Choose a number between 55 and 101 That is multiple of 6,9, and 18

If something is being added to x, subtract it from both sides.
If something is being subtracted from x, add it on both sides.
Same with multiplication and division. If x is being divided, multiply. If x is being multiplied, divide.

We perform inverse operations to<em> combine like terms</em>. This means to get x to one side and everything else on the other.

<h2>Solving the Questions</h2><h3>Question 1</h3>

$5x+7=15$

Because 7 is being added to x, subtract it from both sides:

$5x+7-7=15-7\\5x=8$

Because x is being multiplied by 5, divide both sides by 5:

$\dfrac{5x}{5}=\dfrac{8}{5}\\\\x=\dfrac{8}{5}$

Therefore. $x=\dfrac{8}{5}$ .

<h3>Question 2</h3>

$7x+4=5x-18$

Here, we can group all the x values on the left side of the equation. Subtract 5x from both sides:

$7x+4-5x=5x-18-5x\\2x+4=-18$

To isolate x, subtract 4 from both sides:

$2x+4-4=-18-4\\2x=-22$

Divide both sides by 2:

$\dfrac{2x}{2}=\dfrac{-22}{2}\\\\x=-11$

Therefore, $x=-11$ .

6 0

1 year ago

Describe the behavior of the function ppp around its vertical asymptote at x=-2x=−2x, equals, minus, 2.

insens350 [35]

Answer:

$x->-2^{-}, p(x)->-\infty$ and as $x->-2^{+}, p(x)->-\infty$

Step-by-step explanation:

Given

$p(x) = \frac{x^2-2x-3}{x+2}$ -- Missing from the question

Required

The behavior of the function around its vertical asymptote at $x = -2$

$p(x) = \frac{x^2-2x-3}{x+2}$

Expand the numerator

$p(x) = \frac{x^2 + x -3x - 3}{x+2}$

Factorize

$p(x) = \frac{x(x + 1) -3(x + 1)}{x+2}$

Factor out x + 1

$p(x) = \frac{(x -3)(x + 1)}{x+2}$

We test the function using values close to -2 (one value will be less than -2 while the other will be greater than -2)

We are only interested in the sign of the result

----------------------------------------------------------------------------------------------------------

As x approaches -2 implies that:

$x -> -2^{-}$ Say x = -3

$p(x) = \frac{(x -3)(x + 1)}{x+2}$

$p(-3) = \frac{(-3-3)(-3+1)}{-3+2} = \frac{-6 * -2}{-1} = \frac{+12}{-1} = -12$

We have a negative value (-12); This will be called negative infinity

This implies that as x approaches -2, p(x) approaches negative infinity

$x->-2^{-}, p(x)->-\infty$

Take note of the superscript of 2 (this implies that, we approach 2 from a value less than 2)

As x leaves -2 implies that: $x>-2$

Say x = -2.1

$p(-2.1) = \frac{(-2.1-3)(-2.1+1)}{-2.1+2} = \frac{-5.1 * -1.1}{-0.1} = \frac{+5.61}{-0.1} = -56.1$

We have a negative value (-56.1); This will be called negative infinity

This implies that as x leaves -2, p(x) approaches negative infinity

$x->-2^{+}, p(x)->-\infty$

So, the behavior is:

$x->-2^{-}, p(x)->-\infty$ and as $x->-2^{+}, p(x)->-\infty$

6 0

3 years ago