K METHOD CONTINUED - PUNJAB-CURRICULUM-AND-TEXTBOOK-BOARD-PCTB / 10TH-CLASS / MATHEMATICS / MATHEMATICS-10-10TH-CLASS-NBF-2024 / VARIATIONS / K-METHOD-CONTINUED

Example 2:

If $a:b = c:d$ , then show that $pa + qb:ma - nb = pc + qd:mc - nd$

Solution: Let $\frac{a}{b} = \frac{c}{d} = k$ , then $a = bk$ and $c = dk$

L.H.S = $p a + q b : m a - n b = \frac{p a + q b}{m a - n b} = \frac{p k b + q b}{m k b - n b} = \frac{b (p k + q)}{b (m k - n)} = \frac{p k + q}{m k - n}$

R.H.S = $p c + q d : m c - n d = \frac{p c + q d}{m c - n d} = \frac{p k d + q d}{m k d - n d} = \frac{d (p k + q)}{d (m k - n)} = \frac{p k + q}{m k - n}$

i.e., $pa + qb:ma - nb = pc + qd:mc - nd$

3.4 K-Method (Continued)

Example 2: