How do you solve the system (x+y)2+3(x−y)=30 and xy+3(x−y)=11 ?

1 Answer
George C.
Sep 18, 2016

(x,y) is one of:

(1+√6,−1+√6)

(1−√6,−1−√6)

(5,−2)

(2,−5)

Explanation:

Notice that:

(x+y)2−4xy=(x−y)2

So we can get a quadratic in (x−y) by subtracting 4 times the second equation from the first...

−14=30−4⋅11

−14=((x+y)2+3(x−y))−4(xy+3(x−y))

−14=((x+y)2−4xy)+(3−12)(x−y)

−14=(x−y)2−9(x−y)

Add 14 to both ends to get:

0=(x−y)2−9(x−y)+14

0=((x−y)−2)((x−y)−7)

So x−y=2 or x−y=7


Case x−y=2

From the first given equation, we have:

30=(x+y)2+3(x−y)

30=(2y+2)2+6=4(y+1)2+6

Subtract 6 from both ends and transpose to get:

4(y+1)2=24

Hence:

(y+1)2=6

So:

y+1=±√6

So

y=−1±√6

with corresponding values of x given by x=y+2

So solutions:

(x,y)=(1+√6,−1+√6)

(x,y)=(1−√6,−1−√6)


Case x−y=7

From the first given equation, we have:

30=(x+y)2+3(x−y)

30=(2y+7)2+21

Subtract 21 from both ends and transpose to get:

(2y+7)2=9

Hence:

2y+7=±√9=±3

So:

2y=−7+3=−4

or:

2y=−7−3=−10

Hence y=−2 or y=−5

Then we have corresponding values for x using x=y+7

Hence solutions:

(x,y)=(5,−2)

(x,y)=(2,−5)

graph{((x+y)^2+3(x-y)-30)(xy+3(x-y)-11) = 0 [-7.22, 12.78, -6.36, 3.64]}

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