Question

1) Two integers have a sum of 47 and a difference of 23. Find the product of the numbers.

Answer 1

Given:

Two intergers have a sum of 47 and a difference of 23.

Let's find the product of the two numbers.

Let x and y represent the numbers.

We have:

Two integers have a sum of 47: x + y = 47

Two integers have a difference of 23: x - y = 23

We gave the system of equations:

x + y = 47.......................equation 1

x - y = 23.......................equation 2

Let's solve the system simultaneously using substitution method.

Rewrite equation 1 for x:

x = 47 - y

Substitute (47 - y) for x in equation 2:

(47 - y) - y = 23

47 - y - y = 23

47 - 2y = 23

Subtract 47 from both sides:

47 - 47 - 2y = 23 - 47

-2y = -24

Divide both sides of the equation by -2:

`\begin{gathered} (-2y)/(-2)=(-24)/(-2) \\ \\ y=12 \end{gathered}`

Now, substitute 12 for y in either of the equations.

Let's take equation 1.

x + y = 47

x + 12 = 47

Subtract 12 from both sides:

x + 12 - 12 = 47 - 12

x = 35

Therefore, we have:

x = 35, y = 12

The numbers are 35 and 12.

To find the product of the numbers, let's multiply the numbers:

35 x 12 = 420

Therefore, the product of the numbers is 420.

ANSWER:

420