Which is the number that, multiplied by 3, then increased by three-fourths of the product, divided by 7, diminished by one-third of the quotient, multiplied by itself, diminished by 52, the square root found, addition of 8, division by 10, gives the number 2 ?
Let x = number.
Mult. by 3 : 3x
Increased by 3/4(3x) : 3x + (9/4)x = (21/4)x
Divided by 7: (21/4)x * (1/7) = (3/4)x
Dim. by 1/3 of quot. : (3/4)x - (1/3)(3/4)x = (1/2)x
Mult. by itself: (1/2)x * (1/2)x = (1/4)x^2
Dimin. by 52 : (1/4)x^2 - 52
Square root: SQRT[(1/4)x^2 - 52]
Increased by 8: SQRT[(1/4)x^2 - 52] + 8
Divided by 10: {SQRT[(1/4)x^2 - 52] +8 }/10
So, we have {SQRT[(1/4)x^2 - 52] +8 }
------------------------- = 2
10
Hence, {SQRT[(1/4)x^2 - 52] = 12
and squaring both sides gives:
(1/4)x^2 - 52 = 144 ..>
(1/4)x^2 = 196 ..>
x^2 = 784 ..>
x = 28 or x = -28
Algebraically speaking .....
