# Kaylin throws two dice. the difference in face value between the two dice is three, and the...

## Question:

Kaylin 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 two numbers.

Write and solve an algebraic equation that finds the numbers Kaylin threw.

## Domain:

The domain is the set of numbers a variable can have. There are many ways to write a domain such as interval notation and set notation. In many word problems, the domain is not stated but can be found using a sensible argument of the situation. For example, if the word problem gives that {eq}x {/eq} is the length of a piece of paper then {eq}x {/eq} cannot be a negative number or 0 since length cannot be negative or 0.

Let {eq}x {/eq} be the larger value of the dice rolled and {eq}y {/eq} be the smaller value. Note that both {eq}x {/eq} and {eq}y {/eq} can only take on the values 1, 2, 3, 4, 5, or 6. Translate each part of the sentence to an algebraic expression and write a system of equations.

"the difference in face values between the two dice is three": {eq}x - y = 3 {/eq}

"the product of the numbers thrown is equal to twice the sum of the two numbers": {eq}xy = 2(x+y) {/eq}

Together, these equations form a system of equations given by

{eq}\begin{cases} x - y = 3\\ xy = 2(x+y) \end{cases} {/eq}

To solve, use the substitution method. That is, first solve for {eq}x {/eq} in the first equation.

{eq}x - y = 3 ~~ \implies ~~ x = 3+y {/eq}

Then substitute {eq}3+y {/eq} for {eq}x {/eq} in the second equation, simplify, and solve for {eq}y {/eq}.

{eq}\begin{aligned} xy = 2(x+y) ~~ \implies ~~ (3+y)y & = 2((3+y)+y)\\ 3y + y^2 & = 2(3+2y)\\ 3y + y^2 & = 6+4y\\ 3y + y^2 - 6 - 4y & = 0\\ y^2 - y - 6 & = 0\\ (y - 3)(y+2) &= 0 ~~ \implies ~~ y - 3 = 0 ~ ~\text{or}~~ y+2 = 0~~ \implies ~~ y = 3 ~ ~\text{or}~~ y=-2 \end{aligned} {/eq}

Note that the value for {eq}y {/eq} cannot be negative because the variable represents the face value of the dice. Therefore the only value for {eq}y {/eq} is 3.

To find the associated value of {eq}x {/eq}, substitute 3 in for {eq}y {/eq} into either equation and solve for {eq}x {/eq}.

For {eq}y = 3 {/eq}: {eq}x = 3 + (3) = 6 {/eq} 