Question

The difference between the squares of two numbers is 20. three times the square of the first number increased by the square of the second number is 124. find the numbers

Answer 1

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 $x^{2} -y^2=20$.  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 $3x^2+y^2=124$.  Let's solve the first equation for x^2: $x^2=y^2+20$ and sub that value for x^2 into the second equation.  $3(y^2+20)+y^2=124$ and $3y^2+60+y^2=124$.  Subtract 60 from both sides and combine the y^2 terms to get $4y^2=64$.  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.  $x^2-16=20$, x^2 = 4, so x = 2.  Your solutions are x = 2 and y = 4.  There you go!