How do you solve the system of equations #9x + 8y = 0# and #3x - 8y = - 48#?

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Answer 1 · 2017-04-19T20:56:41

#(-4, 9/2)#

Since there is a #-8y# and a #+8y#, let's add the equations.

#9x+8y=0#
#3x-8y=-48#

#12x=-48#
#x=-4#

Plug this in, and:
#9(-4)+8y=0#

#8y=36#

#y=9/2#

Thus, our answer is #(-4, 9/2)#.

Answer 2 · 2017-04-19T21:02:52

#x=-4" , "y=4.5#

#9x+8y=0" , "(1)#
#3x-8y=-48" , "(2)#

#"let us sum the equation (1) and (2)."#

#9x+cancel(8y)+3x-cancel(8y)=0-48#

#12x=-48#

#"let us divide both sides of equation by 12."#

#(cancel(12)x)/cancel(12)=(-48)/12#

#x=-4#

#"let us write x=-4 in the equation (1)."#

#9*(-4)+8y=0#

#-36+8y=0#

#8y=36#

#(cancel(8)y)/cancel(8)=36/8#

#y=36/8=9/2=4.5#