Question

how to solve 0.01x+0.07y=0.22 and 0.03x-0.05y=0.14 by substitution

Answer 1

0.01x + 0.07y = 0.22
0.03x -0.05y = 0.14

x + 7y = 22
3x - 5y = 14

x = 22 -7y

3*(22-7y) -5y =14
66 - 21y -5y =14
-26y = - 52
y =2

x + 14 = 22

x= 8