If LM $∥$ AB, AL = x - 3, AC = 2x, BM = x - 2, BC = 2x + 3. What is the value of AC?

[2 Marks]

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Solution

It's given LM $∥$ AB,
So, $\frac{C L}{C A} = \frac{C M}{C B}$ [by Basic Proportionality Theorem]
[1 Mark]

$\frac{(x - 3)}{2 x} = \frac{(x - 2)}{(2 x + 3)}$

$(x - 3) (2 x + 3)$ = $(x - 2) 2 x$

$2 x^{2} - 3 x - 9$ = $2 x^{2} - 4 x$
$x = 9$
$S o, A C = 2 x = 18$ [1 Mark]

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