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Algebra / Linear equations in one variable Difficulty: Medium

$3 times, open parenthesis, 2 x minus 6, close parenthesis, minus 11, equals, 4 times, open parenthesis, x minus 3, close parenthesis, plus 6$

If x is the solution to the equation above, what is the value of $x minus 3$ ?

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Explanation

Choice B is correct. Because 2 is a factor of both $2 x$ and 6, the expression $2 x minus 6$ can be rewritten as $2 times, open parenthesis, x minus 3, close parenthesis$ . Substituting $2 times, open parenthesis, x minus 3, close parenthesis$ for $2 x minus 6$ on the left-hand side of the given equation yields $3 times 2, times, open parenthesis, x minus 3, close parenthesis, minus 11, equals, 4 times, open parenthesis, x minus 3, close parenthesis, plus 6$ , or $6 times, open parenthesis, x minus 3, close parenthesis, minus 11, equals, 4 times, open parenthesis, x minus 3, close parenthesis, plus 6$ . Subtracting $4 times, open parenthesis, x minus 3, close parenthesis$ from both sides of this equation yields $2 times, open parenthesis, x minus 3, close parenthesis, minus 11, equals 6$ . Adding 11 to both sides of this equation yields $2 times, open parenthesis, x minus 3, close parenthesis, equals 17$ . Dividing both sides of this equation by 2 yields $x minus 3, equals, the fraction 17 over 2$ .

Alternate approach: Distributing 3 to the quantity $2 x minus 6$ on the left-hand side of the given equation and distributing 4 to the quantity $x minus 3$ on the right-hand side yields $6 x minus 18, minus 11, equals, 4 x minus 12, plus 6$ , or $6 x minus 29, equals, 4 x minus 6$ . Subtracting $4 x$ from both sides of this equation yields $2 x minus 29, equals negative 6$ . Adding 29 to both sides of this equation yields $2 x equals 23$ . Dividing both sides of this equation by 2 yields $x equals, the fraction 23 over 2$ . Therefore, the value of $x minus 3$ is $the fraction 23 over 2, end fraction, minus 3$ , or $the fraction 17 over 2$ .

Choice A is incorrect. This is the value of x, not $x minus 3$ . Choices C and D are incorrect. If the value of $x minus 3$ is $the fraction 15 over 2$ or $negative of the fraction 15 over 2$ , it follows that the value of x is $the fraction 21 over 2$ or $negative of the fraction 9 over 2$ , respectively. However, solving the given equation for x yields $x equals, the fraction 23 over 2$ . Therefore, the value of $x minus 3$ can’t be $the fraction 15 over 2$ or $negative of the fraction 15 over 2$ .