4= square root of (-6-2x) plus square root of (31-3x)

     Original equation 
     4 = √-6-2x+√31-3x

     Isolate
     -√-6-2x = -4+√31-3x

      Tidy up 
     √-6-2x = 4-√31-3x

     Raise both sides to the second power
     (√-6-2x)2 = (4-√31-3x)2

     After squaring 
     -6-2x = 31-3x+16-8√31-3x

     Current equation 
     -6-2x = 31-3x+16-8√31-3x

     Isolate radical on the left hand side
     8√31-3x = 6+2x+31-3x+16

      Tidy up 
     8√31-3x = 53-x

     Raise both sides to the second power
     (8√31-3x)2 = (53-x)2

     After squaring 
     1984-192x = x2-106x+2809

     Rearranged equation
     x2  + 86x  + 825 = 0

     This equation has two rational roots:
       {x1, x2}={-11, -75}
 

     Original equation, root isolated, after tidy up
     √-6-2x = 4-√31-3x

     Plug in  -11 for  x 
      √-6-2•(-11) = 4-√31-3•(-11)

      Simplify
      √16 = -4
      Solution does not check 
      4 ≠ -4 

     Original equation, root isolated, after tidy up
     √-6-2x = 4-√31-3x

     Plug in  -75 for  x 
      √-6-2•(-75) = 4-√31-3•(-75)

      Simplify
      √144 = -12
      Solution does not check 
      12 ≠ -12 

Hopefully this helped You

sine

step-by-step explanation:

option c) graph of line going through 1, 4 and 3, 10

step-by-step explanation:

let y equal to margo's coin collection

let x equal to number of years

at the begining x=0, y=1 → point=(x,y)→point=(0,1)

she adds 3 coins per year, then y=1+3x

at x=1, y=1+3(1)=1+3→y=4→point=(x,y)→point=(1,4)

at x=3, y=1+3(3)=1+9→y=10→point=(x,y)→point=(3,10)

the graph of line going through (1, 4) and (3, 10), then the option c is the correct.