Question

solve the following pair by substitution method(i)0.2x+0.3y=1.3 and 0.4+0.5y=2.3

Answer 1

solution:
0.2x+0.3y=1.3. _eq(1)
0.4x+0.5y=2.3. _eq(2)
from equation(1) find the value of x
0.2x=1.3-0.3y
x=(1.3-0.3y)÷0.2. _(3)
putting the value of x in eq_(2)
0.4(1.3-0.3y)÷0.2)+0.5y=2.3
2(1.3-0.3y)+0.5y=2.3
2.6-0.6y+0.5y=2.3
-0.1y=2.3-2.6
-0.1y=-0.3
1/10y=3/10
y=3/10÷1/10
y=3
putting the value of y in equation (3)
x=[1.3-0.3(3)]÷0.2
x=(1.3-0.9)÷0.2
x=0.4÷0.2
x=2