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in Linear Programming by (15 points)
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\( (a-b) x+(a+b) y=2 a^{2}-2 b^{2} \) \( (a+b)(x+y)=4 a b \)

Solve:

(a – b) x + (a + b) y = 2a– 2band (a + b) (x + y) = 4ab

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2 Answers

0 votes
by (20 points)

The given system of equations are:
(a - b)x + (a + b)y = 2a2 - 2b2
So, (a - b)x + (a + b)y - 2a2 - 2b2 = 0
 (a - b)x + (a + b)y - 2(a2 - b2) = 0 ........(i)
And (a + b)(x + y) = 4ab
So, (a +b)x + (a + b)y - 4ab = 0 ..........(ii)
The given system of equation is in the form of
a1x + b1y - c1 = 0
and a2x + b2y - c2 = 0
Compare (i) and (ii) , we get
a1 = a - b, b1 = a + b, c1 = -2(a2 + b2)
a2 = a + b, b2 = a + b, c2 = -4ab
By cross-multiplication method
x2(a+b)(a2b2+2ab) =y2(ab)(a2+b2) =12b(a+b)
Now, x2(a+b)(a2b2+2ab)=12b(a+b) 

x=2aba2+b2b


And, y2(ab)(a2+b2)=12b(a+b) 
y=(ab)(a2b2)b(a+b)
The solution of the system of equations are 2aba2+b2b and (ab)(a2b2)b(a+b) respectively.

+1 vote
by (35.3k points)

Given that,

(a – b) x + (a + b) y = 2a2 – 2b2

(a + b) (x + y) = 4ab

(a – b) x + (a + b) y – 2(a– b2) = 0

(a + b)x +  (a + b)y – 4ab = 0

On comparing both the equation with the general form we get

 a1 = a – b, b1 = a + b, c1 = -2,

a2 = a + b, b2 = a + b, c2 = -4ab

Now by using cross multiplication we get

x/(b1c– b2c1) = y/(c1a– c2a1) = 1/(a1b2 – a2b1)

⇒ x/(-(a + b)4ab + 2(a + b) (a2 – b2)) = y/(− 2(a2 − b2)(a + b) + 4ab(a – b)) 

= 1/((a − b)(a + b) − (a + b)(a + b))

⇒ x/(2(a + b)(a2 – b2 + 2ab)) = 1/-2b(a + b)

x = (2ab – a2 + b2)/b

and,

= -y/(2(a – b) (a2 + b2) -2b (a + b)) = 1/ -2b(a + b)

y = (a – b)(a2 + b2)/ b(a + b)

Hence, x = (2ab – a2 + b2)/b and y = (a – b)(a2 + b2)/ b(a + b)

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