These equations really have been frustrating me:
2x - (x - 5) / 4 = (3x -1) / 2 + 1
2x -7 / x + 4 = 5 / x + 4 - 2
(The / means over as a fraction) My main problem is just getting the fractions gone, I know I have to multiply but I dont know by what or what to multiply it to. Can someone solve them and tell me how to get rid of these fractions easily step by step?
2006-10-04 14:10:15 · 6 answers · asked by Anonymous in Education & Reference ➔ Other - Education
Hi I am hoping you are trying to find the value of "x" here . The first equation I take it as :
2x - (x-5)/4 = (3x-1)/2 +1
Step : 1 Make one diviser for the equations on both sides
8x -(x -5)/4 = (3x-1) +2/2
Take both of them out of the brackets
8x-x+5/4 = 3x -1 + 2 /2
7x + 5/4 = 3x + 1/2
7x + 5/2 = 3x +1
7x + 5 = 2(3x +1)
7x + 5 = 6x + 2
Get the x values on LHS and numbers on RHS
7x - 6x = 2-5
x = -3
Now you try the other one following the example above...
It was easy but it took me 20 mins to type it up... :)
2006-10-04 14:44:42 · answer #1 · answered by faztboz 2 · 0⤊ 0⤋
Why not, just this once though
8x-x+5=6x-2+4
7x+5=6x+2
x=-3
first multiply all that crap by 4 to get rid of your fractions, then the rest is pretty cut and dry
next one
2x+4=12/x+2
x+2=6/x+1
x+1=6/x
x=2
2006-10-04 14:18:23 · answer #2 · answered by Anonymous · 0⤊ 0⤋
I know how to solve them, but it will do you no good to copy the answer from me. It will just show your teacher that I know how to do it and you're good at copying the answer from me. The first thing I would do with problem #1 is re-write 1 as 2/2. Then combine it with the fraction next to it. Then re-write 2x as 8x/4. Combine it with the fraction next to it. Then multiply both sides of the equation by 4. That will get rid of your fractions and make life easier. Then solve for x.
You could also do the same for #2, except you re-write 4 as 4x/x and -2 as -2x/x. I assume that the x in the 2nd equation is not equal to zero. Combine the fractions on both sides, then multiply both sides by x. That will get rid of the mess. Then combine like terms, and solve for x.
2006-10-04 14:30:42 · answer #3 · answered by salsera 5 · 0⤊ 0⤋
3(2x-(x-5)) = 4(3x-1)
6x-3x+15 = 12x -4
3x+15 = 12x-4
9x =19
x= 19/9
im not really sure its what you wanted but i tried... you just have to cross your multiplication ... and multiply !!
(2x-7)(x+2)=5(x+4)
try to finish that one !! :P
2006-10-04 14:19:07 · answer #4 · answered by Nini69 1 · 0⤊ 1⤋
Yes, I can help you solve these math equations.
2006-10-04 14:11:22 · answer #5 · answered by B-B@!! P!@Y@ 4 · 0⤊ 1⤋
Yes, if i WANTED to i could.......but i've been doing math hw for the past hour, so i'm rrrreeeeaaally sick of it, sry lol
2006-10-04 14:13:08 · answer #6 · answered by Mimi L 3 · 0⤊ 1⤋