3.2 Theorems on Proportions (Continued)

Example 5:

If $m:n = p:q$ Then prove that $3m + 7n:3m - 7n = 3p + 7q:3p - 7q$

Solution: Since $m:n = p:q$ or $\frac{m}{n}=\frac{p}{q}$$\frac{m}{n} = \frac{p}{q}$

Multiplying both sides by $\frac{3}{7}$$\frac{3}{7}$ , we get

$\frac{3m}{7n}=\frac{3p}{7q}$$\frac{3m}{7n} = \frac{3p}{7q}$

Then using componendo- dividend theorem

$\frac{3m+7n}{3m-7n}=\frac{3p+7q}{3p-7q}$$\frac{3m + 7n}{3m - 7n} = \frac{3p + 7q}{3p - 7q}$

Thus $3m + 7n:3m - 7n = 3p + 7q:3p - 7q$

Example 6:

If $5m + 3n:5m - 3n = 5p + 3q:5p - 3q$ then show that $m:n = p:q$

Solution: Given that $5m + 3n:5m - 3n = 5p + 3q:5p - 3q$

or $\frac{5m+3n}{5m-3n}=\frac{5p+3q}{5p-3q}$$\frac{5m + 3n}{5m - 3n} = \frac{5p + 3q}{5p - 3q}$

By componendo- dividend theorem

$\frac{(5m+3n)+(5m-3n)}{(5m+3n)-(5m-3n)}=\frac{(5p+3q)+(5p-3q)}{(5p+3q)-(5p-3q)}$$\frac{(5m + 3n) + (5m - 3n)}{(5m + 3n) - (5m - 3n)} = \frac{(5p + 3q) + (5p - 3q)}{(5p + 3q) - (5p - 3q)}$

$\frac{5m+3n+5m-3n}{5m+3n-5m+3n}=\frac{5p+3q+5p-3q}{5p+3q-5p+3q}$$\frac{5m + 3n + 5m - 3n}{5m + 3n - 5m + 3n} = \frac{5p + 3q + 5p - 3q}{5p + 3q - 5p + 3q}$

$\frac{10m}{6n}=\frac{10p}{6q}$$\frac{10m}{6n} = \frac{10p}{6q}$

Multiplying both sides by $\frac{6}{10}$$\frac{6}{10}$ .

$m:n = p:q$

i.e., $m:n = p:q$