How do solve the following linear system?: 11x + 3y + 7 = 0 , -6x-2y=-8 11x+3y+7=0,6x2y=8?

2 Answers
Jun 20, 2018

x=-19/2,y=65/2x=192,y=652

Explanation:

Multiplying the first equation by 22 and the second equation by 33 and adding both

4x=-384x=38
so
x=-19/2x=192
plugging this in the second equation
-6(-19/2)-2y=-86(192)2y=8
57-2y=-8572y=8
-2y=-652y=65

y=65/2y=652

Jun 20, 2018

x=-19/2x=192 and y=65/2y=652

Explanation:

Given
[1]color(white)("XXX")11x+3y+7=0XXX11x+3y+7=0
[2]color(white)("XXX")-6x-2y=-8XXX6x2y=8

To make this easier to work with I will convert [1] into the same standard form as [2] by subtracting 77 from both sides
[3]color(white)("XXX")11x+3y=-7XXX11x+3y=7

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

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

The least common multiple for the (magnitude of)coefficients of yy in [2] and [3] is 66, so we will convert each of [2] and [3] into equivalent forms with a term +-6y±6y.

Multiplying [2] by 33
[4]color(white)("XXX")-18x-6y=-24XXX18x6y=24
Multiplying [3] by 22
[5]color(white)("XXX")22x+6y=-14XXX22x+6y=14

Now if we add [4] and [5], we can eliminate the yy term:
[6]color(white)("XXX")4x=-38XXX4x=38

Dividing both sides of [6] by 44 we get
[7]color(white)("XXX")x=-19/2XXXx=192

We can now substitute (-19/2)(192) for xx back in one of our original equations. For demonstration purposes I will use equation [2]
[8]color(white)("XXX")-6 * (-19/2)-2y=-8XXX6(192)2y=8

Simplifying
[9]color(white)("XXX")57-2y=-8XXX572y=8

[10]color(white)("XXX")-2y=-65XXX2y=65

[11]color(white)("XXX")y=65/2XXXy=652

Giving the final solution
color(white)("XXX")x=-19/2 " or " -9 1/2XXXx=192 or 912
and
color(white)("XXX")y=65/2" or " 32 1/2XXXy=652 or 3212

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