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 ?
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
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
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
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
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