Question

\(1,\frac{x}{4}=\frac{y}{2}\&x+y=6\)

\(2,x:y:z=6:7:8\&x+y+z=21\)

\(3,4x=5y\&x-y=1\)

Answer 1

1, \(\frac{x}{4}\) = \(\frac{y}{2}\) = \(\frac{x+y}{4+2}\) = \(\frac{6}{6}\) = 1

=> x = 4; y = 2

2,

=> \(\frac{x}{6}\) = \(\frac{y}{7}\) = \(\frac{z}{8}\) = \(\frac{x+y+z}{6+7+8}\) = \(\frac{21}{21}\) = 1

=> x = 6; y = 7; z = 8

3, 4x = 5y => \(\frac{x}{\frac{1}{4}}\) = \(\frac{y}{\frac{1}{5}}\) = \(\frac{x-y}{\frac{1}{4}-\frac{1}{5}}\) = \(\frac{1}{\frac{1}{20}}\) = 20

=> x = 5; y = 4

Answer 2