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dem82 [27]
2 years ago
9

He determines that 6 is an extraneous solution because the difference of the numerators is 6, so the 6s cancel to 0. which best

describes the reasonableness of the student’s solution? his solution for x is correct and his explanation of the extraneous solution is reasonable. his solution for x is correct, but in order for 6 to be an extraneous solution, both denominators have to result in 0 when 6 is substituted for x. his solution for x is correct, but in order for 6 to be an extraneous solution, one denominator has to result in 0 when 6 is substituted for x. his solution for x is incorrect. when solved correctly, there are no extraneous solutions.
Mathematics
1 answer:
seropon [69]2 years ago
5 0

The question is incomplete. Completed question is given below the answer.

His solution for x is correct, but in order for 6 to be an extraneous solution, one denominator has to result in 0 when 6 is substituted for x.

Given  

Then the solution follows thus:

Step 1: 8(x – 4) = 2(x + 2)

Step 2: 4(x – 4) = (x + 2)

Step 3: 4x – 16 = x + 2

Step 4: 3x = 18

Step 5: x = 6

It can be seen that his solution is correct. But 6 is not an extraneous solution.

An extraneous solution is a solution to an equation that emerges from the process of solving the problem but is not a valid solution to the original problem.

When 6 is substituted into the original equation, the original equation holds.

Therefore, his solution for x is correct, but in order for 6 to be an extraneous solution, one denominator has to result in 0 when 6 is substituted for x.

Learn more about extraneous solution here: brainly.com/question/3751209

#SPJ4

Completed question:-

A student solves the following equation for all possible values of x:His solution is as follows:

Step 1: 8(x – 4) = 2(x + 2)

Step 2: 4(x – 4) = (x + 2)

Step 3: 4x – 16 = x + 2

Step 4: 3x = 18

Step 5: x = 6

He determines that 6 is an extraneous solution because the difference of the numerators is 6, so the 6s cancel to 0.

Which best describes the reasonableness of the student’s solution?

His solution for x is correct and his explanation of the extraneous solution is reasonable.

His solution for x is correct, but in order for 6 to be an extraneous solution, both denominators have to result in 0 when 6 is substituted for x.

His solution for x is correct, but in order for 6 to be an extraneous solution, one denominator has to result in 0 when 6 is substituted for x.

His solution for x is incorrect. When solved correctly, there are no extraneous solutions.

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Solution for <em>2x+1=5x-5</em> equation

Simplifying

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Solving

<em>1 + 2x = -5 + 5x </em>

Solving for variable<em> 'x'. </em>

Move all terms containing x to the left, all other terms to the right.

Add <em>'-5x'</em> to each side of the equation.

<em>1 + 2x + -5x = -5 + 5x + -5x </em>

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<em>1 + -3x = -5 + 5x + -5x </em>

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<em>1 + -3x = -5 + 0 </em>

<em>1 + -3x = -5 </em>

Add '-1' to each side of the equation.

<em>1 + -1 + -3x = -5 + -1 </em>

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<em>-3x = -5 + -1 </em>

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<em>-3x = -6 </em>

Divide each side by <em>'-3'. </em>

<em>x = 2 </em>

Simplifying

<em>x = 2</em>

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