# Thread: simple question but i need help

1. ## simple question but i need help

solve the following linear equation

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

thank you

regards

2. Originally Posted by rpatel
solve the following linear equation

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

thank you

regards
x = 4/11; solve for x, that is:

Multiply both sides by x:

x^2 + 4*x - 1 = x^2 - 7*x + 3

Get x's to one side:

Add +7*x to the left side and subtract the x^2's.

11*x - 1 = 3

11*x = 4

x = 4/11.

3. hello thank you for that

7x+2/5 = 4x-1/2

i tried to use similar method to the answer above but it isn't working out.

4. Originally Posted by rpatel
solve the following linear equation

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

thank you

regards
Just in case:
(x-1)/(x+4) = (x+3)/(x-7) (If it IS this use parathesis!!)

Multiply both sides by (x+4)(x-7):
The LHS is
(x-1)(x-7) = x^2 - 8x + 7

The RHS is
(x+3)(x+4) = x^2 + 7x + 12

So
x^2 - 8x + 7 = x^2 + 7x + 12
-8x + 7 = 7x + 12
-15x = 5
x = -1/3.

-Dan

5. Originally Posted by rpatel
hello thank you for that

7x+2/5 = 4x-1/2

i tried to use similar method to the answer above but it isn't working out.
7x + 2/5 = 4x - 1/2
3x = -2/5 - 1/2 = -4/10 - 5/10 = -9/10
x = -3/10

-Dan

6. i think the answers above are incorrect.

7. Originally Posted by rpatel
i think the answers above are incorrect.
Specifically, which posts? You can see numbers at the top right.

-Dan

8. Hello, rpate1!

You really must learn to use parentheses ... (and some spaces wouldn't hurt).

. . . . . . . . . . . . . 2x + 3
You cannot write: --------- .as: .2x + 3/5
. . . . . . . . . . . . . . . 5

It should be obvious that it looks like "2x plus three-fifths".

You must write: .(2x + 3)/5
. . so everyone knows that it is "all over 5".

7x + 2/5 .= .4x - 1/2

Is that really "7x plus two-fifths equals 4x minus one-half" ??

Or did you mean: .(7x + 2)/5 .= .(4x - 1)/2 ?

9. (x-1)/(x+4) = (x+3)/(x-7)

i still don't quite understand how you get the answer for this

for the second one i mean

(7x+2)/5 = (4x-1)/2

10. rpatel sent me a PM stating he thought my solution to (x-1)/(x+4) = (x+3)/(x-7) was wrong. This is my reply to him:

Let's take the LHS of the equation first:
(x - 1)/(x + 4) = ([-1/3] - 1)/([-1/3] + 4)

= ([-1/3] - [3/3])/([-1/3] + [12/3])

= ([-4/3])/([11/3]) = -4/11.

The RHS:
(x + 3)/(x - 7) = ([-1/3] + 3)/([-1/3] - 7)

= ([-1/3] + [9/3])/([-1/3] - [21/3])

= ([8/3])/([-22/3]) = -8/22 = -4/11.

If any of this is unclear, please let me know.

I'm going to post this response in your thread as well, in case anyone else is having the same question.

-Dan

11. Originally Posted by rpatel
(x-1)/(x+4) = (x+3)/(x-7)

i still don't quite understand how you get the answer for this
Let's take it step by step.

The first thing I want to do is get rid of all the fractions. To do this on the LHS I have to multiply by (x + 4), to do it on the RHS I have to multiply by (x - 7). So I'm going to multiply both sides by (x + 4)(x - 7).

LHS:
[(x - 1)/(x + 4)]*(x + 4)(x - 7) Note that the two factors of (x + 4) will cancel, leaving:
(x - 1)(x - 7) = x^2 - 8x + 7.

RHS:
[(x + 3)/(x - 7)]*(x + 4)(x - 7) Note that the two factors of (x - 7) will cancel, leaving:
(x + 3)(x + 4) = x^2 + 7x + 12.

The rest is equating the two quadratic expressions and solving for x.

If you are having troubles with the expansion consider the general case:
(x + a)(x + b). If we distribute the multiplication on the right we get
x(x + b) + a(x + b) = x^2 + bx + ax + ab = x^2 + (a + b)x + ab.

Please let me know if anything is still confusing, and what specifically is confusing you.

-Dan

12. Originally Posted by rpatel
(x-1)/(x+4) = (x+3)/(x-7)

i still don't quite understand how you get the answer for this

for the second one i mean

(7x+2)/5 = (4x-1)/2
Again, start by clearing the fractions by multiplying both sides by 10.
[(7x + 2)/5]*10 = [(4x - 1)/2]*10

(7x + 2)*2 = (4x - 1)*5

14x + 4 = 20x - 5
-6x = -9
x = 3/2.

-Dan

13. thank you topsquark

i have printed out your solution and i am studying it now.

thank you.

all help was appreichated, thanks again.

i haven't had 3 months of maths now. i just had my first maths lesson today. i am going to have a test next week on algebra. thats why i am practicing now.

regards

14. Hello, rpatel

. . . . . . . . x - 1 . . . x + 3
We have: . ------ .= .------
. . . . . . . . x + 4 . . .x - 7

Multiply both sides by the LCD: (x + 4)(x - 7)

. . . . . . . . . . . . x - 1 . . . . . . . . . . . . . . x + 3
. . (x + 4)(x - 7) ------- . = . (x + 4)(x - 7) -------
. . . . . . . . . . . . x + 4 . . . . . . . . . . . . . . x - 7

Reduce: . (x - 7)(x - 1) .= .(x + 4)(x + 3)

. . . . . . . . x² - 8x + 7 .= .x² + 7x + 12

. . . . . . . . . . . . .-15x .= .5

. . . . . . . . . . . . . . .x .= .-1/3

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

. . . . . . . . 7x + 2 . . . 4x - 1
We have: . -------- .= .-------
. . . . . . . . . . 5 . . . . . . 2

Multiply both sides by the LCD: 10

. . . . . 7x + 2 . . . . . 4x - 1
. . . 10 -------- .= .10 -------
. . . . . . . 5 . . . . . . . . .2

Reduce: . 2(7x + 2) .= .5(4x - 1)

. . . . . . . . 14x + 4 .= .20x - 5

. . . . . . . . . . . -6x .= .-9

. . . . . . . . . . . . .x .= .3/2

15. Originally Posted by AfterShock
x = 4/11; solve for x, that is:

Multiply both sides by x:

x^2 + 4*x - 1 = x^2 - 7*x + 3

Get x's to one side:

Add +7*x to the left side and subtract the x^2's.

11*x - 1 = 3

11*x = 4

x = 4/11.
You've done something wrong. Not sure what it is, but this is how I did it

(x-1)/(x+4) = (x+3)/(x-7)

(x-1)/(x+4) - (x+3)/(x-7) = 0

((x-1)(x-7) - (x+4)(x+3)) / ((x+4)(x-7)) = 0

(x-1)/(x-7) - (x+4)/(x+3) = 0

x^2 - 8x + 7 - (x^2 + 7x + 12) = 0

-15x - 5 = 0

-15x = 5

x = -5/15 = -1/3

Page 1 of 2 12 Last