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3 years ago

What is the answer to 7(2e−1)−3=6+6e

1 answer:

8 0

<span>Simplifying
7(2e + -1) + -3 = 6 + 6e

Reorder the terms:
7(-1 + 2e) + -3 = 6 + 6e
(-1 * 7 + 2e * 7) + -3 = 6 + 6e
(-7 + 14e) + -3 = 6 + 6e

Reorder the terms:
-7 + -3 + 14e = 6 + 6e

Combine like terms: -7 + -3 = -10
-10 + 14e = 6 + 6e

Solving
-10 + 14e = 6 + 6e

Solving for variable 'e'.

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

Add '-6e' to each side of the equation.
-10 + 14e + -6e = 6 + 6e + -6e

Combine like terms: 14e + -6e = 8e
-10 + 8e = 6 + 6e + -6e

Combine like terms: 6e + -6e = 0
-10 + 8e = 6 + 0
-10 + 8e = 6

Add '10' to each side of the equation.
-10 + 10 + 8e = 6 + 10

Combine like terms: -10 + 10 = 0
0 + 8e = 6 + 10
8e = 6 + 10

Combine like terms: 6 + 10 = 16
8e = 16

Divide each side by '8'.
e = 2

Simplifying
<span>e=2
</span></span>
(sorry, i went into depth)

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We first expand the binomial squared, so we get

$\large\displaystyle\text{$\begin{gathered}\sf \displaystyle L = \lim_{x \to \infty}{\frac{x^4 - x^2 + 4}{x^3 - 5}} = \infty \end{gathered}$}$

Note that in the numerator we get x⁴ while in the denominator we get x³ as the highest degree terms. Therefore, the degree of the numerator is greater and the limit will be \infty. Recall that when the degree of the numerator is greater, then the limit is \infty if the terms of greater degree have the same sign.

Dividing numerator and denominator by the term of highest degree:

$\large\displaystyle\text{$\begin{gathered}\sf L = \lim_{x \to \infty}\frac{x^{4}-x^{2} +4 }{x^{3}-5 } \end{gathered}$}\\$

$\ \ = \lim_{x \to \infty\frac{\frac{x^{4} }{x^{4} }-\frac{x^{2} }{x^{4}}+\frac{4}{x^{4} } }{\frac{x^{3} }{x^{4}}-\frac{5}{x^{4}} } }$

$\large\displaystyle\text{$\begin{gathered}\sf \bf{=\lim_{x \to \infty}\frac{1-\frac{1}{x^{2} } +\frac{4}{x^{4} } }{\frac{1}{x}-\frac{5}{x^{4} } } \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{1}{0}=\infty } \end{gathered}$}$

Note that, in general, 1/0 is an indeterminate form. However, we are computing a limit when x →∞, and both the numerator and denominator are positive as x grows, so we can conclude that the limit will be ∞.

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