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Sagot :
Let's carefully analyze each given equation to determine which one could have been solved by Melanie.
### A. [tex]\(\frac{4}{x-3} = \frac{\pi}{10}\)[/tex]
- If [tex]\(x = 3\)[/tex], then the denominator of the left fraction becomes zero, making the equation undefined.
- As a result, [tex]\(3\)[/tex] cannot be a solution.
### B. [tex]\(\frac{x-3}{4} = \frac{2x-6}{4x}\)[/tex]
- If [tex]\(x = 3\)[/tex]:
- The numerator of the left fraction becomes [tex]\(0\)[/tex] (since [tex]\(3-3=0\)[/tex]), and the numerator of the right fraction also becomes [tex]\(0\)[/tex] (since [tex]\(2 \cdot 3 - 6 = 0\)[/tex]).
- Let's cross-multiply:
[tex]\[ (x-3) \cdot (4x) = 4 \cdot (2x - 6) \][/tex]
[tex]\[ 4x(x - 3) = 8x - 24 \][/tex]
[tex]\[ 4x^2 - 12x = 8x - 24 \][/tex]
[tex]\[ 4x^2 - 20x + 24 = 0 \][/tex]
[tex]\[ (2x - 6)(2x - 4) = 0 \][/tex]
[tex]\[ x = 3 \text{ and } x = 2 \][/tex]
- Here, [tex]\(x = 3\)[/tex] is a valid solution, not an extraneous one.
### C. [tex]\(\frac{8}{x^2-9} = \frac{5}{2x-6}\)[/tex]
- If [tex]\(x = 3\)[/tex]:
- The denominator of the left fraction becomes [tex]\(0\)[/tex] (since [tex]\(3^2 - 9 = 0\)[/tex]), and the denominator of the right fraction also becomes [tex]\(0\)[/tex] (since [tex]\(2 \cdot 3 - 6 = 0\)[/tex]).
- Let's cross-multiply:
[tex]\[ 8 \cdot (2x - 6) = 5 \cdot (x^2 - 9) \][/tex]
[tex]\[ 16x - 48 = 5x^2 - 45 \][/tex]
[tex]\[ 5x^2 - 16x + 3 = 0 \][/tex]
- Solving for [tex]\(x\)[/tex]:
[tex]\[ x = 3 \text{ and } x = \frac{1}{5} \][/tex]
- Here, [tex]\(x = 3\)[/tex] becomes an extraneous solution because it makes both denominators zero, so it cannot be a valid solution for the original equation.
### D. [tex]\(\frac{8}{x+3} = \frac{12}{4x-3}\)[/tex]
- If [tex]\(x = 3\)[/tex], then the denominators of both fractions are non-zero:
- Left denominator: [tex]\(3 + 3 = 6\)[/tex]
- Right denominator: [tex]\(4 \cdot 3 - 3 = 9\)[/tex]
- Hence, [tex]\(x = 3\)[/tex] does not make either of the fractions undefined.
Given the conditions of the problem where [tex]\(3\)[/tex] is an extraneous solution, we can conclude that the correct equation Melanie could have solved is:
C. [tex]\(\frac{8}{x^2-9} = \frac{5}{2x-6}\)[/tex]
### A. [tex]\(\frac{4}{x-3} = \frac{\pi}{10}\)[/tex]
- If [tex]\(x = 3\)[/tex], then the denominator of the left fraction becomes zero, making the equation undefined.
- As a result, [tex]\(3\)[/tex] cannot be a solution.
### B. [tex]\(\frac{x-3}{4} = \frac{2x-6}{4x}\)[/tex]
- If [tex]\(x = 3\)[/tex]:
- The numerator of the left fraction becomes [tex]\(0\)[/tex] (since [tex]\(3-3=0\)[/tex]), and the numerator of the right fraction also becomes [tex]\(0\)[/tex] (since [tex]\(2 \cdot 3 - 6 = 0\)[/tex]).
- Let's cross-multiply:
[tex]\[ (x-3) \cdot (4x) = 4 \cdot (2x - 6) \][/tex]
[tex]\[ 4x(x - 3) = 8x - 24 \][/tex]
[tex]\[ 4x^2 - 12x = 8x - 24 \][/tex]
[tex]\[ 4x^2 - 20x + 24 = 0 \][/tex]
[tex]\[ (2x - 6)(2x - 4) = 0 \][/tex]
[tex]\[ x = 3 \text{ and } x = 2 \][/tex]
- Here, [tex]\(x = 3\)[/tex] is a valid solution, not an extraneous one.
### C. [tex]\(\frac{8}{x^2-9} = \frac{5}{2x-6}\)[/tex]
- If [tex]\(x = 3\)[/tex]:
- The denominator of the left fraction becomes [tex]\(0\)[/tex] (since [tex]\(3^2 - 9 = 0\)[/tex]), and the denominator of the right fraction also becomes [tex]\(0\)[/tex] (since [tex]\(2 \cdot 3 - 6 = 0\)[/tex]).
- Let's cross-multiply:
[tex]\[ 8 \cdot (2x - 6) = 5 \cdot (x^2 - 9) \][/tex]
[tex]\[ 16x - 48 = 5x^2 - 45 \][/tex]
[tex]\[ 5x^2 - 16x + 3 = 0 \][/tex]
- Solving for [tex]\(x\)[/tex]:
[tex]\[ x = 3 \text{ and } x = \frac{1}{5} \][/tex]
- Here, [tex]\(x = 3\)[/tex] becomes an extraneous solution because it makes both denominators zero, so it cannot be a valid solution for the original equation.
### D. [tex]\(\frac{8}{x+3} = \frac{12}{4x-3}\)[/tex]
- If [tex]\(x = 3\)[/tex], then the denominators of both fractions are non-zero:
- Left denominator: [tex]\(3 + 3 = 6\)[/tex]
- Right denominator: [tex]\(4 \cdot 3 - 3 = 9\)[/tex]
- Hence, [tex]\(x = 3\)[/tex] does not make either of the fractions undefined.
Given the conditions of the problem where [tex]\(3\)[/tex] is an extraneous solution, we can conclude that the correct equation Melanie could have solved is:
C. [tex]\(\frac{8}{x^2-9} = \frac{5}{2x-6}\)[/tex]
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