solve this using substiturion method.
quickly!!!!
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By substituting in eq 1
3x/2-5y/3= -2
3x/2= -2+5y/3
Taking lcm
3x/2= -6+5y/3
So x = -6+5y/3*2/3
X= -12+10y/9
Substituting the value of y in eq 2 we get
-12+10y/27+y/2=13/6
Y/2=13+12-10x/54 (LCM of 6 and 27 is 54)
Canceling 2 of y/2 with 54 lcm we get
Y= 25-10y/27
By cross multiplication we get y=25/37
Putting the value of y in eq 1 we get
X =238/333
3x/2-5y/3= -2
3x/2= -2+5y/3
Taking lcm
3x/2= -6+5y/3
So x = -6+5y/3*2/3
X= -12+10y/9
Substituting the value of y in eq 2 we get
-12+10y/27+y/2=13/6
Y/2=13+12-10x/54 (LCM of 6 and 27 is 54)
Canceling 2 of y/2 with 54 lcm we get
Y= 25-10y/27
By cross multiplication we get y=25/37
Putting the value of y in eq 1 we get
X =238/333
akshatkotnala00:
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Answer:
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