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How do you solve the following system: #-5x + 3y= 6, 8x-3y=3 #?

Answer 1

Isaiah Anthony

#x = 3#
#y = 7#

Add the two equations together to cancel the #3y# and #-3y#:

#" " -5x + 3y = 6#

#"+ " ( 8x - 3y = 3)#

#-> -5x + 8x + 3y + (-3y) = 6 + 3#

# 3x = 9#

# x = 3#

Substitute #x# into one of the equations:

#8x-3y = 3#

#8(3)-3y = 3#

#24 - 3y = 3#

#-3y = -21#

# y = 7#

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Answer 2

Jameson Allen

To solve the system of equations:

-5x + 3y = 6
8x - 3y = 3

You can use the elimination method.

First, add equation 1 and equation 2 together to eliminate ( y ):

( (-5x + 3y) + (8x - 3y) = 6 + 3 )
( \Rightarrow -5x + 8x = 9 )
( \Rightarrow 3x = 9 )

Divide both sides by 3 to solve for ( x ):

( x = 3 )

Now, substitute ( x = 3 ) into equation 1 to solve for ( y ):

( -5(3) + 3y = 6 )
( \Rightarrow -15 + 3y = 6 )
( \Rightarrow 3y = 6 + 15 )
( \Rightarrow 3y = 21 )

Divide both sides by 3 to solve for ( y ):

( y = 7 )

Therefore, the solution to the system of equations is ( x = 3 ) and ( y = 7 ).

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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.