Question
Asked Nov 21, 2019
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solve the system using substituition or elimination 

x2-xy-2y2=0

xy+x+6=0

 

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Expert Answer

Step 1

Given:

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x-y2y20 .(i) .(i) xyx+6= 0

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Step 2

Take the value of y in terms of x from equation (ii) and put in equation (i).

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x-6 y = -x-6 =0 -r- 6 ,(x+36+12.x) x2x 6-2 =0 xx6x2x2 -72-24x 0 xx4x2-24x - 72 = 0 ..(ii)

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Step 3

Now, x = 3 is one of the zeroes of the equation (iii).

Because, 34 + 33 + 4(3)2 – 24(3) – 72 is equal to zero.

So, (x – 3) will be t...

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4x216x24 x-3xx 4x2 - 24x- 72 x-3x3 4x34x2-24x - 72 4x2 -12x2 16x2 24x-72 16x2-48x 24x-72 24x-72 0 +

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Tagged in

Math

Algebra

Equations and In-equations