Mathematics, 28.03.2020 22:42 100154281

A student solves the following equation and
determines that the solution is -2. Is the student
correct? Explain.
3/a+2 - 6a/a^2-4 = 1/a-2

Answers: 3

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Answers

Answer from: Birlem101

Solving the equation, it is found that it has no solutions, thus, the student is not correct.

The equation is:

$\frac{3}{a + 2} - \frac{6a}{a^2 - 4} = \frac{1}{a - 2}$

Applying the subtraction of perfect squares, we have that:

a^2 - 4 = (a - 2)(a + 2)

Then

$\frac{3}{a + 2} - \frac{6a}{(a - 2)(a + 2)} = \frac{1}{a - 2}$

$\frac{3}{a + 2} - \frac{6a}{(a - 2)(a + 2)} - \frac{1}{a - 2} = 0$

$\frac{3(a - 2) - 6a - (a + 2)}{(a - 2)(a + 2)} = 0$

$\frac{3a - 6 - 6a - a - 2}{(a - 2)(a + 2)} = 0$

$\frac{-4a - 8}{(a - 2)(a + 2)} = 0$

In a fraction, the denominator cannot be zero, thus:

$a - 2 \neq 0 \rightarrow a \neq 2$

$a + 2 \neq 0 \rightarrow a \neq -2$

Looking at the numerator for the solution:

-4a - 8 = 0

4a = -8

$a = -\frac{8}{4}$

a = -2

However, a = -2 is a restriction, as it makes the denominator zero, thus the function has no solutions.

A similar problem is given at link

Answer from: dakotagorski05

Since the equation is undefined for -2

Therefore, NO SOLUTION for the given equation.

Step-by-step explanation:

Considering the expression

$\frac{3}{a+2}-6\cdot \frac{a}{-4+a^2}=\frac{1}{a-2}$

$\frac{3}{a+2}-\frac{6a}{-4+a^2}=\frac{1}{a-2}$

$\mathrm{Find\:Least\:Common\:Multiplier\:of\:}a+2,\:-4+a^2,\:a-2:\quad \left(a+2\right)\left(a-2\right)$

$\mathrm{Multiply\:by\:LCM=}\left(a+2\right)\left(a-2\right)$

$\frac{3}{a+2}\left(a+2\right)\left(a-2\right)-\frac{6a}{-4+a^2}\left(a+2\right)\left(a-2\right)=\frac{1}{a-2}\left(a+2\right)\left(a-2\right)$

$\frac{3}{a+2}\left(a+2\right)\left(a-2\right):\quad 3\left(a-2\right)$ $-\frac{6a}{-4+a^2}\left(a+2\right)\left(a-2\right):\quad -6a$ $\frac{1}{a-2}\left(a+2\right)\left(a-2\right):\quad a+2$

so equation becomes

$3\left(a-2\right)-6a=a+2$

-3a-6=a+2

-3a-6+6=a+2+6

-4a=8

$\mathrm{Divide\:both\:sides\:by\:}-4$

$\frac{-4a}{-4}=\frac{8}{-4}$

a=-2

$\mathrm{Verify\:Solutions}$

$\mathrm{Take\:the\:denominator\left(s\right)\:of\:}\frac{3}{a+2}-6\frac{a}{-4+a^2}-\frac{1}{a-2}\mathrm{\:and\:compare\:to\:zero}$

$\mathrm{Solve\:}\:a+2=0:\quad a=-2$

$\mathrm{Solve\:}\:-4+a^2=0:\quad a=2,\:a=-2$

$\mathrm{Solve\:}\:a-2=0:\quad a=2$

So the following points are undefined

$a=-2,\:a=2$

Since the equation is undefined for -2

Therefore, NO SOLUTION for the given equation.

Answer from: Quest

∠1≅∠2 by the alternate exterior angles theorem.

step-by-step explanation:

given, a ∥ b and ∠1 ≅ ∠3 .we have to prove that e ∥ f

we know that ∠1≅∠3 and that a || b because they are given. we see that by the alternate exterior angles theorem. therefore, ∠2≅∠3 by the transitive property. so, we can conclude that e || f by the converse alternate exterior angles theorem.

we have to fill the missing statement.

transitivity property states that if a = b and b = c, then a = c.

now, given ∠1≅∠3 and by transitivity property ∠2≅∠3 .

hence, to apply transitivity property one angle must be common which is not in result after applying this property which is ∠1.

the only options in which ∠1 is present are ∠1 and ∠2, ∠1 and ∠4

∠1 and ∠4 is not possible ∵ after applying transitivity we didn't get ∠4.

hence, the missing statement is ∠1≅∠2.

so, ∠1≅∠2 by the alternate exterior angles theorem.

Answer from: Quest

Ineed with this same question!
In a circle with a radius of 36.9 m, an arc is intercepted by a central angle of 8π5 radians.<

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