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