Algebra II (check)

Solve: sqrt3x+6 + 4 =< 7

sqrt3x+6 + 4 =< 7
subtract 4 from both sides
sqrt3x+6 =< 3
subtract 6 from both sides
3x/3x =< -3x/3x
x =< -1 (answer)

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  1. Your solution makes no sense.

    First of all I will assume you meant
    √(3x+6) + 4 ≤ 7 , then
    √(3x+6) ≤ 3
    Now at this point we should realize that
    3x+6 ≥ 0 or else the square root term is undefined
    so before we do anything else x ≥ -2

    now back to √(3x+6) ≤ 3
    squaring both sides gives us

    3x+6 ≤ 9
    3x ≤ 3
    x ≤ 1

    so finally -2 ≤ x ≤ 1

    check by taking a value of x outside that domain on both sides and subbing it in the original, it will not work.

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