We need a system of equations to solve this. "Difference" is to subtract, and we are taking this difference of the 2 squared unknowns to be 20. That equation is [tex] x^{2} -y^2=20[/tex]. Our "first number is x^2, so 3 times that is 3x^2. "Increased by" is adding to that first number. What we are adding is the second number. The second equation is [tex]3x^2+y^2=124[/tex]. Let's solve the first equation for x^2: [tex]x^2=y^2+20[/tex] and sub that value for x^2 into the second equation. [tex]3(y^2+20)+y^2=124[/tex] and [tex]3y^2+60+y^2=124[/tex]. Subtract 60 from both sides and combine the y^2 terms to get [tex]4y^2=64[/tex]. Divide both sides by 4 to get y^2 = 16 and y = 4. Let's go back now and solve for x. We will use the fact that y^2 = 16 to solve for x^2 and then take the square root of it. [tex]x^2-16=20[/tex], x^2 = 4, so x = 2. Your solutions are x = 2 and y = 4. There you go!
