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Complete the tables and determine if the two equations are equivalent.

Mathematics

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Question 1

1.1
To complete the table for y=2x−10y = 2x - 10, substitute each input value of xx into the equation.

  1. Calculate yy for each xx value.

    • For x=1x=1: y=2(1)−10=2−10=−8y = 2(1) - 10 = 2 - 10 = -8
    • For x=2x=2: y=2(2)−10=4−10=−6y = 2(2) - 10 = 4 - 10 = -6
    • For x=3x=3: y=2(3)−10=6−10=−4y = 2(3) - 10 = 6 - 10 = -4
    • For x=4x=4: y=2(4)−10=8−10=−2y = 2(4) - 10 = 8 - 10 = -2
    • For x=5x=5: y=2(5)−10=10−10=0y = 2(5) - 10 = 10 - 10 = 0

    The completed table is:

    Input values (xx)12345
    y=2x−10y = 2x - 10-8-6-4-20

    1.2
    To complete the table for y=6x+25−4x−35y = 6x + 25 - 4x - 35, first simplify the expression, then substitute each input value of xx.

  2. Simplify the expression for yy.

    y=6x+25−4x−35y = 6x + 25 - 4x - 35 y=(6x−4x)+(25−35)y = (6x - 4x) + (25 - 35) y=2x−10y = 2x - 10
  3. Calculate yy for each xx value using the simplified expression.

    • For x=1x=1: y=2(1)−10=2−10=−8y = 2(1) - 10 = 2 - 10 = -8
    • For x=2x=2: y=2(2)−10=4−10=−6y = 2(2) - 10 = 4 - 10 = -6
    • For x=3x=3: y=2(3)−10=6−10=−4y = 2(3) - 10 = 6 - 10 = -4
    • For x=4x=4: y=2(4)−10=8−10=−2y = 2(4) - 10 = 8 - 10 = -2
    • For x=5x=5: y=2(5)−10=10−10=0y = 2(5) - 10 = 10 - 10 = 0

    The completed table is:

    Input values (xx)12345
    y=6x+25−4x−35y = 6x + 25 - 4x - 35-8-6-4-20

    1.3
    Is 2x−102x - 10 equivalent to 6x+25−4x−356x + 25 - 4x - 35? Explain your answer.

  4. Simplify the second expression.

    6x+25−4x−35=(6x−4x)+(25−35)=2x−106x + 25 - 4x - 35 = (6x - 4x) + (25 - 35) = 2x - 10
  5. Compare the simplified expression with the first expression.

    The simplified expression 2x−102x - 10 is identical to the first expression 2x−102x - 10.

    Yes, the expressions are equivalent. This is because when the expression 6x+25−4x−356x + 25 - 4x - 35 is simplified by combining like terms, it results in 2x−102x - 10, which is the same as the first expression.

    Question 2

    2.1.1
    Substitute m=1m = 1 in A and B respectively.

    A) For m(2−m)m(2 - m):
    Substitute m=1m=1:

    1(2−1)=1(1)=11(2 - 1) = 1(1) = \boxed{1}

    B) For m(−m+2)m(-m + 2):
    Substitute m=1m=1:

    1(−1+2)=1(1)=11(-1 + 2) = 1(1) = \boxed{1}

    2.1.2
    Do you get the same answer?

    Yes, both expressions evaluate to 1 when m=1m=1.

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