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3 years ago

How can one third x − 2 = one fourth x + 11 be set up as a system of equations?

Mathematics

2 answers:

timurjin [86]3 years ago

4 0

Answer: (x-2) = (x+11)

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

3 4

Step-by-step explanation:

First part is x-2 divided by on third. So the x-2 acts a the 1 so you put x-2 over 3. And you get:

(x-2)

--------

For the other side of the equation it is x +11 divide by 1/4. So the x+11 will act a the 1 in the fraction. And you will get:

(x+11)

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UNO [17]3 years ago

3 0

For this case we have the following expression:

$\frac {1} {3} (x-2) = \frac {1} {4} (x + 11)$

So, we have two equations of the form: $y = \frac {1} {3} (x-2)\\y = \frac {1} {4} (x + 11)$

Rewriting equation 1:

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

If we multiply by 3 on both sides of the equation we have:

$9y-3x = -6$

Rewriting equation 2:

$y = \frac {1} {4} (x + 11)\\4y = x + 11\\4y-x = 11$

If we multiply by 4 on both sides of the equation:

$16y-4x = 44$

Answer:

$9y-3x = -6\\16y-4x = 44$

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*Asymptotes*<br> g(x) =2x+1/x-3 <br><br> Give the domain and x and y intercepts

Nataly [62]

Answer: Assuming the function is $g(x)=\frac{2x+1}{x-3}$ :

The x-intercept is $(\frac{-1}{2},0)$ .

The y-intercept is $(0,\frac{-1}{3})$ .

The horizontal asymptote is $y=2$ .

The vertical asymptote is $x=3$ .

Step-by-step explanation:

I'm going to assume the function is: $g(x)=\frac{2x+1}{x-3}$ and not $g(x)=2x+\frac{1}{x}-3$ .

So we are looking at $g(x)=\frac{2x+1}{x-3}$ .

The x-intercept is when y is 0 (when g(x) is 0).

Replace g(x) with 0.

$0=\frac{2x+1}{x-3}$

A fraction is only 0 when it's numerator is 0. You are really just solving:

$0=2x+1$

Subtract 1 on both sides:

$-1=2x$

Divide both sides by 2:

$\frac{-1}{2}=x$

The x-intercept is $(\frac{-1}{2},0)$ .

The y-intercept is when x is 0.

Replace x with 0.

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

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

$y=\frac{0+1}{-3}$

$y=\frac{1}{-3}$

$y=-\frac{1}{3}$ .

The y-intercept is $(0,\frac{-1}{3})$ .

The vertical asymptote is when the denominator is 0 without making the top 0 also.

So the deliminator is 0 when x-3=0.

Solve x-3=0.

Add 3 on both sides:

x=3

Plugging 3 into the top gives 2(3)+1=6+1=7.

So we have a vertical asymptote at x=3.

Now let's look at the horizontal asymptote.

I could tell you if the degrees match that the horizontal asymptote is just the leading coefficient of the top over the leading coefficient of the bottom which means are horizontal asymptote is $y=\frac{2}{1}$ . After simplifying you could just say the horizontal asymptote is $y=2$ .