Stacey throws two dice. The difference in face value between the two dice is three, and the product of the numbers thrown is equal to twice the sum of the two numbers. Show and solve an alebraic equation that finds the numbers Stacey threw.
Let a be the larger number.
a - b = 3
a*b = 2(a + b)
a*(a-3) = 2a + 2(a-3)
a^2 -3a = 2a + 2a -6
a^2 -7a +6 = 0
(a - 6)(a -1) = 0
a = 6 or 1
Forget the a = 1 solution, because b cannot be negative
a = 6 and b = 3
To solve this problem, let's first assign variables to the numbers Stacey threw. Let's say the first number is x and the second number is y. We're given two conditions:
1. The difference in face value between the two dice is three. This can be represented as |x - y| = 3.
2. The product of the numbers thrown is equal to twice the sum of the two numbers. This can be expressed as x * y = 2(x + y).
From the first condition, we know that |x - y| = 3. This means that either (x - y) = 3 or (y - x) = 3.
Case 1: x - y = 3
In this case, we can rewrite the second condition using x - y = 3 as follows:
x * y = 2(x + y)
x * y = 2x + 2y
We can rearrange this equation to get quadratic form and solve for x:
x * y - 2x - 2y = 0
xy - 2x - 2y = 0
By factoring, we can rewrite this equation as:
y(x - 2) - 2(x - 2) = 0
(y - 2)(x - 2) = 4
From here, we can see that there are multiple solutions for x and y that satisfy this equation.
Case 2: y - x = 3
Similarly, if we start with y - x = 3 and follow the same steps, we will end up with the equation:
(y - 2)(x - 2) = -4
Again, there are multiple solutions for x and y that satisfy this equation.
Therefore, there are multiple pairs of numbers Stacey could have thrown that fulfill the given conditions.