Question
If a, b, c and d are in proportion, prove that: (5a + 7b) (2c – 3d) = (5c + 7d) (2a – 3b).
✓
Answer
It is given that
$a, b, c, d$ are in proportion
Consider $\frac{a}{b}=\frac{c}{d}=k$
$a = bk _{ l } c = dk$
LHS $=(5 a+7 b)(2 c-3 d)$
LHS $=(5 b k+7 b)(2 d k-3 d) \ldots$ [Substituting the values]
LHS $=b(5 k+7) d(2 k-3) \quad \ldots[$ Taking out the common terms]
LHS $=b d(5 k+7)(2 k-3)$
LHS $=$ bd $[5 k(2 k-3)+7(2 k-3)]$
LHS $=b d\left(10 k^2-15 k+14 k-21\right)$LHS $=b d\left(10 k^2-k-21\right) \ldots(I)$
$
\text { RHS }=(5 c+7 d)(2 a-3 b)
$
RHS $=(5 dk +7 d )(2 bk -3 b ) \quad \ldots$ [Substituting the values]
RHS $=d(5 k+7) b(2 k-3) \ldots[$ Taking out the common terms]
$
\begin{aligned}
& \text { RHS }=b d(5 k+7)(2 k-3) \\
& \text { RHS }=\text { bd }[5 k(2 k-3)+7(2 k-3)] \\
& \text { RHS }=\text { bd }\left(10 k^2-15 k+14 k-21\right) \\
& \text { LHS }=b d\left(10 k^2-k-21\right) \ldots \text { (II) }
\end{aligned}
$From (I) and (II),Therefore, LHS $=$ RHS.
$a, b, c, d$ are in proportion
Consider $\frac{a}{b}=\frac{c}{d}=k$
$a = bk _{ l } c = dk$
LHS $=(5 a+7 b)(2 c-3 d)$
LHS $=(5 b k+7 b)(2 d k-3 d) \ldots$ [Substituting the values]
LHS $=b(5 k+7) d(2 k-3) \quad \ldots[$ Taking out the common terms]
LHS $=b d(5 k+7)(2 k-3)$
LHS $=$ bd $[5 k(2 k-3)+7(2 k-3)]$
LHS $=b d\left(10 k^2-15 k+14 k-21\right)$LHS $=b d\left(10 k^2-k-21\right) \ldots(I)$
$
\text { RHS }=(5 c+7 d)(2 a-3 b)
$
RHS $=(5 dk +7 d )(2 bk -3 b ) \quad \ldots$ [Substituting the values]
RHS $=d(5 k+7) b(2 k-3) \ldots[$ Taking out the common terms]
$
\begin{aligned}
& \text { RHS }=b d(5 k+7)(2 k-3) \\
& \text { RHS }=\text { bd }[5 k(2 k-3)+7(2 k-3)] \\
& \text { RHS }=\text { bd }\left(10 k^2-15 k+14 k-21\right) \\
& \text { LHS }=b d\left(10 k^2-k-21\right) \ldots \text { (II) }
\end{aligned}
$From (I) and (II),Therefore, LHS $=$ RHS.
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