Q. Solve for all values of x.x+8x−7=x−1Answer: x=
Set Up Equation: First, we need to set up the equation given by the problem: (x−7)/(x+8)=(−1)/(x). To solve for x, we will cross-multiply to eliminate the fractions.
Cross-Multiply: Cross-multiplying gives us: x - \(7) \times x = (−1) \times (x + 8)\
Expand Equation: Expanding both sides of the equation, we get: x2−7x=−x−8.
Combine Like Terms: Next, we will move all terms to one side of the equation to set it equal to zero: x2−7x+x+8=0.
Factor or Use Formula: Combining like terms, we have: x2−6x+8=0.
Plug Values: Now, we need to factor the quadratic equation, if possible, to find the values of x. However, the quadratic x2−6x+8 does not factor nicely, so we will use the quadratic formula instead. The quadratic formula is x=2a−b±b2−4ac, where a=1, b=−6, and c=8.
Simplify Square Root: Plugging the values into the quadratic formula, we get: x=2⋅1−(−6)±(−6)2−4⋅1⋅8.
Calculate Solutions: Simplifying inside the square root, we have: x=26±36−32.
Check Validity: Further simplifying, we get: x=26±4.
Correct Cross-Multiplication: Since 4 is 2, we have two possible solutions for x: x=2(6+2) or x=2(6−2).
Correct Cross-Multiplication: Since 4 is 2, we have two possible solutions for x: x=(6+2)/2 or x=(6−2)/2. Calculating these, we find: x=8/2 or x=4/2.
Correct Cross-Multiplication: Since 4 is 2, we have two possible solutions for x: x=(6+2)/2 or x=(6−2)/2. Calculating these, we find: x=8/2 or x=4/2. Therefore, the two solutions are x=4 and x=2. However, we must check these solutions against the original equation to ensure they do not make any denominator zero, as that would invalidate the solution.
Correct Cross-Multiplication: Since 4 is 2, we have two possible solutions for x: x=(6+2)/2 or x=(6−2)/2. Calculating these, we find: x=8/2 or x=4/2. Therefore, the two solutions are x=4 and x=2. However, we must check these solutions against the original equation to ensure they do not make any denominator zero, as that would invalidate the solution. Checking x=4: The original equation is 20. If x=4, then the denominators 22 and 23 are both non-zero, so x=4 is a valid solution.
Correct Cross-Multiplication: Since 4 is 2, we have two possible solutions for x: x=(6+2)/2 or x=(6−2)/2. Calculating these, we find: x=8/2 or x=4/2. Therefore, the two solutions are x=4 and x=2. However, we must check these solutions against the original equation to ensure they do not make any denominator zero, as that would invalidate the solution. Checking x=4: The original equation is 20. If x=4, then the denominators 22 and 23 are both non-zero, so x=4 is a valid solution. Checking x=2: If x=2, then the denominators 27 and 2 are both non-zero, so x=2 is a valid solution.
Correct Cross-Multiplication: Since 4 is 2, we have two possible solutions for x: x=(6+2)/2 or x=(6−2)/2. Calculating these, we find: x=8/2 or x=4/2. Therefore, the two solutions are x=4 and x=2. However, we must check these solutions against the original equation to ensure they do not make any denominator zero, as that would invalidate the solution. Checking x=4: The original equation is 20. If x=4, then the denominators 22 and 23 are both non-zero, so x=4 is a valid solution. Checking x=2: If x=2, then the denominators 27 and 2 are both non-zero, so x=2 is a valid solution. However, upon re-evaluating the cross-multiplication step, we realize that there was a mistake. The correct cross-multiplication should give us x0, which simplifies to x1. Bringing all terms to one side gives us x2, which is correct. But when we apply the quadratic formula, we should have x3, which simplifies to x4. This gives us x5, which results in x=4 or x=2. However, we must discard x=2 because it would make the original equation undefined (as it would result in division by zero). Therefore, the only solution is x=4.
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