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  1. #1
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    questions

    help on this one Solve: 4 6(x + 2)> or = to 5x + 36
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  2. #2
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    Quote Originally Posted by RashanHoliday View Post
    help on this one Solve: 4 6(x + 2)> or = to 5x + 36
    Hello,

    expand the bracket and collext all x on the RHS of the inequality, the constants on the LHS:

    4-6(x+2) \geq 5x+36

    4-6x-12 \geq 5x+36

    -44 \geq 11x. Divide by 11.

    -4 \geq x

    EB
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  3. #3
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    Question ?

    hey,

    how did you get x is more than or equal to 4? Shouldn't it be x is less than or equal to 4? Here's my reasoning.
    1. Distribute the x
    2.Add 12 to both sides
    3. subtract 4 from both sides
    4.subtract 5x from both sides
    5. divide by negative 11 (you flip the sign when you divide by a negative)
    6. You wind up getting x is less than or equal to negative 4.
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  4. #4
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    Quote Originally Posted by mscalculus View Post
    hey,

    how did you get x is more than or equal to 4? Shouldn't it be x is less than or equal to 4? Here's my reasoning.
    start with:

    4-6(x+2) \ge 5x+36

    1. Distribute the x
    I presume you mean distribute the 6

    4-6x-12 \ge 5x+36

    2.Add 12 to both sides
    4-6x \ge 5x+48

    3. subtract 4 from both sides
    -6x \ge 5x+44

    4.subtract 5x from both sides
    -11x \ge 44

    5. divide by negative 11 (you flip the sign when you divide by a negative)
    x \le -4

    6. You wind up getting x is less than or equal to negative 4.
    Yes, but that is what earboth has:

    x \le -4 and -4 \ge x are the same thing.

    RonL
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  5. #5
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    Quote Originally Posted by mscalculus View Post
    hey,

    how did you get x is more than or equal to 4? Shouldn't it be x is less than or equal to 4? Here's my reasoning.
    1. Distribute the x
    2.Add 12 to both sides
    3. subtract 4 from both sides
    4.subtract 5x from both sides
    5. divide by negative 11 (you flip the sign when you divide by a negative)
    6. You wind up getting x is less than or equal to negative 4.
    Hello,

    sorry for this late reply:

    1. if you read my answer from right to left you'll get exactly your answer

    2. I collected the x at the RHS to avoid the division by a negative number. As you have pointed out you have to change the \geq-sign into a \leq-sign, which I don't have to do.

    EB
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