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Worksheet: Solving Rational Equations Using the Least Common Denominator

Q1:

What value of π‘₯ solves the equation π‘₯ βˆ’ 5 4 βˆ’ 1 = π‘₯ 2 ?

Q2:

Solve π‘₯ βˆ’ 3 4 + 1 3 = 2 π‘₯ + 3 2 for π‘₯ .

  • A βˆ’ 2 4 9
  • B 2 3 9
  • C 2 4 9
  • D βˆ’ 2 3 9
  • E βˆ’ 1 4 9

Q3:

Find the solution set of the equation

  • A { 5 , βˆ’ 5 }
  • B { 5 }
  • C { 2 }
  • D βˆ…
  • E { βˆ’ 4 }

Q4:

What value of π‘₯ solves π‘₯ + 1 2 βˆ’ π‘₯ βˆ’ 1 3 = π‘₯ ?

Q5:

Given that 𝑛 ( π‘₯ ) = ο€Ό π‘₯ + 5 π‘₯ + 4 π‘₯ βˆ’ 5 + 6 π‘₯ βˆ’ 7 π‘₯ βˆ’ 1  Γ— π‘₯ βˆ’ 1 π‘₯ + π‘₯ + 1 2 3 2 and 𝑛 ( π‘₯ ) = 5 , find the value of π‘₯ .

  • A1
  • B βˆ’ 1 6
  • C 2 3
  • D 1 1 6
  • E11

Q6:

Given that 2 √ π‘₯ βˆ’ √ 1 1 = 2 √ π‘₯ + √ 1 1 + 2 √ 1 1 , find the value of π‘₯ .

  • A22
  • B11
  • C0
  • D33
  • E 5 5 2

Q7:

Given that 1 1 √ π‘₯ βˆ’ 2 √ 3 = 1 1 √ π‘₯ + 2 √ 3 + 9 2 0 √ 3 , find the value of π‘₯ .

  • A 4 7 6 3
  • B 8 4 4 3
  • C 4 0 4 3
  • D 9 1 6 3
  • E 8 8 4 3

Q8:

Given that 7 √ π‘₯ βˆ’ 2 √ 2 = 7 √ π‘₯ + 2 √ 2 + 5 6 √ 2 , find the value of π‘₯ .

  • A 2 0 8 5
  • B 2 9 6 5
  • C 1 2 8 5
  • D 3 7 6 5
  • E 3 4 4 5

Q9:

Simplify the function 𝑛 ( π‘₯ ) = 5 π‘₯ βˆ’ 1 5 π‘₯ π‘₯ + 2 π‘₯ βˆ’ 1 5 π‘₯ βˆ’ 3 6 βˆ’ π‘₯ π‘₯ βˆ’ π‘₯ βˆ’ 3 0 2 4 3 2 2 2 , then find the solution set of the equation 𝑛 ( π‘₯ ) = 0 .

  • A 𝑛 ( π‘₯ ) = π‘₯ βˆ’ 1 π‘₯ , solution set = { 1 }
  • B 𝑛 ( π‘₯ ) = π‘₯ + 1 π‘₯ , solution set = { 0 }
  • C 𝑛 ( π‘₯ ) = π‘₯ βˆ’ 1 π‘₯ , solution set = { 0 }
  • D 𝑛 ( π‘₯ ) = π‘₯ + 1 π‘₯ , solution set = { βˆ’ 1 }
  • E 𝑛 ( π‘₯ ) = ( π‘₯ + 1 ) π‘₯ ( π‘₯ βˆ’ 1 ) 2 , solution set = { βˆ’ 1 }

Q10:

Simplify the function 𝑛 ( π‘₯ ) = 7 π‘₯ βˆ’ 5 6 π‘₯ π‘₯ βˆ’ 7 π‘₯ βˆ’ 8 π‘₯ βˆ’ 6 4 βˆ’ π‘₯ π‘₯ βˆ’ 7 π‘₯ βˆ’ 8 2 4 3 2 2 2 , then find the solution set of the equation 𝑛 ( π‘₯ ) = 0 .

  • A 𝑛 ( π‘₯ ) = π‘₯ βˆ’ 7 π‘₯ , solution set = { 7 }
  • B 𝑛 ( π‘₯ ) = π‘₯ + 7 π‘₯ , solution set = { 0 }
  • C 𝑛 ( π‘₯ ) = π‘₯ βˆ’ 7 π‘₯ , solution set = { 0 }
  • D 𝑛 ( π‘₯ ) = π‘₯ + 7 π‘₯ , solution set = { βˆ’ 7 }
  • E 𝑛 ( π‘₯ ) = ( π‘₯ + 7 ) π‘₯ ( π‘₯ βˆ’ 7 ) 2 , solution set = { βˆ’ 7 }

Q11:

Simplify the function 𝑛 ( π‘₯ ) = βˆ’ 4 π‘₯ + 2 0 π‘₯ π‘₯ βˆ’ π‘₯ βˆ’ 2 0 π‘₯ βˆ’ 9 βˆ’ π‘₯ π‘₯ + π‘₯ βˆ’ 1 2 2 4 3 2 2 2 , then find the solution set of the equation 𝑛 ( π‘₯ ) = 0 .

  • A 𝑛 ( π‘₯ ) = π‘₯ + 1 π‘₯ , solution set = { βˆ’ 1 }
  • B 𝑛 ( π‘₯ ) = π‘₯ βˆ’ 1 π‘₯ , solution set = { 0 }
  • C 𝑛 ( π‘₯ ) = π‘₯ + 1 π‘₯ , solution set = { 0 }
  • D 𝑛 ( π‘₯ ) = π‘₯ βˆ’ 1 π‘₯ , solution set = { 1 }
  • E 𝑛 ( π‘₯ ) = ( π‘₯ βˆ’ 1 ) π‘₯ ( π‘₯ + 1 ) 2 , solution set = { 1 }

Q12:

Simplify the function 𝑛 ( π‘₯ ) = βˆ’ 1 2 π‘₯ + 1 0 8 π‘₯ π‘₯ βˆ’ 1 1 π‘₯ + 1 8 π‘₯ βˆ’ 1 6 βˆ’ π‘₯ π‘₯ βˆ’ 6 π‘₯ + 8 2 4 3 2 2 2 , then find the solution set of the equation 𝑛 ( π‘₯ ) = 0 .

  • A 𝑛 ( π‘₯ ) = π‘₯ βˆ’ 6 π‘₯ , solution set = { 6 }
  • B 𝑛 ( π‘₯ ) = π‘₯ + 6 π‘₯ , solution set = { 0 }
  • C 𝑛 ( π‘₯ ) = π‘₯ βˆ’ 6 π‘₯ , solution set = { 0 }
  • D 𝑛 ( π‘₯ ) = π‘₯ + 6 π‘₯ , solution set = { βˆ’ 6 }
  • E 𝑛 ( π‘₯ ) = ( π‘₯ + 6 ) π‘₯ ( π‘₯ βˆ’ 6 ) 2 , solution set = { βˆ’ 6 }

Q13:

Solve 3 π‘₯ + 1 + 6 π‘₯ βˆ’ 1 = 2 for π‘₯ .

  • A π‘₯ = βˆ’ 1 3 , π‘₯ = 1
  • B π‘₯ = 1 2 , π‘₯ = 3
  • C π‘₯ = 1 5 , π‘₯ = 5
  • D π‘₯ = βˆ’ 1 2 , π‘₯ = 5
  • E π‘₯ = βˆ’ 1 2 , π‘₯ = 5