Can you help me on an Algebra 1 math problem?

Question

How do you solve a promblem where it says: The second of there #s is 8 more than the first, and the third # is 3 less than 3 times the first. If the third # is 15 more than the second, find the three #s.

Answer 1

The second of there #s is 8 more than the first, and the third # is 3 less than 3 times the first. If the third # is 15 more than the second, find the three #s.

This is an excellent question and something you will trip over several times in algebra I, but you will master it at some time, given you have a good mentor and you pay attention to wording. Let's do this one by one. 

Second # (call this Y)

Step 1: 8 more than first # (call this X)

Step 2: Think of Y as 8 more jelly beans than X. You know that this means addition.

           Y = 8 + X 

Third # (call this Z)

Step 1: Fifteen more than the second # (call this Y)

Step 2: Think of this as 15 more jelly beans than Y (which we know is 8 more jellybeans than X)

           Z = 15 + Y = 15 + (8 + X)

           Z               = 23 + X

Third # (call this Z)

Step 1: 3 less than 3 times the first (call this X)

Step 2: 3 times the first is a number. Times says to multiply so the number is 3X. Now let's read the rest of the phrase. "3 less than 3X". Less than implies subtraction. So we have 3X jellybeans and we take away 3 from that amount. This means that 3X comes first cause we're taking away from it and so the third # is                                

            Z = 3X - 3 

So we have our definitions for Y and Z in terms of X. Note that we have two equations for Z because we were given information on Z based on the second number and the first number but because the second number is defined in terms of the first number X as well, we had converted that equation in terms of X as well. In order to solve for X, we substitute X + 23 in for Z in the equation Z = 3X -3. 

           X + 23 = 3X - 3

                 26 = 2X

                 13 = X

Now we have the first number X solved. Since Z and Y are both expressed in terms of X, so here is the easy part. 

For the third #, you can choose any of the two equations to plug in X = 13. You'll receive same answer.

Z = 3(13) - 3 = 36

Y = 8 + 13 = 21

 

First number = 13

Second number = 21

Third number = 36

Answer 2

The first step to solving a word problem is defining the variables.  For this problem we will define the variables as:

x = First number

y = Second number

z = Third number

Since the second numbers 8 more than the first number:

y = x + 8

Since the third number is 3 less than 3 times the first:

z = 3x -3

Since the third number is 15 more than the second:

z = y + 15

Now that I have three equations, I can solve with either substitution or elimination method.  I am going to use substituion method.

z = y + 15                            Equation 3

z = (x + 8) + 15                    Substitution of the value of y from equation 1 into equation 3

z = x + 23                            Simplify (8 + 15 = 23)

3x - 3 = x + 23                     Substitution of the value of z from equation 2 into equation 3

3x - 3 -x = 23                       Subtract x from each side

3x - x = 23 + 3                     Add 3 to each side

2x = 26                                Simplify (3x - x = 2x, 23 + 3 = 26)

x = 26/2                               Divide each side by 2

x = 13                                  Simplify (26/2 = 13)

y = x + 8                              Equation 1

y = 13 + 8                            Substitution

y = 21                                Simplify (13 + 8 = 21)

z = 3x - 3                             Equation 2

z = 3(13) - 3                        Substitution

z = 39 - 3                            Simplify (3 X 13 = 39)

z = 36                                 Simplify (39 - 3 = 36)

The three numbers are 13, 21, and 36.