK!

So why did I write x=5/6 earlier? That was wrong. Let me correct all answers properly now.

Actually — hold on — maybe I misread the original equation?

Looking again at Problem 1:
It says:
 1/x = 6/(5x) + 1

Yes, that’s what I solved. So answer should be x = -1/5.

But let me check if there's another way — perhaps I can subtract first.

Alternative approach:

Subtract 6/(5x) from both sides:

  1/x - 6/(5x) = 1
→ (5/5x - 6/5x) = 1
→ (-1)/(5x) = 1
→ -1 = 5x → x = -1/5 → same result.

So my initial “final answer” list had an error. Let me restart and solve ALL problems correctly with verification.

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✔ Re-solving all problems carefully:

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Problem 1:
  1/x = 6/(5x) + 1

Multiply both sides by 5x:
  5 = 6 + 5x
→ 5x = -1 → x = -1/5

Check: Left = 1/(-1/5) = -5; Right = 6/(5*(-1/5)) + 1 = 6/(-1) + 1 = -6+1=-5 → ✔️ Valid.

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Problem 2:
  1 = 2/r² - 1/r

Multiply both sides by r² (assuming r ≠ 0):
  r² = 2 - r
→ r² + r - 2 = 0
Factor: (r+2)(r-1)=0 → r = -2 or r = 1

Check both:

If r = -2:
Right side: 2/((-2)^2) - 1/(-2) = 2/4 + 1/2 = 0.5 + 0.5 = 1 → ✔️

If r = 1:
2/(1)^2 - 1/1 = 2 - 1 = 1 → ✔️

Wait — but earlier I said r = -1? That was wrong.

Actually, factoring r² + r - 2 = 0 → roots are r = [-1 ± √(1+8)]/2 = [-1±3]/2 → r=1 or r=-2.

So two valid solutions? But the problem may expect only one? Or maybe I need to see context.

Wait — looking back at user’s image — Problem 2 is written as:

  1 = 2/r² - 1/r

Yes — so solutions are r = 1 and r = -2.

But perhaps the worksheet expects listing both? Or maybe I missed something.

Wait — let me check if either makes denominator zero — no, r≠0, which is fine.

So both are valid.

But in many textbooks, they might list both. However, since the instruction says “solve”, we give all real solutions.

But let me confirm calculation again:

From: 1 = 2/r² - 1/r
Multiply by r²: r² = 2 - r → r² + r - 2 = 0 → (r+2)(r-1)=0 → r=-2, r=1 → both work.

So answer should be r = -2 or r = 1.

But earlier I wrote r=-1 — definitely wrong.

This means I need to re-solve ALL problems accurately.

To avoid confusion, I’ll solve each one systematically below.

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📌 Corrected Step-by-Step Solutions:

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Problem 1:
 1/x = 6/(5x) + 1

Multiply both sides by 5x:
5 = 6 + 5x → 5x = -1 → x = -1/5
✔ Check: LHS = -5, RHS = 6/(5*(-1/5)) + 1 = 6/(-1)+1 = -5 → Good.

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Problem 2:
  1 = 2/r² - 1/r

Multiply by r²:
r² = 2 - r → r² + r - 2 = 0 → (r+2)(r-1)=0 → r = -2 or r = 1
✔ Both checked above — valid.

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Problem 3:
  (2/3)x - 5/6 = 3/4

Add 5/6 to both sides:
(2/3)x = 3/4 + 5/6

Find common denominator (12):
3/4 = 9/12, 5/6 = 10/12 → sum = 19/12

So: (2/3)x = 19/12
Multiply both sides by 3/2:
x = (19/12)*(3/2) = (19*3)/(12*2) = 57/24 = 19/8

✔ Check: (2/3)(19/8) = 38/24 = 19/12; minus 5/6 = 10/12 → 19/12 - 10/12 = 9/12 = 3/4 → ✔️

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Problem 4:
  x/(x-1) - 1/(x-2) = 11/(x² - 3x + 2)

Note: x² - 3x + 2 = (x-1)(x-2)

So multiply entire equation by (x-1)(x-2), assuming x ≠ 1, 2:

Left: x(x-2) - 1(x-1) = 11
Expand: x² - 2x - x + 1 = 11 → x² - 3x + 1 = 11
→ x² - 3x -10 = 0
Factor: (x-5)(x+2)=0 → x=5 or x=-2

Now check for extraneous:

x=5: denominators x-1=4, x-2=3 → ok
Plug into original:
LHS: 5/4 - 1/3 = (15-4)/12 = 11/12
RHS: 11/(25-15+2)=11/12 → ✔️

x=-2: x-1=-3, x-2=-4 → ok
LHS: (-2)/(-3) - 1/(-4) = 2/3 + 1/4 = 8/12 + 3/12 = 11/12
RHS: 11/(4 +6 +2)=11/12 → ✔️

Wait — so both are valid? Then why did I say "no solution" before? Mistake again.

Actually, both x=5 and x=-2 are valid.

But let me double-check the algebra:

After multiplying:
[x/(x-1)]*(x-1)(x-2) = x(x-2)
[-1/(x-2)]*(x-1)(x-2) = -1(x-1)
So total left: x(x-2) - (x-1) = x² -2x -x +1 = x² -3x +1
Set equal to 11 → x² -3x +1 =11 → x² -3x -10=0 → yes.

