Question

I'm trying to figure out an algebraic verbal problem:
The four oldest people in Golden City have lived a total of 384 years put together. The difference in ages for the youngest and the second oldest is 14. The second youngest is 3 years older than the youngest. The oldest is 20 years older than the average of the second oldest and youngest. What are their ages?
So I'm letting a(oldest), b(second oldest), c(second youngest), and d(youngest) be the ages of the four people. According to the problem, it should be this:
A + B + C + D = 384.
B -D = 14
D + 3 = C
B+ D/2 + 20 = A

I don't really know where to start. Could I have some help?

Answer 1

Well if you set up the problem to be a system of equations with the age of each person being A, B, C, D in order from oldest to youngest, you will get the equations:
A+B+C+D = 384
A-B = 14
B-D=3
A-20=(B+D)/2 or A-.5B-.5D=20
Similar to what you have.
Now we can use matrices to solve for each variable.

Our variable matrix ([A]):
[1 1 1 1]
[1 -1 0 0]
[0 1 0 -1]
[1 -.5 0 -.5]
Our solution matrix([B]):
[384]
[14]
[3]
[20]

Then we just do \[[A]^-1 [B]\]
and we will get the result:
[0]
[-3920]
[-218.5]
[4090]
However, this solution isn't correct if we plug in any of the values into the algebraic equations we found, leading me to believe that the problem is written strangely or I can't read.

Answer 2

I'm not really understanding what you just did

Answer 3

That last equation should be:

(B+ D)/2 + 20 = B/2 + D/2 + 20 = A

You can't assume parentheses if they are not there. It will mess you up every time.

The easiest and quickest way to solve is Gaussian elimination. First, get all the variables on the left side of the equations. Then, pick an equation with a variable and add or subtract a suitable multiple of that equation to the other 3 equations so that the chosen variable drops out. For example, if you start with equation #2 and variable "B", you can subtract that from equation one and B will drop out of equation one. Do a similar operation to equation 3 and 4. Equations 1, 3, and 4 will be altered and equation 2 will still be the same . But now, equation 2 is "used" and so is variable "B". Pick another of the (altered) equations and another variable. Do 2 more times so that all variables are isolated.

Answer 4

So I have to use the equations to solve each other?

Answer 5

Yes, but don't be discouraged or overwhelmed, they get progressively easier very quickly. When you are about 2/3 of the way through, you will start to see the solution even before you get the final solution.

Answer 6

If you add the negative of equation #2 (both sides made negative) to equation #1, you will see "B" drop out of equation #1.

Answer 7

So when trying the example you gave me, I get
A + B + C = 384
B - D = 14
2 + D/2 + 20 = A
D + 3 = C
but somehow this doesnt seem right to me

Answer 8

You have to add left side to left side and right side to right side. And you work with adding a multiple of equation #2 to equations #1 and #4, all for the purpose of eliminating "B" from equation #1 and #4. Equation 3 has no "B" in it, so it is untouched in the first pass.

Answer 9

what do you mean, adding a multiple of equation #2?

Answer 10

A multiple of an equation is multiplying both sides of the equal sign by the same non-zero number so as to preserve the equality.

If you multiply equation #2 by negative 1/2 and add it to my correction of your equation #4, you will see "B" drop out of equation #4. At this point, you have an altered equation #1 and #4 and an original equation #2 and #3. However, equation #2 is "used" and so is "B", so it stops being a candidate for being added or subtracted from equations anymore. Now you have a new system and you pick another variable and equation an repeat the process.

Answer 11

and subtrating B from equation #4, (b/2 - b) would it just turn to 2?

Answer 12

sorry I'm just getting more confused as we go...

Answer 13

You have to work with coefficients of "B". Get them to be the same number but of opposite sign. Equation #2 has an implied "1" as a coefficient. You multiply the WHOLE of equation #2 by NEGATIVE 1/2. That means the whole equation. All terms and both sides. Now, you add the left side of equation #2 to the left side of equation #4 and the right side of equation #2 to the right side of equation #4. You will see the "B" TERM (coefficient AND "B") drop out of equation #4. When you are done with this first pass, you keep equation #2 as it is originally, for the time being, until you do another pass with another variable and another equation.

Once you start doing this and get rolling, it is super-easy, just a lot of steps, but it is really really easy.

Answer 14

I'm going to start doing the problem for you and using step-by-step as the example. I have already outlined the whole method, so if you ever want to be able to do these problems on your own in the future, I suggest you go back and read all of this. Otherwise, you'll be stuck relying on others for answers. Here goes:

Answer 15

A + B + C + D = 384
B - D = 14
- C + D = -3
-A + B/2+ D/2 = -20

That's the starting point, after getting all variables over to the left side and with the correction on your equation #4.

Answer 16

Multiply equation #2 by -1 to get -B + D = -14 and add it to equation #1:

A + C + 2D = 370
B - D = 14
- C + D = -3
-A + B/2+ D/2 = -20

Notice how we keep equation 2 in tis original form but use the multiple to do arithmetic with equation #1. Equation #1 is altered. That's the only change so far. We will next be treating equation #4 by adding a multiple of equation #2 to equation #4.

Answer 17

We multiply equation #2 by -1/2 to get -B/2+ D/2 = -7 and we add that to equation #4:

A + C + 2D = 370
B - D = 14
- C + D = -3
-A + D = -27

At this point, we have an altered equation #1 and #4. Equations #2 and #3 are original but equation #2 is "used" (and so is variable "B") because it was "used" in adding a multiple of itself to other equations. At this point, we pick another equation and variable to "use" like we just did with "B". Let's pick equation #4 and variable "A".

Answer 18

This step wil be quick because equation #1 is the only other equation with a variable "A". AND equation #4 ALREADY happens to have the same coefficient for "A" as equation #1 does , but as opposite sign like we want. So, we add equation #4 to equation #1:

C + 3D = 343
B - D = 14
- C + D = -3
-A + D = -27

Before we leave this step, because we will be leaving "A" forever, let's rewrrite equation #4 so that "A" is left positive.

C + 3D = 343
B - D = 14
- C + D = -3
A - D = 27

So, at this time we now have TWO "used" equations (2 and 4) and 2 "finished" variables (Aand B). It's already looking simpler. Let's pick equation #3 and "C".

Answer 19

@craftshark98 all done now you can come back.

Answer 20

We'll add equation #3 to #1 and then make equation #3 have a positive "C". I'm combining steps now so we move faster.

4D = 340
B - D = 14
C - D = 3
A - D = 27

We are almost done, but you can see the system getting progressively easier. It's almost trivial now. We have "used" A, B, and C and equations #2, #3, and #4. All that is left now is equation #1 and "D". Let's first rewite equation #1 by diving the equation by 4:

D = 85
B - D = 14
C - D = 3
A - D = 27

Now we just add equation #1 to equations #2, #3, and #4:

D = 85
B = 99
C = 88
A = 112

There's your newest system and the solution. Many steps but very very easy.