Setting up a linear equations for a word problem

Question

If Xavier were 2 years younger, then five years ago, he would have been a third of Yasmine's age. In three years time, Yasmine will be 8 years older than twice Xavier's age. How old is Xavier?

Answer 1

I got something a little different. The first part of the question seems to focus on Xavier's age 5 years ago, not Yasmine's (at least from the way I interpreted it), so I read it as:

let x = xavier;

x - 2 - 5 = (1/3)y thus;

x - 7 = (1/3)y

x = (1/3)y + 7

To get 'y' in slope form (no fractional), multiplied each side by 3 and moved y to the left side of the equation:

= -y = -3x + 21 *multiply whole equation by -1 so that leading coefficient is non-negative

= y = 3x - 21

Next, focusing on Yasmine, the question states that 'in three years time, Yasmine will be 8 years older than twice Xavier's age.'

So, we set up an equation for Yasmine, which we can denote with y:

y + 3 = 2x + 8;

y = 2x + 5

Since these equations are related and both have to be true, we can set them up as a system of equations:

{y = 3x - 21}

{y= 2x + 5}

*multiply top equation by -1 to cancel out the y's (essentially find x)

-1{y = 3x - 21}

{y= 2x + 5}

-y + y = -3x + 21 + 2x + 5;

0 = -3x + 21 + 2x + 5 --> combine terms;

0 = -x + 26 add x to left side;

x = 26

Use second equation to find y (y = 2x+5):

y = 2(26) + 5

y = 57

test it:

X = 26

1) If Xavier were 2 years younger (24), then 5 years ago (19), he would have been 1/3 of Yasmine's age (which I interpreted to mean current age):

19 * 3 = 57 (57 * 1/3 = 19)

Y = 57

2) In three years time (y =60), Yasmine will be 8 years older than twice Xavier's age (X = 52):

60 - 52 = 8

Answer 2

The solution is a bit long, but I wanted to go over every non-obvious step:

1. Let x be Xavier's age as of today and y be Yasmine's age, also as of today.
2. Note that since we have two variables, we'll try to come up with two equations to solve this problem.
3. If Xavier were 2 years younger, he would be (x-2) today
4. Five years ago, Xavier would be (x-5) years old. But we want to know how old the younger Xavier would have been 5 years ago. So we subtract 5 from (x-2) (and not from x) to get (x-2)-5 = x-7
5. Five years ago, Yasmine would have been y-5 years old
6. I prefer not to work with fractions as they are a bit more prone to errors. So we'll restate younger Xavier being a third of Yasmine's age as Yasmine being 3 times (the younger) Xavier's age
7. So we get: 3 (x-7) = y - 5. This is one equation.
8. In 3 years, Xavier would be (x+3) years old. You've have to be careful not to do (x-2)+3 as we are talking about Xavier, not his younger avatar.
9. In 3 years, Yasmine would be (y+3) years old
10. Twice Xavier's age is 2(x+3) so 8 more than twice Xavier's age is 2(x+3)+8
11. So we get: 2(x+3)+8 = y + 3. This is the second equation.

There are a few ways to solve these 2 equations (also called simultaneous equations).

We'll use substitution:

We note from the first equation (line 7 above) that:

y - 5 = 3 (x-7)

By adding 5 to both sides we get:

y = 3x - 21 + 5 = 3x - 16

That's the first step in substitution (writing one of the variables in terms of the other)

For the second step, replace every y in equation 2 (line 11 above) with 3x - 16 to get:

2(x+3)+8 = 3x - 16 +3

Now we have an equation in a single variable! :)

2x + 14 = 3x - 13

Grouping all x terms to the right side the and numbers to the left side (if we did the opposite, we'd get a -x on the left, which I don't like as much), we get:

14 + 13 = 3x - 2x = x,

So x = 27

Then we solve for y:

Pick either equation (line 7 or line 11). For demonstration, we'll pick the one from line 7 (you should try it out with the one from line 11 to make sure that works too)

Equation: 3(x-7) = y- 5

But x = 27

So 3(27-7) = y-5

60 = y - 5

Or y = 65

Solution: Xavier is 27 years old and Yasmine is 65 years old.

To verify:

Xavier 2 years younger (25), but 5 years ago, would have been 20

Yasmine, 5 years ago, was 60

Indeed 60 = 3 * 20

In 3 years, Xavier would be 30, and Yasmine would be 68. Two times Xavier's age (60) plus 8 is indeed 68.

The takeaways:

By breaking the problem into tiny steps, we are able to solve even a problem that might have sounded confusing at first.

Even after many years at this, I still made and error (or two) when setting up the equations! Checking the answer to make sure that it works is an important step to solving a problem (correctly!) :)

With practice, you will be able to combine several of the steps leading to a quicker solution. But when in doubt, slow down, break up the problem, and proceed carefully.

You've got this!

Answer 3

Let X be Xavier's current age and Y be Yasmine's current age.

First condition: "If Xavier were 2 years younger, then five years ago, he would have been a third of Yasmine's age."

1. Five years ago, Xavier’s age was X − 2 − 5 = X − 7 [we subtract an additional 2 because of the "if Xavier were 2 years younger" phrase]
2. Five years ago, Yasmine’s age was Y − 5.
3. The condition says X − 7 = 1/3 * (Y − 5)
4. So, the first equation simplified is Y = 3X - 16

Second condition: "In three years, Yasmine will be 8 years older than twice Xavier’s age."

1. In three years, Xavier’s age will be X + 3, and Yasmine’s age will be Y + 3.
2. The condition says Y + 3 = 2(X + 3) + 8
3. So, the second equation simplified is Y = 2X + 11

Set the first and second equations equal to each other since they're both stating what Y is equal to:

3X - 16 = 2X + 11

X = 27

Solve for Y by plugging X = 27 back into either the first or second equation.

Y = 2(27) + 11

Y = 65

Xavier is 27 years old; Yasmine is 65 years old.

Hope this was helpful.