Solve for x: 3 over 3 x plus 1 over quantity x plus 4 equals 10 over 7x .

2 answers:

8 0

3 1 10
------ + --------------- = ----------
3x x +4 7x

1 1 10
------ + ------------ = ----------
x x +4 7x

x + 4 + x 10
------------------ = ----------
   x(x +4) 7x

2x + 4 10
------------------ = ----------
   x(x +4) 7x

7x(2x + 4) = 10x(x+4)

14x^2 + 28x = 10x^2 + 40x

4x^2 - 12x = 0
4x(x - 3) = 0
4x = 0
x = 0
x - 3 = 0
x = 3

answer x = 0 and x = 3

EleoNora [17]4 years ago

7 0

<h2>Answer:</h2>

The value of x is:

<h2>Step-by-step explanation:</h2>

We are asked to solve:

$\dfrac{3}{x}+\dfrac{1}{x+4}=\dfrac{10}{7x}$

on taking least common multiple in the left hand side of the equation we get:

$\dfrac{3(x+4)+3x}{3x(x+4)}=\dfrac{10}{7x}\\\\\\\dfrac{3x+12+3x}{3x(x+4)}=\dfrac{10}{7x}\\\\\\\dfrac{6x+12}{3x(x+4)}=\dfrac{10}{7x}\\\\\\7x(6x+12)=10(3x(x+4))\\\\\\x(42x+84)=x(30x+120)\\\\\\x(42x+84-30x-120)=0\\\\\\x=0\ or\ 12x-36=0\\\\\\x=0\ or\\\\\\12x=36\\\\\\i.e.\\\\\\x=0\ or\ x=3$

But x≠0

because if x=0 then,

$\dfrac{1}{3x}$ will not be defined.

Hence, the value of x is:

You might be interested in

PLEASE HELP

lisov135 [29]

This college cost and savings calculator is the ideal tool for your college planning needs. It combines a college cost calculator with a 529 college savings calculator. Obtain a personalized projection of your future college costs by entering your child's age, the type of college you're saving for, and your household income. This calculator shows you how much to save for college in a 529 college savings plan. You can adjust the monthly contribution to see how much monthly contributions can compound over time and the percentage of total college costs your savings will cover.

4 0

3 years ago

If A+B+C=<img src="https://tex.z-dn.net/?f=%5Cpi" id="TexFormula1" title="\pi" alt="\pi" align="absmiddle" class="latex-formula"

seraphim [82]

Answer:

$a + b + c = \pi \\ = > c= \pi - a - b \\ < = > \tan(c) = \tan(\pi - a - b) = -\tan(a + b)$

Step-by-step explanation:

we have:

$\tan(a) + \tan(b) + \tan(c) \\ = \tan(a) + \tan(b) - \tan(a + b) \\ = \tan( a) + \tan(b) - \frac{ \tan(a) + \tan(b) }{1 - \tan(a) \tan(b) } \\ = \frac{ ( \tan(a) + \tan(b) ) \tan(a) \tan(b) }{ \tan(a) \tan(b) - 1 } (1)$

we also have:

$\tan(a) \tan(b) \tan(c) \\ = - \tan(a) \tan(b) \tan(a + b) \\ = \frac{ -(\tan( a ) + \tan(b) ) \tan(a) \tan(b) }{1 - \tan(a) \tan(b) } \\ = \frac{( \tan(a) + \tan(b)) \tan(a) \tan(b) }{ \tan(a) \tan(b) - 1 } (2)$