Question

Joshua is a year older than his sister McKay. The product of their ages is 33 more than 7 times Joshua’s age. How old are Joshua and McKay?

Answer 1

Let the age of Joshua be x

And, the age of McKay be y

According to Question,

$\rm x= y + 1 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \rm \cdots(1) \\ \\ \rm{}xy = 33 + 7x\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \rm \cdots(2)$

Putting Equation 1 in Equation 2,

$\rm{}(y + 1)y = 33 + 7(y + 1) \\ \\ \to \tt{ {y}^{2} + y } = 33 + 7y + 7 \\ \to \tt {y}^{2} + y - 7y = 33 + 7 \\ \to \tt {y}^{2} - 6y = 40 \\ \to \tt {y}^{2} - 6y - 40 = 0 \\ \to \tt{y}^{2} - 10y + 4y - 40 = 0 \\ \to \tt{}y(y - 10) + 4(y - 10) = 0 \\ \to \tt (y - 10)(y + 4) = 0 \\ \therefore \bf \: y =10 \: \: \: or \: \: - 4$

Since,

• y is the value of age of McKay, so it can't be negative, thus, only positive value is considered.

So,

• Value of y is 10

Now, Putting value of y in Equation 1,

$\bf{}x = y + 1 \\ \to \tt x = 10 + 1 \\ \to \tt x = 11$

Hence,

• Age of Joshua (x) = 11 years
• Age of McKay (y) = 10 years