We can set up a system of equations like this, where x is the first number and y is the second:
[tex]\left \{ {{y= \sqrt{x} -4} \atop {x+y=38}} \right[/tex]
Since we know what y is equal to, we can plug that into the second equation for y to solve for x:
[tex]\\ x+ \sqrt{x} -4=38 \\ x+ \sqrt{x} -42 =0 \\ \sqrt{x} = 42-x \\ (\sqrt{x})^{2} = (42-x)^{2} \\ x = 1764 - 84x + x^{2} \\ x^{2} -85x+1764 = 0 \\ (x-36)(x-49)=0 \\ x = 36, 49[/tex]
x can equal either 36 or 49. If x equals 36:
[tex]36+y=38 \\ y=2[/tex]
If x equals 49:
[tex]49+y = 38 \\ y = -11[/tex]
Although both are solutions, x=36 and y=2 or x=49 and y=-11, I would use x=36 and y=2 because plugging the 49 and -11 in causes a predicament:
[tex]-11 = \sqrt{49}-4 \\ -11=7-4 \\ -11 \neq 3[/tex]
*Note that the negative solution of √49 does work
Plugging in 36 and 2 turns out fine:
[tex]2 = \sqrt{36} -4 \\
2 = 6-4 \\
2 = 2[/tex]
