Rational equations...?

Question

Hey, I'm doing some algebra stuff, but I don't know how to do what I'm working on. Here's some examples of what I'm dealing with:

(2a - 3)/6 = (4a)/3 + 2/3

(3x)/5 + 3/4 = (6x)/10

(2b - 3)/7 - b/2 = (b + 3)/14

Could anyone tell me how to solve these problems and others like them?

Answer 1

1.(2a-3)/6=(4a+2)/3 i took LCD for the RHS
now cross multiolying 3(2a-3)=6(4a+2)
removing brackets 6a-9=24a+12
now transposing 6a-24a=12+9
-18a=21
a=-21/18=>-7/6
2.3x/5+3/4=6x/10
reduce the LHS to a common denominator
(12x+15)/20=6x/10
cross multiplying
10(12x+15)=20*6x
since the x es are getting cancelled out thiscannot be solved
3.(2b-3)/7-b/2=(b+3)/14
again reducing the LHS tocommon denominator
[2(2b-3)-7b]/14=(b+3)/14
since both sides have the same denominator they cancel out
4b-6-7b=b+3
transposing -4b=9
b=-9/4

Answer 2

I haven't doubled checked any of these.

Problem 1
(2a - 3)/6 = (4a)/3 + 2/3
Multiply both sides by 6
6((2a - 3)/6) = 6((4a)/3 + 2/3)
Distribute the 6 through the parentheses
6(2a - 3)/6 = 6(4a)/3 + 6(2/3)
Reduce the fractions
2a - 3 = 8a + 4
Subtract 2a from both sides
3 = 6a + 4
Subtract 4 from both sides
-1 = 6a
Divide by 6 on both sides
-1/6 = a

Next Problem
Multiple by the least common multiply on both sides, which is 20
20((3x)/5 + 3/4) = 20(6x)/10
Distribute 20 through the parentheses
20(3x)/5 + 20(3/4) = 20(6x)/10
Reduce the fractions
12x + 15 = 12x
Subtract 12x on both sides
15 = 0
Since 15 is not equal to 0, this problem has no solution.

Next Problem
(2b - 3)/7 - b/2 = (b + 3)/14
Multiple by the least common multiply on both sides, which is 14
14((2b - 3)/7 - b/2) = 14(b + 3)/14
Distribute the 14 through the parentheses
14(2b - 3)/7 - 14b/2 = 14(b + 3)/14
Reduce the fractions
2(2b - 3) - 7b = b + 3
Distribute the 2 through the parentheses
4b - 6 - 7b = b + 3
Add like terms
-3b - 6 = b + 3
Subtract b from both sides
-4b - 6 = 3
Add 6 to both sides
-4b = 9
Divide by -4 on both sides
b = -9/4

Answer 3

In each of these, find the Lowest Common Multiple (LCM) of the denominators of each of the terms of the equation. Then, multiply the entire equation through by that amount to get a normal equation without any fractions.

For example, in the first problem:

(2a - 3)/6 = (4a)/3 + 2/3

The LCM of 6, 3, and 3 is 6. After multiplying through by 6, you have:

2a - 3 = 2(4a) + 2(2)
2a - 3 = 8a + 4
-6a = 7
a = -7/6

Answer 4

(2a - 3)/6 = ((4a)/3) + (2/3)
(2a - 3)/6 = (4a + 2)/3
cross multiply
3(2a - 3) = 6(4a + 2)
6a - 9 = 24a + 12
-18a = 21
a = (-21/18)
a = (-7/6)

((3x)/5) + (3/4) = ((6x)/10)
((3x)/5) + (3/4) = ((3x)/5)
Multiply everything by 20
4(3x) + 5(3) = 4(3x)
12x + 15 = 12x
0x = 15
x = (15/0)
x = Undefined.

((2b - 3)/7) - (b/2) = ((b + 3)/14)
Multiply everything by 14
2(2b - 3) - 7b = b + 3
4b - 6 - 7b = b + 3
-3b - 6 = b + 3
-4b = 9
b = (-9/4)

Answer 5

... 10/(x-a million) + a million = 20/(x²-a million) ... x ? ±a million ? Else equation will change into undefined or 10/(x-a million) + a million = 20/[ (x-a million)(x+a million) ] or 10(x+a million) + (x-a million)(x+a million) = 20 or 10x + 10 + x² - a million = 20 or x² + 10x - 11 = 0 or (x - a million) (x + 11) = 0 or x = a million ....... ? Reject with the aid of initial constraint or x = -11 .... ? very last answer

Answer 6

Parathesis
Exponets
Multiplication
Division
Addition
Subtraction

In that order. Please Excuse My Dear Aunt Sally