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

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Solution

It's given LM $∥$ AB,
Therefore, $\frac{A L}{A C} = \frac{C M}{C B}$
$∴ \frac{x - 3}{2 x} = \frac{x - 2}{2 x + 3}$

$o r (2 x + 3) (x - 3) = (x - 2) (2 x)$
$o r 2 x^{2} - 6 x + 3 x - 9 = 2 x^{2} - 4 x$
$o r - 3 x + 4 x = 9$
$o r x = 9$
Now, AC = 2 $x$
Therefore, AC = $2 \times 9 = 18$

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

It's given LM ∥ AB,Therefore, ALAC=CMCB ∴x−32x=x−22x+3 or(2x+3)(x−3)=(x−2)(2x)or2x2−6x+3x−9=2x2−4xor−3x+4x=9orx=9Now, AC = 2x Therefore, AC = 2×9=18

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

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It's given LM $∥$ AB,
Therefore, $\frac{A L}{A C} = \frac{C M}{C B}$
$∴ \frac{x - 3}{2 x} = \frac{x - 2}{2 x + 3}$

$o r (2 x + 3) (x - 3) = (x - 2) (2 x)$
$o r 2 x^{2} - 6 x + 3 x - 9 = 2 x^{2} - 4 x$
$o r - 3 x + 4 x = 9$
$o r x = 9$
Now, AC = 2 $x$
Therefore, AC = $2 \times 9 = 18$