Question

ALGUEM ME AJUDA NESSA QUESTÃO???PRFVR!!!!

Answer 1

Explicação passo-a-passo:

a) x + y = 108

$\frac{x}{y} = \frac{5}{4} \\ \\ x = \frac{5y}{4}$

$\frac{5y}{4} + y = 108 \\ \\ \frac{5y + 4y}{4} = 108 \\ \\ 9y = 4 \times 108 \\ 9y = 432 \\ y = \frac{432}{9} \\ \\ y = 48$

x + 48 = 108

x = 108 - 48

x = 60

b) x - y = 54

$\frac{x}{y} = \frac{10}{7} \\ \\ x = \frac{10y}{7}$

$\frac{10y}{7} - y = 54 \\ \\ \frac{10y - 7y}{7} = 54 \\ \\ 3y = 7 \times 54 \\ 3y = 378 \\ y = \frac{378}{3} \\ \\ y = 126$

x - 126 = 54

x = 54 + 126

x = 180

c) x + y = 80

$\frac{x}{9} = \frac{y}{7} \\ \\ x = \frac{9y}{7}$

$\frac{9y}{7} + y = 80 \\ \\ \frac{9y + 7y}{7} = 80 \\ \\ 16y = 7 \times 80 \\ 16y = 560 \\ y = \frac{560}{16 } \\ \\ y = 35$

x + 35 = 80

x = 80 - 35

x = 45

d) a - b = 55

$\frac{a}{11} = \frac{b}{6} \\ \\ a = \frac{11b}{6}$

$\frac{11b}{6} - b = 55 \\ \\ \frac{11b - 6b}{6} = 55 \\ \\ 5b = 6 \times 55 \\ 5b = 330 \\ b = \frac{330}{5} \\ \\ b = 66$

a - 66 = 55

a = 55 + 66

a = 121