1 2 2x 3y 12 7 3x 2y 1 2
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Solve for x and y:
1/2(2x+3y)+12/7(3x–2y)=½
7/(2x+3y)+4/(3x–2y)=2. For 2x+3y≠0 and 3x–2y≠0.
Answered by admin @
SOLUTION:
1/2(2x+3y)+12/7(3x–2y)=1/2
7/(2x+3y)+4/(3x–2y)=2
Let 1/(2x+3y)= u & 1/(3x-2y)= v
Then the given system of equations becomes
u/2 + 12v/7= 1/2
(7u + 12×2v) / 14= 1/2
(7u + 24v) =( ½) × 14
7u +24v = 7……………..(1)
7u + 4v = 2 ………………(2)
Subtracting equation 2 from eq 1.
7u +24v = 7
7u + 4v = 2 [elimination method]
(-) (-) (-)
--------------------
20 v = 5
v = 5/20= ¼
Putting v = ¼ in equation 1.
7u +24v = 7
7u + 24 (¼) = 7
7u + 6= 7
7u = 7-6
7u = 1
u = 1/7
Now,
u = 1/7 = 1/(2x+3y)
(2x+3y) = 7……………(3)
v = ¼ = 1/(3x-2y)
(3x-2y) = 4……………..(4)
Multiply equation 3 by 2 an equation 4 by 3,
4x + 6y = 14……………(5)
9x - 6y =12……………..(6)
Adding Equations 5 and 6,
4x + 6y = 14
9x - 6y =12
----------------
13x = 26
x = 26/13
x = 2
Putting x= 2 in question 5,
4x + 6y = 14
4(2) +6y = 14
8 +6y = 14
6y = 14 -8
6y = 6
y = 6/6= 1
y = 1
Hence, x= 2 ,y= 1 is the solution of the given system of equations.
HOPE THIS WILL HELP YOU...
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