Question

26. Positive integers x, y, satisfy xy + z = 160. Compute the smallest possible value of
x+yz​

Answer 1

Answer:

The smallest possible value of x + yz is 50.

Step-by-step explanation:

We know that, xy +z = 160

According to the Question, x, y, z are positive integer. So, x, y, and z can not be zero.  

Let, ( x*y) = 158 [ x = 158 and y = 1 ] and z = 2  

xy + z = 160 ……….(i)

x+ yz  …….. (ii)

Now, we have to put the value of x, y, and z in equation (i)  

And we get, (158 * 1) + 2 = 160

x + yz = 158 + (1*2) = 160

79*2 + 2 = 160

x + yz = 79 + (2*2) = 83

156 + 4 = 160

156*1 + 4 = 160

x + yz = 156 + (1*4) = 160

78*2 + 4 = 160

x + yz = 78 + (2*4) = 86

52*3 + 4 = 160

x + yz =52 + (3*4) = 64

39*4 + 4 = 160

x + yz =39 + (4*4) = 55

26*6 + 4 = 160

x + yz =26 + (6*4) = 50

13*12 + 4 = 160

x + yz =13 + (12*4) = 61

The smallest possible value of x + yz is 50