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3 years ago

11

Solve by using any method of substitution4 +8x = y 6x - y = 2

1 answer:

Bumek [7]3 years ago

8 0

Answer:

(-3,-20)

Step-by-step explanation:

4 +8x = y ...... equation (1)

6x - y = 2 ........equation(2)

from equation 2

6x - y = 2

6x -(4+8x) =2

6x - 4 + 8x = 2

+4 +4

6x - 8x = 6

-2x=6

divide by -2

x= -3

put x in equation (1) to solve for y

4 + 8x = y

4+8(-3)=y

4-24=y

-20=y

y= -20

solution = (-3,-20)

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3 years ago

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3 years ago

1. Hoover Dam spans the Colorado

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3 years ago

<img src="https://tex.z-dn.net/?f=f%28x%29%20%3D%20%5Cfrac%7Bx-5%7D%7B2x%5E%7B2%7D-5x-3%20%7D" id="TexFormula1" title="f(x) = \f

Vedmedyk [2.9K]

i) The given function is

$f(x)=\frac{x-5}{2x^2-5x-3}$

The factored form is

$f(x)=\frac{x-5}{(x-3)(2x+1)}$

The domain are the values of x for which the function is defined.

$(x-3)(2x+1)\ne 0$

$(x-3)\ne0,(2x+1)\ne 0$

$x\ne3,x\ne-\frac{1}{2}$

ii) To find the vertical asymptotes, equate the denominator to zero.

$(x-3)(2x+1)=0$

$(x-3)=\ne0,(2x+1)=0$

$x=3,x=-\frac{1}{2}$

iii) To find the roots, equate the numerator to zero.

$x-5=0$

The root is $x=5$

iv) To find the y-intercept, put $x=0$ into the function.

$f(0)=\frac{0-5}{(0-3)(2(0)+1)}$

$f(0)=\frac{-5}{(-3)(1)}$

$f(0)=\frac{5}{3}$

The y-intercept is $\frac{5}{3}$

v) The horizontal asymptote is given by;

$lim_{x\to \infty}\frac{x-5}{2x^2-5x-3}=0$

The horizontal asymptote is $y=0$

vi) The function is not reducible. There are no holes.

vii) The given function is a proper rational function.

Proper rational functions do not have oblique asymptotes.

3 0

3 years ago

Read 2 more answers