CPhill

Yep....that's correct....!!!

xover2-2x over3=4x over 5 +2

x/2 - (2x)/3 = (4x)/5 + 2      multiply through by the common denominator of 2,3 and 5  = 30

15x - 20x  = 24x  + 60

-5x = 24x + 60    subtract 24x from both sides

-29x  = 60    divide by -29 on both sides

x = -60 / 29

1/4(5x-1)=3-2x    multiply both sides by 4

5x - 1  = 12 - 8x      add 1, 8x to both sides......

13x  = 13      divide by 13 on both sides

x = 1

I assume this might be......

log₂(x^2 + 2x) = 3     if so, we can write

2^3  = x^2 + 2x    →  8 = x^2 + 2x        rearrange

x^2 + 2x - 8 = 0      factor

(x + 4) (x - 2)  = 0      and setting each factor to 0, we have that x = -4   and x = 2

We must reject x = 2, because it leads to taking a log of 0  in the original problem, and this is undefined

x = -4  is the only solution

$15 / 100  =  15 cents

By a trig identity, we have that.......

sin Θ  = cos (90 - Θ)     so

sin 30°  = cos (90 - 30)°  =  cos 60°

I think geno made a slight error here.......taking it from this point, we have

x^2 + x = 5x + 10 - 5

x^2 + x = 5x + 5     subtract 5x, 5 from both sides

x^2 - 4x - 5  = 0     factor

(x - 5) (x + 1) = 0    and setting each factor to 0 we have that x = 5  or x = -1

Reject  x= -1

So x = 5 ......x + 1 = 6 .... and x + 2 = 7

And......as a check...... (x)(x + 1)  + 5 = 5(x + 3)  →  (5)(6) + 5  = 5(7)  

And 35 = 35

(1/2)log₅ 16 - 3log₅ x + 4 log₅ y    and by a property of logs, we have

log₅ 16^1/2 - log₅ x³ +  log₅ y⁴      =

log₅ 4 - log₅ x³ +  log₅ y⁴      rearrange  as

log₅ 4 + log₅ y⁴  -  log₅ x³    and by another log property, we can write

log₅  [ ( 4y⁴ ) /  x³ ]

Cross-multiply .......this gives........

(x - y) dy = (x + y) dx      integrate both sides

∫ ( x- y) dy  = ∫ (x + y)  dx   

[On the left side, x is considered to be a constant......same with y on the right side]   

xy  - y^2/2  =  x^2/2 + xy + C      subtract xy from each side

-y^2/2 =  x^2/2 + C₁

(And C1 is just a constant )

y^2  = 2C₁ - x^2   let 2C₁  = C₂ 

y = ±√[C₂ - x^2)

I believe this might be correct......Alan....will you check it????

There's nothing to "solve"......we have have to have an "equal sign" somewhere....