Drew is twice as old as his sister joy.The quotient of their ages five years ago is seven less than Joy's present age. Find out how old they are at present.

present age of Joy --- x

present age of Drew --- 2x

five years ago:
Joy = x-5
Drew = 2x-5

(x-5)(2x-5) = x-7
2x^2 -15x + 25 = x-7
2x^2 - 16x + 32 = 0
x^2 - 8x + 16 = 0
(x-4)^2 = 0
x = 4

Joy is now 4, and Drew is 8

check:
5 years ago:
Joy = -1 ??
Drew = 3
product of their ages = -3
7 less than Joy's present age = 4-7 = -3

notice that "mathematically" the answer is correct, but in reality the question is bogus.

There are 2 answers though. It could also be 10 and 20. I saw this when I solved it by making the quotient of their ages 5 years ago be equal to Joy's age-7 in one formula.

(2x-5)/(x-5)=x-7
multiply both sides by x-5, move everything to one side, and factor the quadratic you get (x-10)(x-4), and 2 answers to the problem are revealed.

Let's assume Drew's current age is D and Joy's current age is J.

According to the given information, we can form two equations:

1. Drew is twice as old as Joy:
D = 2J

2. The quotient of their ages five years ago is seven less than Joy's present age:
(D-5)/(J-5) = J - 7

Now, let's solve these equations step-by-step:

From equation 1, we have:
D = 2J

Substituting D = 2J in equation 2, we get:
(2J-5)/(J-5) = J - 7

Cross-multiplying, we get:
2J - 5 = (J - 5)(J - 7)

Expanding the right side:
2J - 5 = J^2 - 7J - 5J + 35

Simplifying and rearranging, we get:
J^2 - 14J + 40 = 0

We can factorize this quadratic equation as follows:
(J - 10)(J - 4) = 0

Setting each factor to zero, we have two possible solutions:
J - 10 = 0 or J - 4 = 0

So, either J = 10 or J = 4

If J = 10, then using equation 1, we find:
D = 2(10) = 20

If J = 4, then using equation 1, we find:
D = 2(4) = 8

Therefore, Drew is either 20 or 8 years old, and Joy is either 10 or 4 years old.

To solve this problem, we need to set up a system of equations based on the given information.

Let's assume that Drew's present age is represented by "D" and Joy's present age is represented by "J".

From the information provided, we can write the following equation:

1. Drew is twice as old as his sister Joy: D = 2J

Next, we can set up an equation based on the quotient of their ages five years ago. To do this, we need to determine Joy's age five years ago and Drew's age five years ago.

Joy's age five years ago would be J - 5, and Drew's age five years ago would be D - 5.

According to the given information, the quotient of their ages five years ago is (J - 5)/(D - 5). And this quotient is seven less than Joy's present age, which is J.

Therefore, we can write the equation:

2. (J - 5)/(D - 5) = J - 7

Now, we have a system of two equations (Equations 1 and 2), which we can solve simultaneously to find the ages of Drew and Joy.

Let's substitute the value of D from Equation 1 into Equation 2:

(J - 5)/(2J - 5 - 5) = J - 7

Simplifying, we have:

(J - 5)/(2J - 10) = J - 7

Cross-multiplying, we get:

J - 5 = (2J - 10)(J - 7)

Expanding and simplifying further, we have:

J - 5 = 2J^2 - 14J - 20J + 70

Combining like terms, we get:

J - 5 = 2J^2 - 34J + 70

Rearranging the equation, we have:

2J^2 - 35J + 75 = 0

Now, we can solve this quadratic equation using factoring, completing the square, or the quadratic formula. In this case, let's use the quadratic formula:

J = (-b ± √(b^2 - 4ac)) / (2a)

By substituting the values a = 2, b = -35, and c = 75 into the formula, we can calculate J.

J = (-(-35) ± √((-35)^2 - 4(2)(75))) / (2(2))

Simplifying further, we have:

J = (35 ± √(1225 - 600)) / 4

J = (35 ± √625) / 4

J = (35 ± 25) / 4

This gives us two possible values for J:

J₁ = (35 + 25) / 4 = 60 / 4 = 15

J₂ = (35 - 25) / 4 = 10 / 4 = 2.5 (not possible as it represents a fraction)

Since we cannot have a fractional or decimal age, we discard the value J₂ = 2.5.

Therefore, Joy's present age is J = 15.

Now, we can substitute this value back into Equation 1 to find Drew's age D:

D = 2J
D = 2(15)
D = 30

Hence, the present age of Drew is 30, and the present age of Joy is 15.