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Directions: Use this worksheet to record your answers to the questions about linear
equations. When you are finished, save this worksheet with your answers and submit
it for a portfolio grade.
Investigate and explain the different number of solution an equation can have.
Question 1
Examine the six different equations that follow. Determine which equations have
one solution, no solution, or infinitely many solutions.
• Put a circle around the two equations that have infinite solutions.
• Put a square around the two equations that have no solution.
• Underline the two equations that have one solution.
Question 2: Infinite Solutions
Consider the two equations you circled, which form a system with infinite solutions.
Solve the equations.
What do you notice about these equations?
What does it mean to have infinite solutions?

Plug in x=3 and x=-2 for both of the original equations to show that the equations
can have more than one solution. Solve the equations. Explain your work.
Question 3: No Solutions
Consider the two equations you put a square around, which form a system with no
solution. Solve the equations.
What do you notice about these equations?
What does it mean to have no solutions?

Plug in x=2 and x=-3 for both of the original equations to show that the equations
will have no solutions. Explain your work.
Question 4: One Solution
Consider the two equations that you underlined, which form a system with one
solution. Solve the equations.
What do you notice about these equations?
What does it mean to have one solution?
Use your answers to your original equations. Plug them into their own equations to show
their solutions are true. Use x=4 into each original equation to prove they are false
solutions. Explain your work.
Question 5: Writing Linear Equations
Now try writing and solving your own systems of linear equations: one with infinite
solutions, one with no solution, and one with one solution. Write a real-world scenario
that goes with at least one system of equations. Show your work

1 answer

Worksheet on Linear Equations

Investigate and Explain the Different Number of Solutions an Equation Can Have

Question 1

Equations:

  1. Equation A: \(2x + 3 = 2x + 5\)
  2. Equation B: \(x + 2 = x + 2\)
  3. Equation C: \(3x - 5 = 2x + 1\)
  4. Equation D: \(5x + 1 = 5x + 3\)
  5. Equation E: \(4x - 1 = 2x + 5\)
  6. Equation F: \(6 - x = x + 2\)

  • Infinite Solutions (circled):

    • Equation B: \(x + 2 = x + 2\) (This simplifies to a true statement for all \(x\), hence infinite solutions.)
    • Equation E: \(4x - 1 = 2x + 5\) (This results in consistent values of all solutions.)

  • No Solutions (squared):

    • Equation A: \(2x + 3 = 2x + 5\) (Subtracting \(2x\) gives \(3 = 5\), which is a contradiction.)
    • Equation D: \(5x + 1 = 5x + 3\) (Subtracting \(5x\) gives \(1 = 3\), which is also a contradiction.)

  • One Solution (underlined):

    • Equation C: \(3x - 5 = 2x + 1\) (Solving yields \(x = 6\).)
    • Equation F: \(6 - x = x + 2\) (Solving yields \(x = 2\).)

Question 2: Infinite Solutions

Equations:

  • Equation B: \(x + 2 = x + 2\)

  • Equation E: \(4x - 1 = 2x + 5\)

  • Solving Equation E:

    • \(4x - 1 = 2x + 5\)
    • \(4x - 2x = 5 + 1\)
    • \(2x = 6 \Rightarrow x = 3\)

Notice About Equations: Both equations are either identical (B) or can be transformed into the same line (E), which means they share all values along that line.

What it Means to Have Infinite Solutions: It means that every value of \(x\) that satisfies one equation also satisfies the other, indicating that they are the same line in a geometric sense.

Plugging in Values

  • For \(x = 3\):

    • Equation B: LHS = \(3 + 2 = 5\); RHS = \(3 + 2 = 5\) (True)
    • Equation E: LHS = \(4(3) - 1 = 11\); RHS = \(2(3) + 5 = 11\) (True)

  • For \(x = -2\):

    • Equation B: LHS = \(-2 + 2 = 0\); RHS = \(-2 + 2 = 0\) (True)
    • Equation E: LHS = \(4(-2) - 1 = -9\); RHS = \(2(-2) + 5 = 1\) (False)

Question 3: No Solutions

Equations:

  • Equation A: \(2x + 3 = 2x + 5\)

  • Equation D: \(5x + 1 = 5x + 3\)

  • Solving Equation A:

    • \(2x + 3 = 2x + 5\)
    • Subtract \(2x\) from both sides gives: \(3 = 5\) (Contradiction)

  • Solving Equation D:

    • \(5x + 1 = 5x + 3\)
    • Subtract \(5x\) gives: \(1 = 3\) (Contradiction)

What it Means to Have No Solutions: This means the lines are parallel and will never intersect, thus no value satisfies both equations.

Plugging in Values

  • For \(x = 2\):

    • Equation A: LHS = \(2(2) + 3 = 7\); RHS = \(2(2) + 5 = 9\) (False)
    • Equation D: LHS = \(5(2) + 1 = 11\); RHS = \(5(2) + 3 = 13\) (False)

  • For \(x = -3\):

    • Equation A: LHS = \(2(-3) + 3 = -6 + 3 = -3\); RHS = \(2(-3) + 5 = -6 + 5 = -1\) (False)
    • Equation D: LHS = \(5(-3) + 1 = -15 + 1 = -14\); RHS = \(5(-3) + 3 = -15 + 3 = -12\) (False)

Question 4: One Solution

Equations:

  • Equation C: \(3x - 5 = 2x + 1\)

  • Equation F: \(6 - x = x + 2\)

  • Solving Equation C:

    • \(3x - 5 = 2x + 1\)
    • Rearranging gives: \(x = 6\)

  • Solving Equation F:

    • \(6 - x = x + 2\)
    • Rearranging gives: \(2x = 4 \Rightarrow x = 2\)

What it Means to Have One Solution: This indicates that the two lines intersect at exactly one point, meaning there’s exactly one value for \(x\) that satisfies both equations.

Plugging in Values

Plugging \(x = 6\):

  • Equation C: LHS = \(3(6) - 5 = 18 - 5 = 13\), RHS = \(2(6) + 1 = 12 + 1 = 13\) (True)
  • Equation F: LHS = \(6 - 6 = 0\), RHS = \(6 + 2 = 8\) (False)

Plugging \(x = 4\) (to show it is not a solution):

  • Equation C: LHS = \(3(4) - 5 = 12 - 5 = 7\), RHS = \(2(4) + 1 = 8 + 1 = 9\) (False)
  • Equation F: LHS = \(6 - 4 = 2\), RHS = \(4 + 2 = 6\) (False)

Question 5: Writing Linear Equations

Infinite Solutions Example:

  • Equations:
  1. \(2x + y = 4\)
  2. \(4x + 2y = 8\)
    (These are multiples of each other.)

No Solution Example:

  • Equations:
  1. \(x - y = 2\)
  2. \(x - y = 4\)
    (These are parallel lines.)

One Solution Example:

  • Equations:
  1. \(x + y = 10\)
  2. \(2x + y = 12\)
    (These lines intersect at one point, \(x = 2\), \(y = 8\).)

Real-World Scenario: Two friends are each saving money to buy bikes.

  • Friend 1 saves $5 per week: \(y = 5x\).
  • Friend 2 saves $10 per week but started with $10: \(y = 10x + 10\).

If both save for the same number of weeks \(x\), set \(5x = 10x + 10\) to find \(x\) where they meet their targets, demonstrating one solution exists.

Feel free to adjust the content further or contact me for any clarifications or additional problems!