Question

Considere a fórmula: f= a+3b/5, determine:
f,se a=1 e b=-2

b,se a=6 e f=-3

Answer 1

f se a= 1 b = -2
$f \: = 1 + 3 \times - 2 \div 5 \\ f = \frac{1}{1} - \frac{ - 6}{5} \\ f = \frac{5 - ( - 6)}{5} \\ f = \frac{11}{5}$

b, se a = 6 f = -3
$- 3 = 6 + 3b \div 5 \\ - 3 - 6 = 3b \div 5 \\ -9 \times 5 = 3b \\ \frac{ - 45}{3} = b \\ b = - 15$