# Thread: linear equation with fraction

1. ## linear equation with fraction

Hello, I am stuck on a couple of linear equations.
Could someone please tell me the answer, then I will be able to see the method.

Thanks kindly

2/x + 1/2x = 7

4/x+2 - 6/2x = 5/2x+4

2. Originally Posted by fran1942
Hello, I am stuck on a couple of linear equations.
Could someone please tell me the answer, then I will be able to see the method.

Thanks kindly

2/x + 1/2x = 7

4/x+2 - 6/2x = 5/2x+4
As with all equations where there is a fraction you need to find a common denominator. For the first problem you have denominators x and 2x. What is the common denominator for these?

-Dan

3. 1. $\displaystyle\frac{2}{x}+\frac{1}{2x}=7$

$\displaystyle\frac{4}{2x}+\frac{1}{2x}=7$

$\displaystyle\frac{5}{2x}=7$

$14x=5$

$\displaystyle{x=\frac{5}{14}}$.

2. $\displaystyle\frac{4}{x+2}-\frac{6}{2x}=\frac{5}{2x+4}$

$\displaystyle\frac{8x-6(x+2)}{2x(x+2)}=\frac{5}{2x+4}$

$\displaystyle\frac{2(x-6)}{2x(x+2)}=\frac{5}{2x+4}$

$\displaystyle\frac{x-6}{x(x+2)}=\frac{5}{2x+4}$

$5x(x+2)=2(x-6)(x+2)$

since, $x+2\not=0$

$5x=2(x-6)$

$\therefore x=-4$.

4. Originally Posted by lanierms
...

2. $\displaystyle\frac{4}{x+2}-\frac{6}{2x}=\frac{5}{2x+4}$

$\displaystyle\frac{8x-6(x+2)}{2x(x+2)}=\frac{5}{2x+4}$

$\displaystyle\frac{2(x+1)}{2x(x+2)}=\frac{5}{2x+4}$

...
I don't want to pick at you but the last quoted line is definitely wrong:

$\displaystyle\frac{8x-6(x+2)}{2x(x+2)}=\frac{5}{2x+4}~\implies~\displays tyle\frac{8x-6x-12}{2x(x+2)}=\frac{5}{2x+4}$

By the way: You should have allowed the OP to do a little bit of the question by himself.

5. Originally Posted by lanierms
$5x(x+2)=2(x+1)(x+2)$
EASIER this way:
at this point, divide both sides by (x + 2):
5x = 2x + 2
3x = 2
x = 2/3

6. Originally Posted by earboth
I don't want to pick at you but the last quoted line is definitely wrong:

$\displaystyle\frac{8x-6(x+2)}{2x(x+2)}=\frac{5}{2x+4}~\implies~\displays tyle\frac{8x-6x-12}{2x(x+2)}=\frac{5}{2x+4}$

By the way: You should have allowed the OP to do a little bit of the question by himself.
Whoops. Sorry, made a mistake. ............. I edited my solution.

7. Originally Posted by fran1942
Hello, I am stuck on a couple of linear equations.
Could someone please tell me the answer, then I will be able to see the method.

Thanks kindly

2/x + 1/2x = 7

4/x+2 - 6/2x = 5/2x+4
the best hint for this problem would be to assume 1/x as X and 1/2x as Y. so therefore the equation now becomes as 2X + Y = 7
and 4X + 2 - 6Y = 5Y + 4

the second equation can be further simplified as 4X - 11Y = 2 .
let the first equation 2X + Y = 7 be (1)
and second equation 4X - 11Y = 2 be (2)
now (1) x 4 and (2) x 2 , and the equation becomes
8X + 4Y = 28
8X + 22Y = 4
subtracting these two equations we get -18Y = 24
therefore Y is - 24/18 = - 4/3
applying Y = - 4/3 in (1) we get 8X - 16/3 = 28
8X = 28 + 16/3
= (84 + 16)/3 = 100/3
X= 100/ (3x8)
= 25/12

8. Originally Posted by krishnan
the best hint for this problem would be to assume 1/x as X and 1/2x as Y. so therefore the equation now becomes as 2X + Y = 7
and 4X + 2 - 6Y = 5Y + 4
the second equation can be further simplified as 4X - 11Y = 2 .
let the first equation 2X + Y = 7 be (1)
and second equation 4X - 11Y = 2 be (2)
No, not a "system", but 2 different problems.

9. Originally Posted by Wilmer
No, not a "system", but 2 different problems.
oh! then it is much simpler. the problem is 2/x + 1/2x = 7 . take 1/x as X , so the equation now becomes as 2X + X/2 = 7
now make the denominator common on the left hand side of the equation ie. (4X + X)/2 = 7
now this becomes (4X + X) = 14 . so therefore 5X = 14 , X = 14/5
as a result X = 14/5 = 1/x , x= 5/14
the same procedure can be applied for the second problem too.

10. Originally Posted by Wilmer
EASIER this way:
at this point, divide both sides by (x + 2):
5x = 2x + 2
3x = 2
x = 2/3
Provided, of course, that x+ 2 is not 0. Which means there is another solution.