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December 21, 2014

December 21, 2014

Posted by **True** on Sunday, July 22, 2012 at 6:37pm.

|5x - 2y| | 8|

|x + 6y| = |10|

A.16

B.-1

C.4

D.1

- more math.. -
**MathMate**, Monday, July 23, 2012 at 8:25amI 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.. -
**MathMate**, Monday, July 23, 2012 at 8:26amI 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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