Question

Write a system of equations where the product of two numbers is 10 and the first number is 3 more than the second.
Can you determine the solution? Guess and check.

Answer 1

write short without / '[;/

Answer 2

Let's denote the two numbers as \( x \) and \( y \). Based on the problem you provided, we can set up the following system of equations:

1. The product of the two numbers is 10:
\[
xy = 10
\]

2. The first number is 3 more than the second number:
\[
x = y + 3
\]

Now, we can substitute the second equation into the first equation to solve for \( y \):

Substituting \( x = y + 3 \) into \( xy = 10 \):

\[
(y + 3)y = 10
\]

Expanding this gives:

\[
y^2 + 3y = 10
\]

Rearranging the equation gives:

\[
y^2 + 3y - 10 = 0
\]

Next, we can factor the quadratic equation:

To factor \( y^2 + 3y - 10 \), we look for two numbers that multiply to \(-10\) and add up to \(3\). The numbers \(5\) and \(-2\) work because:

\[
5 \cdot (-2) = -10 \quad \text{and} \quad 5 + (-2) = 3
\]

So we can factor the equation as:

\[
(y + 5)(y - 2) = 0
\]

Setting each factor equal to zero gives us:

1. \( y + 5 = 0 \) ⟹ \( y = -5 \)
2. \( y - 2 = 0 \) ⟹ \( y = 2 \)

Now we can find the corresponding \( x \) values using \( x = y + 3 \):

1. If \( y = -5 \):
\[
x = -5 + 3 = -2
\]
So one solution is \( (x, y) = (-2, -5) \).

2. If \( y = 2 \):
\[
x = 2 + 3 = 5
\]
So another solution is \( (x, y) = (5, 2) \).

**Solutions:**
1. \( (x, y) = (5, 2) \)
2. \( (x, y) = (-2, -5) \)

Both pairs satisfy the original conditions of the problem.