230k views
4 votes
1) Two integers have a sum of 47 and a difference of 23. Find the product of the numbers.

1 Answer

5 votes

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

User Ashterothi
by
8.2k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories