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3 years ago

Solve for x 3x+3/x-4 = 3x+2/x+4

Mathematics

2 answers:

Lerok [7]3 years ago

5 0

Answer:

here you go! with step by step so you can do it next time

VashaNatasha [74]3 years ago

3 0

Answer:

-7/8

Step-by-step explanation:

cross multiply first then expand the equations when you cross multiply it will be

(3x+3)(x+4)=(3x+2)(x-1)

3x(x+4)+3(x+4)=3x(x-1)+2(x-1)

3x²+12x+3x+12=3x²-3x+2x-2

3x²+15x+12=3x²-x-2

3x²-3x²+15x+x=-2-12

16x/16=-14/16

x=-14/16

simplified to

-7/8

I hope this helps

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Find the solution of y = –x – 3 for x = 1.

nikklg [1K]

Answer:

A. :)

Step-by-step explanation:

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3 years ago

3w - 11 = 25 can someone answer this plz

adoni [48]

Answer:

w = 12

Step-by-step explanation:

3w = 11 + 25

3w = 36

w = 12

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3 0

3 years ago

Read 2 more answers

The graph below shows the function f(x)=x-3/x^2-2x-3 which statement is true

Ugo [173]

Answer:

The correct option is A.

Step-by-step explanation:

Domain:

The expression in the denominator is x^2-2x-3

x² - 2x-3 ≠0

-3 = +1 -4

(x²-2x+1)-4 ≠0

(x²-2x+1)=(x-1)²

(x-1)² - (2)² ≠0

∴a²-b² =(a-b)(a+b)

(x-1-2)(x-1+2) ≠0

(x-3)(x+1) ≠0

x≠3 for all x≠ -1

So there is a hole at x=3 and an asymptote at x= -1, so Option B is wrong

Asymptote:

x-3/x^2-2x-3

We know that denominator is equal to (x-3)(x+1)

x-3/(x-3)(x+1)

x-3 will be cancelled out by x-3

1/x+1

We have asymptote at x=-1 and hole at x=3, therefore the correct option is A....

4 0

3 years ago

Read 2 more answers

If p(x) = x2 – 1 and 4(x) = 5(x-1), which expression is equivalent to (p -9)(x)?

jenyasd209 [6]

Answer:

5 times 5 24

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

Someone please be awesome and help me please :(

solong [7]

Answer:

$(x+\frac{b}{2a})^2+(\frac{4ac}{4a^2}-\frac{b^2}{4a^2})=0$

$(x+\frac{b}{2a})^2=\frac{b^2-4ac}{4a^2}$

$x=\frac{-b}{2a} \pm \frac{\sqrt{b^2-4ac}}{2a}$

Step-by-step explanation:

$x^2+\frac{b}{a}x+\frac{c}{a}=0$

They wanted to complete the square so they took the thing in front of x and divided by 2 then squared. Whatever you add in, you must take out.

$x^2+\frac{b}{a}x+(\frac{b}{2a})^2+\frac{c}{a}-(\frac{b}{2a})^2=0$

Now we are read to write that one part (the first three terms together) as a square:

$(x+\frac{b}{2a})^2+\frac{c}{a}-(\frac{b}{2a})^2=0$

I don't see this but what happens if we find a common denominator for those 2 terms after the square. (b/2a)^2=b^2/4a^2 so we need to multiply that one fraction by 4a/4a.

$(x+\frac{b}{2a})^2+\frac{4ac}{4a^2}-\frac{b^2}{4a^2}=0$

They put it in ( )

$(x+\frac{b}{2a})^2+(\frac{4ac}{4a^2}-\frac{b^2}{4a^2})=0$

I'm going to go ahead and combine those fractions now:

$(x+\frac{b}{2a})^2+(\frac{-b^2+4ac}{4a^2})=0$

I'm going to factor out a -1 in the second term ( the one in the second ( ) ):

$(x+\frac{b}{2a})^2-(\frac{b^2-4ac}{4a^2})=0$

Now I'm going to add (b^2-4ac)/(4a^2) on both sides:

$(x+\frac{b}{2a})^2=\frac{b^2-4ac}{4a^2}$

I'm going to square root both sides to rid of the square on the x+b/(2a) part:

$x+\frac{b}{2a}=\pm \sqrt{\frac{b^2-4ac}{4a^2}}$

$x+\frac{b}{2a}=\pm \frac{\sqrt{b^2-4ac}}{2a}$

Now subtract b/(2a) on both sides:

$x=\frac{-b}{2a} \pm \frac{\sqrt{b^2-4ac}}{2a}$

Combine the fractions (they have the same denominator):

$x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$

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3 years ago