If the Least Common Multiple (LCM) of (12, 15 | vedclass

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Question

(A) Step $1$: Find the prime factorization of each number:$12 = 2^2 	imes 3^1$$15 = 3^1 	imes 5^1$$21 = 3^1 	imes 7^1$Step $2$: The $LCM$ is the pr

Vedclass · Accepted Answer

$-40$

Vedclass · Answer

(A) Step $1$: Find the prime factorization of each number:$12 = 2^2 	imes 3^1$$15 = 3^1 	imes 5^1$$21 = 3^1 	imes 7^1$Step $2$: The $LCM$ is the product of the highest powers of all prime factors involved:$LCM(12, 15, 21) = 2^2 	imes 3^1 	imes 5^1 	imes 7^1 = 4 	imes 3 	imes 5 	imes 7 = 420$Step $3$: Solve the given equation for $x$:$420 = 500 + 2x$$2x = 420 - 500$$2x = -80$$x = -40$

If the Least Common Multiple $(LCM)$ of $(12, 15, 21) = 500 + 2x$, then $x = $ . . . . . .

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