Question

Question 3

Answer 1

Answer:

A=5, B=10 and C=15

Step-by-step explanation:

-We are given that:

$A+B+C=30\\\\A+B=C\\\\\therefore 2C=30\\\\C=15\\\\A+B=30-15=15\ \ \ \ ...i$

-Also, given that:

$10A+12B+15C=395, \ C=15\\\\10A+12B=395-15\times 15\\\\10A+12B=170\ \ \ \ \ \ ...ii$

From equation i, we have that:

$A=15-B$

#we substitute A=15-B in equation ii to solve for B:

$10A+12B=170\\\\10(15-B)+12B=170\\\\150A-10B+12B=170\\\\2B=20\\\\B=10\\\\\therefore A=15-10=5$

Hence, the values of our unknowns are A=5, B=10 and C=15