$$(1-y)- \frac{ 5 }{ 45 } = \frac{ 3 }{ 45 }$$
$y=\frac{37}{45}\approx 0.822222222$
$$1-y-\frac{1}{9}=\frac{3}{45}$$
$$\frac{9}{9}-y-\frac{1}{9}=\frac{3}{45}$$
$$\frac{9-1}{9}-y=\frac{3}{45}$$
$$\frac{8}{9}-y=\frac{3}{45}$$
$$\frac{8}{9}-y=\frac{1}{15}$$
$$-y=\frac{1}{15}-\frac{8}{9}$$
$$-y=\frac{3}{45}-\frac{40}{45}$$
$$-y=\frac{3-40}{45}$$
$$-y=-\frac{37}{45}$$
$$y=\frac{37}{45}$$
Show Solution
Hide Solution
