Reply
Fri 3 Jun, 2005 12:03 am
How to solve?!
a. -4(5x 11)=128
b. x 16 2x - 21 = 40
c. 41= 3x - 17 = 5x = 24
d. 4 (3x - 7) 25 = -39
I think you might wanna check your grammar cause something seems to be written incorrectly (particularly c which makes no sense)
However as written
a. -4(5x 11)=128
this reads -210x=128 then x=-128/210=-64/105
b. x 16 2x - 21 = 40
32x^2-21=40
32x^2=61
x^2=61/32
x=+/-sqrt(61/32)
c. 41= 3x - 17 = 5x = 24
??? This one makes no sense
d. 4 (3x - 7) 25 = -39
100(3x-7)=-39
300x-700=-39
300x=661
x=661/300
Rap
You're going to have to transcribe your homework better than that if you want some help. In "a", for example, is that (5x + 11) or (5x - 11), because just plain (5x 11) doesn't make any sense, except for "55".
For "d", let's assume what you meant was 4(3x - 7) + 25 = -39. The first thing to do is multiply "4" by both terms inside the parenthesis. This gives you:
((4 times 3x) + (4 times negative 7)) + 25 = -39
Now simplify:
12x - 28 +25 = -39
Now simplify again
12x -3 = -39
Now add +3 to each side of the "equals" sign:
12x -3 +3 = -39+3
And simplify again
12x = -36
and divide each side by 12 to ger x=-3.
Finally, check your work (ALWAYS ALWAYS ALWAYS check your work)
4 ((3 times -3)-7) +25 = -39 ?
4 (-9-7) +25 = -39?
4(-16) + 25 = -39?
-64+25 = -39?
-39 = -39 QED
The same logic and steps can be used to solve any linear (first order, meaning all exponents are one, meaning no squared or cubed terms) algebra problems. Once they've been properly written down, that is.