more math..

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Find the value of x for which the equation below is true.
|5x - 2y| | 8|
|x + 6y| = |10|

A.16
B.-1
C.4
D.1

  • more math.. -

    I do not know if there is a typo, notably on the RHS of the first equation.
    I assume the question reads as follows. If not, the same strategy can be used.

    |5x - 2y|= |18| ...(1)
    |x + 6y| = |10| ...(2)

    The use of absolute value on the right-hand side (RHS) of the equality sign is unnecessary, but probably helps to underline the situation.

    The two equations can be reduced to the equivalent version:

    5x - 2y = ±54 ...(1A)
    x + 6y = ±10 ...(1B)

    Multiply (1A) by 3 and add to (1B):
    15x-6y+x+6y = ±18±10
    16x = ±54±10
    so
    16x = 64, or 44, or -44, or -64

    From which only the first case x=4 appears on one of the choices (C).

    If there is no typo, or if the equations are different from (1) and (2), you can use the same solution strategy.

  • more math.. -

    I do not know if there is a typo, notably on the RHS of the first equation.
    I assume the question reads as follows. If not, the same strategy can be used.

    |5x - 2y|= |18| ...(1)
    |x + 6y| = |10| ...(2)

    The use of absolute value on the right-hand side (RHS) of the equality sign is unnecessary, but probably helps to underline the situation.

    The two equations can be reduced to the equivalent version:

    5x - 2y = ±18 ...(1A)
    x + 6y = ±10 ...(1B)

    Multiply (1A) by 3 and add to (1B):
    15x-6y+x+6y = ±54±10
    16x = ±54±10
    so
    16x = 64, or 44, or -44, or -64

    From which only the first case x=4 appears on one of the choices (C).

    If there is no typo, or if the equations are different from (1) and (2), you can use the same solution strategy.

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