Roots: [3±√(9+40)]/2 = [3±7]/2 → x=5 or x=-2 → both valid.

So answer is x=5 or x=-2.

Earlier I incorrectly thought one was extraneous — but neither is.

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Problem 5:
 (p-4)/(5p) = 1/(5p) + 1

Subtract 1/(5p) from both sides:
[(p-4) - 1]/(5p) = 1 → (p-5)/(5p) = 1

Multiply both sides by 5p:
p - 5 = 5p → -5 = 4p → p = -5/4

Wait — let me do it differently.

Original:
(p-4)/(5p) = 1/(5p) + 1

Multiply both sides by 5p:
p - 4 = 1 + 5p
→ p - 5p = 1 + 4 → -4p = 5 → p = -5/4

Check:
LHS: (-5/4 - 4)/(5*(-5/4)) = (-5/4 - 16/4) / (-25/4) = (-21/4) / (-25/4) = 21/25
RHS: 1/(5*(-5/4)) + 1 = 1/(-25/4) + 1 = -4/25 + 25/25 = 21/25 → ✔️

So p = -5/4

But earlier I said p=3 — completely wrong.

I must have misread the problem.

Looking back:
Problem 5: (p-4)/(5p) = 1/(5p) + 1 → yes.

So correct answer is p = -5/4.

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Problem 6:
 x/(x+4) = 3 - 4/(x+4)

Combine right side:
3 - 4/(x+4) = [3(x+4) - 4]/(x+4) = (3x + 12 - 4)/(x+4) = (3x+8)/(x+4)

So equation becomes:
x/(x+4) = (3x+8)/(x+4)

Since denominators same and x≠-4, set numerators equal:
x = 3x + 8 → -2x = 8 → x = -4

But x = -4 makes denominator zero → extraneous!

So no solution.

✔ Correct.

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Problem 7:
 x + 6/(x-3) = 2x/(x-3)

Subtract x from both sides:
6/(x-3) = 2x/(x-3) - x

Better: bring all terms to left:

x + 6/(x-3) - 2x/(x-3) = 0
→ x + [6 - 2x]/(x-3) = 0

Or multiply entire equation by (x-3), assuming x≠3:

x(x-3) + 6 = 2x
→ x² - 3x + 6 = 2x
→ x² - 5x + 6 = 0
→ (x-2)(x-3)=0 → x=2 or x=3

But x=3 makes denominator zero → extraneous.

x=2: check
LHS: 2 + 6/(2-3) = 2 + 6/(-1) = 2 - 6 = -4
RHS: 2*2/(2-3) = 4/(-1) = -4 → ✔️

So only x=2 is valid.

Earlier I said x=6 — wrong.

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Problem 8:
 1/x² + 4/x = 3/x²

Subtract 1/x² from both sides:
4/x = 2/x²

Multiply both sides by x² (x≠0):
4x = 2 → x = 2/4 = 1/2

Wait — let me do it properly.

From: 4/x = 2/x²
Multiply both sides by x²: 4x = 2 → x = 1/2

Check:
LHS: 1/(1/2)² + 4/(1/2) = 1/(1/4) + 8 = 4 + 8 = 12
RHS: 3/(1/2)² = 3/(1/4) = 12 → ✔️

But earlier I said x=1 or x=-1 — that was for a different equation.

Wait — let me re-express:

Original: 1/x² + 4/x = 3/x²
Subtract 1/x²: 4/x = 2/x²
Then: 4/x - 2/x² = 0 → (4x - 2)/x² = 0 → numerator=0 → 4x-2=0 → x=1/2

Only solution.

But wait — is there another way? Suppose I multiply everything by x²:

x²*(1/x²) + x²*(4/x) = x²*(3/x²)
→ 1 + 4x = 3 → 4x = 2 → x=1/2 → same.

So only x=1/2.

But earlier I thought x=±1 — that would be if equation was different.

So final corrected answers:

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✔ Final Correct Answers After Verification:

1. x = -1/5
2. r = -2 or r = 1
3. x = 19/8
4. x = 5 or x = -2
5. p = -5/4
6. No solution (x = -4 is extraneous)
7. x = 2
8. x = 1/2

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But wait — let me cross-verify Problem 8 once more because sometimes students miss cases.

Equation: 1/x² + 4/x = 3/x²
Bring all to left: 1/x² + 4/x - 3/x² = 0 → -2/x² + 4/x = 0
Multiply by x²: -2 + 4x = 0 → 4x=2 → x=1/2 → yes.

No other solutions.

Similarly, Problem 2 has two solutions — both valid.

Problem 4 also has two valid solutions.

So the accurate final answers are as above.

However, since the user’s interface likely expects single-line answers per problem, and given that some problems have multiple solutions, I will present them clearly.

But note: In many school contexts, especially if not specified, they might expect all solutions listed.

Given the instructions, I will provide the final answers as computed.

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Final Answer:
1. \( x = -\frac{1}{5} \)
2. \( r = -2 \) or \( r = 1 \)
3. \( x = \frac{19}{8} \)
4. \( x = 5 \) or \( x = -2 \)
5. \( p = -\frac{5}{4} \)
6. No solution
7. \( x = 2 \)
8. \( x = \frac{1}{2} \)