How do you solve the following system: #5x + 8y = -2, 6x+3y=-12#?

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Answer 1 · 2017-10-20T01:00:50

The solution for the system is #(-4.18, 2.36)#.

First, we want to make either x or y by itself, so that later we can plug it back into the equation.

So the 2 equations are #5x + 8y = -2# and #6x + 3y = -12#.

Let's solve for #y# first using the #6x + 3y = -12# equation

#2x + y = -6 # (divide everything by 3)

#y = -6 - 2x#

Now, we plug in the value we just got for y back into the first equation, #5x + 8y = -2#. We solve for x here.

#5x + 8(-6 - 2x) = -2# (plug in value for y)

#5x - 48 - 16x = -2# (distribute)

#-11x = 46# (simplify, put all unknowns on one side, everything else on right side of equation)

#x = -4.1818... ~~ -4.18# (rounded to hundredth's place)

Since we now know the value of x, we can plug that value back into the equation, #y = -6 - 2x# to solve for y.

#y = -6 - 2(-4.18)# (plug in value for x

#y = -6 + 8.36#

#y = 2.36#

The solution is the #(x, y)# coordinate.
We know #x ~~ -4.18# and #y ~~ 2.36#, so:
Final answer: The solution for the system is #(-4.18, 2.36)#.