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

Let's define the two numbers as \( x \) and \( y \). According to the problem, we can outline 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:
\[
x = y + 3
\]

Now, we can substitute the second equation into the first to find solutions for \( x \) and \( y \).

Substituting \( x = y + 3 \) into \( xy = 10 \), we get:
\[
(y + 3)y = 10
\]
Expanding this:
\[
y^2 + 3y - 10 = 0
\]

Next, we can solve this quadratic equation using the quadratic formula:
\[
y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
where \( a = 1 \), \( b = 3 \), and \( c = -10 \).

Calculating the discriminant:
\[
b^2 - 4ac = 3^2 - 4 \cdot 1 \cdot (-10) = 9 + 40 = 49
\]
Now applying the quadratic formula:
\[
y = \frac{-3 \pm \sqrt{49}}{2 \cdot 1} = \frac{-3 \pm 7}{2}
\]

Calculating the two possible values for \( y \):
1. \( y = \frac{-3 + 7}{2} = \frac{4}{2} = 2 \)
2. \( y = \frac{-3 - 7}{2} = \frac{-10}{2} = -5 \)

Now, we can find \( x \) for both values of \( y \):

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

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

Thus, the pairs that satisfy the conditions are:
1. \( (5, 2) \)
2. \( (-2, -5) \)

To verify:
- For \( (5, 2) \):
- Product: \( 5 \times 2 = 10 \)
- The first number is 3 more: \( 5 = 2 + 3 \) (true).

- For \( (-2, -5) \):
- Product: \( -2 \times -5 = 10 \)
- The first number is 3 more: \( -2 = -5 + 3 \) (true).

Both pairs satisfy the system of equations.