How do solve the following linear system?: # 11x + 3y + 7 = 0 , -6x-2y=-8 #?

2 Answers
Jun 20, 2018

Answer:

#x=-19/2,y=65/2#

Explanation:

Multiplying the first equation by #2# and the second equation by #3# and adding both

#4x=-38#
so
#x=-19/2#
plugging this in the second equation
#-6(-19/2)-2y=-8#
#57-2y=-8#
#-2y=-65#

#y=65/2#

Jun 20, 2018

Answer:

#x=-19/2# and #y=65/2#

Explanation:

Given
[1]#color(white)("XXX")11x+3y+7=0#
[2]#color(white)("XXX")-6x-2y=-8#

To make this easier to work with I will convert [1] into the same standard form as [2] by subtracting #7# from both sides
[3]#color(white)("XXX")11x+3y=-7#

We need to eliminate one of the variables (either #x# or #y#) by combining these two equations.
For example, we might decide to eliminate the term containing the variable #y#.

To do this we need to convert equations [2] and [3] into equivalent forms with identical magnitude coefficients for #y#.

The least common multiple for the (magnitude of)coefficients of #y# in [2] and [3] is #6#, so we will convert each of [2] and [3] into equivalent forms with a term #+-6y#.

Multiplying [2] by #3#
[4]#color(white)("XXX")-18x-6y=-24#
Multiplying [3] by #2#
[5]#color(white)("XXX")22x+6y=-14#

Now if we add [4] and [5], we can eliminate the #y# term:
[6]#color(white)("XXX")4x=-38#

Dividing both sides of [6] by #4# we get
[7]#color(white)("XXX")x=-19/2#

We can now substitute #(-19/2)# for #x# back in one of our original equations. For demonstration purposes I will use equation [2]
[8]#color(white)("XXX")-6 * (-19/2)-2y=-8#

Simplifying
[9]#color(white)("XXX")57-2y=-8#

[10]#color(white)("XXX")-2y=-65#

[11]#color(white)("XXX")y=65/2#

Giving the final solution
#color(white)("XXX")x=-19/2 " or " -9 1/2#
and
#color(white)("XXX")y=65/2" or " 32 1/2#

[Since we used [2] to develop the value for #y# we should check this answer using [1] to verify that our results are correct].