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3 years ago

Use the elimination method to solve the following system of equations.

1 answer:

3 0

The answer
2x - y + z = -3 (1)
2x + 2y + 3z = 2 (2)
3x - 3y - z = -4 (3)

 (1) - (2) 3y +2z =5
3 .(1) - 2.(2) 3y+5z = -1

let 's solve
3y +2z =5
3y+5z = -1
z= - 2, and 3y =5+4=9, y=3, so 2x - 3 -2 = -3 implies x= 1

finally x=1, y=3 and z= -2
proof
2x - y + z = -3?? 2.1 - 3 -2 = -3 (true)

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*Asymptotes* g(x) =2x+1/x-3 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 $y=2$ .