# 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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