Help with solving for x?

Qu 1

Will have to GUESS that you mean :-

1/x = 1/a + 1/b

1/x = ( b + a ) / (ab)

x = ab / ( a + b )

Qu 2

y ( 5x - 4 ) = 3 - 2x

y ( 5x) - 4y = 3 - 2x

(5y + 2 ) x = 3 + 4y

x = ( 3 + 4y ) / ( 5y + 2 )

1) x = a = b (reciprocals)

Doesn't seem like sense to me though so I'm going to assume the second equals is actually a plus.

1/x = 1/a + 1/b.

1/x = (a + b) / ba

Invert:

x = ba / (a + b)

2) y = 3 - 2x

-------------------

5x - 4

Cross-multiplication:

y(5x - 4) = 3 - 2x.

5xy - 4y = 3 - 2x.

5xy + 2x = 3 + 4y.

x (5y + 2) = 3 + 4y.

x = (3 + 4y)/(5y + 2)

Solve for x:

1/x = 1/a+1/b

= b/b* 1/a + a/a * 1/b

1/x = ( b+a) / ab

INVERT BOTH SIDES

so x = ab / ( a+b)

Solve for x:

(3-2x) / (5x-4) = y

3 -2x = y(5x -4)

3 - 2x = 5xy -4y

-2x -5xy = -4y -3

2x +5xy = 4y+3

x(2 +5y) = (4y+3)

x = (4y+3) / ( 2 + 5y)

Solve for x:

1/x = 1/a=1/b

x = a = b

(3-2x) / (5x-4) = y

3 -2x = y(5x -4)

3 - 2x = 5xy -4y

-2x -5xy = -4y -3

2x +5xy = 4y+3

x(2 +5y) = (4y+3)

x = (4y+3) / ( 2 + 5y)

x = a = b

If you meant

1/x = 1/a + 1/b

Multiply both sides by xab to get rid of the fractions

ab = xb + xa

ab = x(a+b)

x = ab/(a+b)

y=(3-2x)/(5x-4)

y(5x-4) = 3-2x

5xy - 4y = 3-2x

5xy +2x = 4y+3

x(5y+2) = 4y+3

x = (4y+3)/(5y+2)

1/x=1/a gives x=a

similarly

x=b

y=3-2x/5x-4

x=4y+3/2-5y

1/x = 1/a = 1/b

1/x = 1/a

x = a

1/x = 1/b

x = b

therefore x = a = b