Question

The difference between the squares of two numbers is 24. Three times the square of the first number increased by the square of the second number is 76. Find the numbers

Answer 1

Answer:

Step-by-step explanation:

So the first step is to simply set up the problem based on what we are given. So here have two numbers, we are going to call the first number x and the second one y. With that now addressed, we can now proceed with the setup.

So the difference between the squares of the numbers is 24. So we have:

$x^{2} -y^{2} = 24$

Then it says that three times the square of the first number (which we said was x) increased by the square of the second number is 76. So:

$3x^{2} + y^{2} = 76$

Now we can see that this is simply a system of equations and we can use elimination to solve this! We even have the setup already as the coefficients in front of our y are opposite in sign and are equal. So:

$x^{2} -y^{2} = 24\\3x^{2} +y^{2} = 76$

We can cancel our y squared terms out and that leaves, when we add the equations together:

$4x^{2} = 100$

We can then solve for x by diving by four and taking the square root of the result.

$x^{2} = 25$

Therefore, x = ±5

We have both negative and positive answers because if we squared -5 or +5 they would both give us 25. So we cant rule a negative answer out yet.

So now we can plug in x = -5 or +5 to either equation to solve for y as so:

$5^{2} - y^{2} = 24\\25 - y^{2} = 24\\-y^{2} = -1 \\y^{2} = 1$

So y = ±1

In this case both negative and positive versions of our answer work (you can also double check), so we are left with:

x = ±5 and y = ±1