Question

use substitution method

(1) 0.3x + 0.2y = 5 (2) 0.5x + 0.4y = 11

Answer 1

(1) 0.3x +0.2y =5
x=(5-0.2y)/0.3


(2) 0.5x + 0.4y = 11

0.5((5-0.2y)/3)+0.4y = 11
(2.5-0.10y)/3 +0.4y = 11
2.5/3 - 0.10/3y + 0.4y = 11
0.4y - 0.10/3y = 11 - 2.5/3
y = (11-2.5/3)/(0.4-0.10/3)

Answer 2

thank you so much