# Lesson Tutor : Algebra : Grade 9 : Using Equations to Solve Puzzles

Basic Algebra – Lesson 7
Using Equations to Solve Puzzles
by Elaine Ernst Schneider

Objective(s): By the end of this lesson the student will be able to:
set up an equation that “matches” in numbers what the puzzle expresses in words.

Pre Class Assignment:  Completion or review of Basic Algebra – Lesson 6

Resources/Equipment/Time Required:

Outline:
In the last lesson, you learned to balance an equation, making what is on one side of the equal sign equal to what is on the other side. In other words, the left member and the right member of the equation “balance.”

Examples of equations:

5 + 7 – 2 = 2(5) (the answer on the left side is 10 and the answer on the right side is 10)

100/25 = 347 – 343 (the answer is 4 on the left of the equal sign and 4 on the right also)

You also learned to isolate variables. Equations involving unknown variables are solved by balancing the left member and the right member. In the equation, y + 5 = 12, the proper procedure in solving for y is to “isolate” y. This means we want y to stand by itself on one side of the equal sign.

Now, keep in mind that we want every thing to “balance” on either side of the equal sign. This means whatever I do to the left member, I must do to the right member. So, to isolate y and solve the equation, I must “move” the 5 to the other side of the equal sign. To do this, I must make it zero on the left side of the equation by subtracting 5:

y + 5 = 12

y + 5 – 5 = 12 – 5

y + 0 = 12 – 5

y = 7

Equations can also be used to solve puzzles. You must set up an equation that “matches” in numbers what the puzzle expresses in words. Then, solve for the variable. That will be the answer to the puzzle.

Why don’t you try a few?

1. Farmer Brown told Bob and Sue that they could pick apples from his tree, but that neither of them could take more than 20. They worked for a while, and then Bob asked Sue, “Have you picked your limit yet?”
Sue replied, “Not yet. But if I had twice as many as I have now, plus half as many as I have now, I would have my limit.” How many did Sue have?

2. A little boy was told not to eat the grapes from the vine for fear that he would eat too many and get a stomachache. Sneaking out to the grape arbor when his mother wasn’t looking, the little boy ate grapes for five days, each day eating 6 more than the day before. In fact, after five days, the little boy was so sick that he had to confess to his mother that he had eaten 100 grapes. How many grapes did the little boy eat on EACH of the five days?

3. How high is a tree that is 15 feet shorter than a pole three times as tall as the tree?

CUT HERE_______________________________________________________

1. Let x = the number of apples she had.

2x + 1/2 x = 20

(2x as a fraction with 2 as the denominator would be written 4x/2.)

4x/2 + 1/2 x = 20

5x/2 = 20

5x/2 X 2 = 20 X 2

5x = 40

5x/5 = 40/5

x = 8 apples

2. Let x = number of grapes the little boy ate the first day

x + 6= number of grapes eaten the second day

x + 6 + 6 = number of grapes eaten the third day

x + 6 + 6 + 6 = number of grapes eaten the fourth day

x + 6 + 6 + 6 + 6 = number of grapes eaten the fifth day

Five days’ worth of grapes = 100 in all. Therefore, the equation to set up is:

x + (x + 6) + (x + 6 + 6) + (x + 6 + 6 + 6) + (x + 6 + 6 + 6 + 6) = 100

x + x + 6 + x + 12 + x + 18 + x + 24 = 100

5x + 60 = 100

5x + 60 – 60 = 100 – 60

5x = 40

5x/5 = 40/5

x = 8 grapes eaten on the first day

(Now put 8 in place of x in all the other expressions.)

x + 6 = 14 grapes eaten the second day

x + 6 + 6 = 20 grapes eaten the third day

x + 6 + 6 + 6 = 26 grapes eaten the fourth day

x + 6 + 6 + 6 + 6 = 32 grapes eaten the fifth day

(To check, add 8, 14, 20, 26, ad 32. They equal 100.)

3. Let y = number of feet in height of tree

3y would equal the number of feet in height of the pole

Set up the equation:

y = 3y – 15

y – 3y = 3y – 3y – 15

-2y = -15

-2y/-2 = -15/-2

y = 7.5 feet

3y = 22.5 feet

Pre-Requisite To: Basic Algebra Lesson 8