Question

0.2x+0.3y=13 and 0.4x+0.5y=2.3 by substitution method​

Answer 1

0.2x+0.3y=13 ..............1

0.4x+0.5y=2.3..............2

multiply eq1 ×10 and eq2 ×10

2x+3y=130.............3

4x+5y=23................4

form eq1

2x = 130-3y

x= 130-3y/2

put x in eq2

4(130-3y/2) +5y=23

2(130-3y) +5y = 23

260-6y +5y = 23

-y = 23 -260

-y = -137

y= 137

put y in x

x= 130 -3×137 /2

x= 130- 411/2