To solve this problem, we have to build two equations with the given information. Using x and y to represent the two numbers:
• Equation 1
[tex]x\times y=592[/tex]• Equation 2
[tex]x+y=53[/tex]Now that we have to equations, we have to isolate one variable from one equation and replace it in the other.
[tex]x=53-y[/tex]Then, we will replace this value of x in Equation 1:
[tex](53-y)\cdot y=592[/tex]Solving for y we get:
[tex]53y-y^2=592[/tex][tex]-y^2+53y-592=0[/tex]As we got this expression, we will have to use the General Quadratic Formula. With the help of a calculator, we get both values:
[tex]y_1=16[/tex][tex]y_2=37[/tex]Finally, we have to replace these values in Equation 1 to evaluate which meets the condition:
[tex]x_1=\frac{592}{y_1}[/tex][tex]x_1=\frac{592}{16}=37[/tex][tex]x_2=\frac{592}{y_2}[/tex][tex]x_2=\frac{592}{37_{}}=16[/tex]We have to evaluate the values in each equation:
[tex]\begin{gathered} 37+16=53 \\ 53=53 \end{gathered}[/tex][tex]37\cdot16=592[/tex]The first numbers meet the condition.
Answer: 37 and 16
