(4x^2 + 6x) / (2x) = 2x + 6?

Question

can anyone solve this and tell me in steps how they did it????

Answer 1

Multiply both sides of the equation by 2x

Then you will have:

4x^2+6x=2x(2x+6)
4x^2+6x=4X^2+12X
6x=12X

x=0

Answer 2

Since one of the side has fraction, so use cross multiply to get rid of the fraction.
Multiply (2x) with (2x+6). So, the equation will be like this:
(4x^2 + 6x) = (2x)(2x +6)

Expand the 2x and 2x+6
4x^2 + 6x = 4x^2 + 12x
4x^2 - 4x^2 + 6x - 12x = 0 (make one side = 0)
-6x = 0
x = 0

I think there's something wrong with the question. Because the answer cannot be x=0.

Answer 3

It appears that you have a mistake in the equation you have posted here, because it is NOT an equality. There appear to be two ways to solve this problem which contradict each other:
--------------
Method 1:

(4x^2 + 6x)/2x = 2x + 6

divide (4x^2 + 6x) by 2x to get 2x + 3

2x+3 = 2x + 6

subtract 2x from both sides to get

3 ≠ 6
--------------
Method 2:

(4x^2 + 6x)/2x = 2x + 6

multiply both sides by 2x

4x^2 + 6x = 2x*(2x + 6)

4x^2 + 6x = 4x^2 + 12x

subtract 4x^2 from both sides to get

6x = 12 x

subtract 6x from both sides to get

0 = 6x, therefore x = 0
---------------
I am at a loss to explain why I get the different answers. Both methods are valid. this reminds me of the mathematical conundrum whereby you can show that 1=2.

Good luck.

Answer 4

(4x^2 + 6x) / (2x)

You can split this up like

4x^2/(2x) + 6x/(2x)

As long as there is only 1 term in the denominator (bottom) of the fraction so (A+B)/C = A/C + B/C

Once you have that then the rest should be staight forward

4x^2/(2x) + 6x/(2x)

divide out similar factors (4/2=2, 6/2=3, x^2/x=x and x/x=1) so

4x^2/2x = 2x and 6x/2x=3 and

(4x^2 + 6x) / (2x) = 2x+3

You might want to check your result.

Answer 5

This cannot be done. You've made a mistake somewhere!

4x^2 + 6x = 2x(2x +6 )
4x^2 + 6x = 4x^2 + 12x
6x = 12x???

Answer 6

(4x^2 + 6x) / (2x) = 2x + 6
[ cross multiply]
=> 4x^2 + 6x = 4x ^2 + 12x
=> 12x - 6x = 0
=> 6x = 0
=> x = 0

now if u wanted to prove the equation...
(4x^2 + 6x)/2x
= (4x^2)/2x + 6x/2x
= 2x + 3 ( not equal to R.H.S.)
therefore, the equation is false

Answer 7

Are you sure the problem's right?

(4x^2+6x)=(2x)(2x+6)
4x^2+6x=4x^2+12x
Subtract 4x^2 from both sides and divide both sides by "x"
6=12

If you're looking for a value for "x," I don't think any exists that makes this equation true.

Answer 8

Fundamental rules of Alkgebra:

if "this" = "that" then "this" + "anything" = "that" + "anything" (eq. 1)

"this" times "anything" = "that" times "anything" (eq. 2)

"anything"/"anything" = 1.000 (eq. 3)

Therefore, if "anything" = "2x" then:

(2X)(4X^2 + 6X)/(2x) = (2X)(2X + 6) "using eq. 2"

(4X^2 + 6X) = (2X)(2X + 6) "using eq. 3"

(4X^2 + 6X) - (2X)(2X + 6) = 0 "using eq. 1"

Next, factor the result using ordinary math:

4X^2 + 6X - 4X^2 -12X = 0
6X - 12X = 0
-6X = 0

It appears X = 0

of course, substituting "0" in the original equation yields :

0/0 = 6

What makes this problem rather peculiar is the fact the "4X^2" terms cancel out. If they did not, one would have a second degree polynominal whose answers ("roots") could be calculated.

Answer 9

a) Are you looking for the value of x?
b) Or you want to know if the equation is correct?

If your question is a, the value of x does not exist because if you substitute 0 to the value of x (as the answer of some answerers) the equation becomes===>
(4 (0)^2 + 6(0) / 2 (0) = 2 (0) + 6
(0 + 0)/0 = 0 + 6
0/0 = 6
0 = 6


If your question is b, the answer is NO.

Answer 10

(4x² + 6x) / 2x = 2x + 6
2x + 3 = 2x + 6
Cannot be solved---error in question?