Question

Given equations are: ( frac{2 x+1}{3}+frac{3 y+2}{5}=2 )
and ( frac{2(2 x+1)}{3}-frac{3(3 y+2)}{5}=-1 )
Let ( frac{2 x+1}{3}=u ) and ( frac{3 y+2}{5}=v )
Then the equations become ( u+v=2 )
( 2 u-3 v=-1 )
Multiplying (iii) by 3 , ( 3 u+3 v=6 )
Adding (iv) and (v), ( 5 u=5 Rightarrow u=1 ) Substituting this value of ( u ) in (iii), ( 1+v=2 Rightarrow v=2-1=1 )
Then ( frac{2 x+1}{3}=u=1 quad ) and ( quad frac{3 y+2}{5}=v=1 )
( Rightarrow quad 2 x+1=3 quad ) and ( quad 3 y+2=5 )
( Rightarrow quad 2 x=3-1=2 quad ) and ( quad 3 y=5-2=3 )
( Rightarrow quad x=1 )
and ( quad y=1 )

# Solve the equations: 2x+1 3y + 2 3 5 2(2x+1) 3(3y + 2) ON 5 ==-1

Solution