Question

A positive real number is 2 more than another. If the sum of the squares of the two numbers is 42, find the numbers.

Answer 1

Answer:

Step-by-step explanation:

From the first sentence x (the number) is 2 more than another number (y). So, x+2=y. The sum of the squares of the two numbers is 42, meaning that $\\ (x+2)^{2}$+$y^{2}$=42. First, we square root the ENTIRE equation to get rid of the square. If you do one thing to one side, you have to do it to the other. After this, we have x+2+y=$\sqrt{42}$. Now, we subtract 2 from the other side to isolate the variable. You can isolate either one but normally I go for x. Now we have x+y=$\sqrt{42}$+2. Now, we want to get y to the other side so we subtract it from the left side and the right side which gives us x=-y+$\sqrt{42}$+2. now we know what X is equal to. Now we plug in x for $\\ (x+2)^{2}$+$y^{2}$=42, which gives us          $(-y+\sqrt42}+2+2)^{2}$+$y^{2}$=42. Now we only have one variable which is what we want. Next, we square root the entire thing to get rid of the squares, which gives us -y+$\sqrt{42}$+4+y=$\sqrt{42}$. I got lost in my work. I must have done something wrong I cant find out. If anyone wants to pick up where I left off go ahead but there's a timer and I have one minute left. I cant finish the problem in a minute. I'm sorry and hopefully I lead you somewhere.