How do you solve #((4, 8), (2, 5))((x), (y))=((0),(6))#?

Answer 1

#((x),(y))=((-12),(6))#

To solve #((4,8),(2,5))((x),(y))=((0),(6))#, we need to first find inverse of the matrix #((4,8),(2,5))#.
The general form for inverse of matris #((a,b),(c,d))# is #1/(ad-bc)((d,-b),(-c,a))#.
Hence inverse of #((4,8),(2,5))# is #1/(4*5-8*2)((5,-8),(-2,4))# or #1/4((5,-8),(-2,4))#. Multiplying both sides of #((4,8),(2,5))((x),(y))=((0),(6))# by this gives
#((x),(y))=1/4((5,-8),(-2,4))((0),(6))# .............(A)
as #1/4((5,-8),(-2,4))((4,8),(2,5))=((1,0),(0,1))#

Solving (A) by simple matrix multiplication gives us

#((x),(y))=1/4((-48),(24))=((-12),(6))#
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Answer 2

#(x,y)=(-12,6)#

(this is just an alternative method to that provided by Shwetank Mauria that doesn't require the use of an inverse matrix)

#((4,8),(2,5))xx((x),(y)) = ((4x+8y),(2x+5y))# (by standard matrix multiplication)
Therefore #color(white)("XXX")((4,8),(2,5))xx((x),(y))= ((0),(6))# is equivalent to [1]#color(white)("XXX")4x+8y=0# [2]#color(white)("XXX")2x+5y=6#
If we multiply equation [2] by #2# and subtract the result form equation [1], we get [3]#color(white)("XXX")-2y=-12# and from this [4]#color(white)("XXX")y=6#
Substituting #6# for #y# back in equation [1] gives [5]#color(white)("XXX")cancel(4)x+cancel(8)^2xx(6)=0# and [6]#color(white)("XXX")x=-12#
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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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