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Question
If a, b, c and d are in proportion, prove that: (5a + 7b) (2c – 3d) = (5c + 7d) (2a – 3b).
Solution
It is given that
a, b, c, d are in proportion
Consider `a/b = c/d = k`
a = bk, c = dk
LHS = (5a + 7b)(2c – 3d)
LHS = (5bk + 7b)(2dk – 3d) ...[Substituting the values]
LHS = b(5k + 7) d(2k – 3) ...[Taking out the common terms]
LHS = bd (5k + 7)(2k - 3)
LHS = bd [5k (2k - 3) + 7(2k - 3)]
LHS = bd (10k2 - 15k + 14k - 21)
LHS = bd (10k2 - k - 21) ... (I)
RHS = (5c + 7d)(2a – 3b)
RHS = (5dk + 7d)(2bk – 3b) ...[Substituting the values]
RHS = d(5k + 7) b(2k – 3) ...[Taking out the common terms]
RHS = bd (5k + 7)(2k – 3)
RHS = bd [5k(2k - 3) + 7(2k - 3)]
RHS = bd (10k2 - 15k + 14k - 21)
LHS = bd (10k2 - k - 21) ... (II)
From (I) and (II),
Therefore, LHS = RHS.
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