Question

( begin{array}{ll}text { Sol. } & 2 x+3 y=10end{array} )
( y ) ( v= )
( Rightarrow quad 3 y=10-2 x )
( Rightarrow quad-y=4-3 )
( Rightarrow quad y=frac{10-2 x}{3} )
( Rightarrow quad y=3 x-4 )
When ( x=5, quad y=frac{10-2(5)}{3}=0 )
When ( x=0, y=3(0)-4=-4 )
When ( x=8, y=frac{10-2(8)}{3}=-2 )
When ( x=1, y=3(1)-4=-1 )
When ( x=11, quad y=frac{10-2(11)}{3}=-4 )
When ( x=3, y=3(3)-4=5 )
Table for ( 2 x+3 y=10 ) is Table for ( 3 x-y=4 ) is
begin{tabular}{|c|c|c|c|}
hline( x ) & 5 & 8 & 11
hline( y ) & 0 & -2 & -4
hline
end{tabular}
From the graph, the solution is the point of intersection of the lines, i.e., ( x=2, y=2 )

# Solve the systems of equations graphically: 2x + 3y = 10 3.x - y=4

Solution