Question 62523
 1)We have 2 equations with us
            3x-2y=3 ---------(1)
            x+4y=29 ---------(2)

 From equation (2) we first find what is x
 subtracting 4y on both sides
==> x+4y-4y=29-4y
==> x=29-4y -----(3)
 We substitute this value of x in equation (1)
we get,  3(29-4y)-2y=3
      ==> 87-12y-2y=3
      Subtracting 87 on both sides we get,
      87-87-12y-2y=3-87
  ==> -12y-2y=-84 
  ==> -14y=-84
 Now dividing both sides by -14,
   (-14/-14)y=(-84/-14)
  ==> y=+6
  From (3) we substitute the value oy y=6 to find x
==>  x=29-4y
==>  x=29-(4*6)
  ==> x=29-24
==> x=+5
The values of x and y are +5 and +6.


2) We have 2 equations with us
            4x-y=12---------(1)
            3x-2y=14 ---------(2)

 From equation (1) we find what y is
 subtracting 4x on both sides, we get
   -4x+4x-y=12-4x
==> -y=12-4x
Dividing by -1 on both sides we get
 -y/-1=(12-4x)/-1
==> y=-12+4x ----(3)
Substituting this value of y in equation (2)

     3x-2(-12+4x)=14
==>  3x+24-8x=14
==>  -5x+24=14
subtracting -24 on both sides, we get
    
     -24+24-5x=14-24
  ==> -5x=-10
Dividing on both sides by -5,
 -5x/-5=-10/-5
==>  x=2
Substituting this value of x in equation (3)

  ==> y=-12+4x
 ==> x=-12+4*2
==> x=-12+8
  ==>x=-4
The values of x and y are 2 and -4.


3) 2x+5=24 where 2x is to power 3

  2x^3+5=24
 subtracting 5 on both sides 
 2x^3+5-5=24-5
==> 2x^3=19
 dividing by 2 on both sides we get,
  (2x^3)/2=19/2
==> x^3=9.5
 Finding the cube root of 9.5 we get the value of x
==> x=2.1