Maths algebra help – equations?

2)
1/(x – 2) + 3/(x + 6) = 1/2
common denominators:
2(x + 6)/2(x + 6)(x – 2) + 3(2)(x – 2)/2(x + 6)(x – 2) = (x – 2)(x + 6)/2(x + 6)(x – 2)
multiply through by 2(x + 6)(x – 2):
2(x + 6) + 6(x – 2) = (x – 2)(x + 6)
distribute:
3x + 12 + 6x – 12 = x^2 + 6x – 2x – 12
9x = x^2 + 4x – 12
x^2 – 5x – 12 = 0
quadratic formula:
x = [5 +/- SQR(25 – 4(1)(-12))]/2(1)
x = [5 +/- SQR(73)]/2

3)
x + 2/5 + x – 1/2 =
2x + 4/10 – 5/10 =
2x – 1/10

– .–

the trick to any algebra equation is first rewriting the equation so that you combine all like letters (variables), and then rearranging it so that you have “letter = number”.

(1) with fractions, they can be added into 1 main fraction if the bottoms are equal. for this problem, it would be easiest if both fractions had 3x on the bottom. we can rewrite 1/x as 3/3x the same way we can say 1/2 = 2/4 = 3/6 = 32/64. now we have

3/3x + 3/3x = 2

combing fractions, we get

6/3x = 2 -> [we can simplify 6/3x because 6/3 = 2]
2/x = 2 -> [multiply both sides by x, and the x/x turns into a 1 on the left, and we get 2x on the right]
2 = 2x -> [divide both sides by 2]
1 = x

and there you have it.

(2) this one is a bit tricky, but the rule you need to remember for this one is that with any number, if you multiply it by x/x, it stays the same. the same way 2*(5/5) = 2, and 2*(143/143) = 2. the denominators of our fractions are (x-2) and (x+6), and somehow we need to make them both the same. well lets pretend they were 3 and 5 instead. we know 3*5=15, right? and 3 goes into 15 5 times, and 5 goes into 15 3 times, so by multiplying the bottoms together we can get our common denominator. but if we multiply the bottom of the first fraction by (x+6), we also have to multiply the top as well. doing this, we get:

(x+6)/[(x-2)(x+6)] + 3(x-2)/[(x-2)(x+6)] = 1/2

hopefully that doesn’t look too confusing. but now that the bottoms are the same, we can combine them and get

[(x+6) + 3(x-2)]/[(x-2)(x+6)] = 1/2

multiplying everything all together, we can rewrite this as

[4x – 4]/[x^2 + 4x – 12] = 1/2 -> [if we multiply everything by the left denominator, we can get rid of it on the left and bring it to the right, and the same with the 2 on the right to bring it to the left. doing this we get:]

2[4x – 4] = [x^2 + 4x – 12]
8x – 8 = x^2 + 4x – 12 -> [by bringing all the terms to one side, we get:]
0 = x^2 – 4x – 4

from here you can factor and solve for x. for (3) there is no equation, so i cant help you with that one, sorry. Also, dont trust my math, for I almost for certain missed something, but i hope that the theory atleast makes sense

(1) and (2), multiply both sides by x and solve for x
Example (1)
1/x + 3/3x = 2… I’m going to presume 3/3x means 3/(3x) and not (3/3)x
(x)(1/x) + (x)[3/(3x)] = (x)(2)
1+3/3=2x
1+1=2x
2=2x
1=x… or x=1 if you prefer.
Example (2) is done exactly the same y

In (3), there’s no equation.