What you should learn

To solve equations with the variable on both sides

To solve equations containing grouping symbols

NCTM Curriculm Standards 2 - 4, 6 - 9

 

In doing this the teacher wants to make sure that the following words are incorporated into the introductory lesson:

Identity

Introduction: In the 1928 Olympics, the winner of the men's 800-meter run, Douglas Lowe of Great Britain, won the race in 1 minute 51.8 seconds, or 111.8 seconds. In that same year, the winner of the women's 800-meter run, Lina Radke of Germany, won the race in 2 minutes 16.8 seconds, or 136.8 seconds. Over the next 64 years, the men's winning times decreased an average of 0.127 second per year, and the women's winning times decreased an average of 0.332 second per year. Suppose the times continued to decrease at these rates. When will men and women have the same winning times in the 800-meter run?

After x years, the men's winning times can be represented by 111.8 - 0.127x.

After x years, the women's winning times can be represented by 136.8 - 0.332x.

The times would be the same when the two expressions are equal.

Many equations contian variables on each side. To solve these types of equations, first use the addition or subtraction property of equality to write an equivalent equation that has all of the variables on one side. Then solve the equation.

Method 1:

111.8 - 0.127x = 136.8 - 0.332x

111.8 - 0.127x + 0.332x = 136.8 - 0.332x + 0.332x

111.8 + 0.205x = 136.8

111.8 - 111.8 + 0.205x = 136.8 - 111.8

0.205x = 25

(0.205x)/0.205 = 25/0.205

x = 122 (rounded)

At these rates, the men's and women's winning times will be equal about 122 years after 1928, or in the year 2050.

 

Method 2:

Since two of the decimals involve thousandths, another way to solve the equation would be to multiply each side by 1000 to clear the decimals.

111.8 - 0.127x = 136.8 - 0.332x

(1000)111.8 - (1000)0.127 = (1000)136.8 - (1000)0.332x

111,800 - 127x = 136,800 - 332x

111,800 + 205x = 136,800

205x = 25,000

x = 122 (rounded)

Exercise 1: Solve (3/8) - (1/4)x = (1/2)x - (3/4)

(3/8) - (1/4)x = (1/2)x - (3/4)

8[(3/8) - (1/4)x] = 8[(1/2)x - (3/4)]

8(3/8) - 8(1/4)x = 8(1/2)x - 8(3/4)

3 - 2x = 4x - 6

3 = 6x - 6

9 = 6x

3/2 = x

The solution is 3/2.

When solving equaitons that contain grouping symbols, first use the distributive property to remove the grouping symbols.

Exercise 2: One angle of a triangle measures 10 degrees more than the second. The measure of the third angle is twice the sum of the first two angles. Find the measure of each angle. (Hint: Start by drawing a diagram).

Some equations with the variable on both sides may have no solution. That is, there is no value of the variable that will result in a true equation.

Exercise 3: Solve 5n + 4 = 7(n + 1) - 2n

5n + 4 = 7(n + 1) - 2n

5n + 4 = 7n + 7 - 2n

5n + 4 = 5n + 7

4 = 7

Since 4 = 7 is a false statement, this equation has no solution

 

 

An equation that is true for every value of the variable is called an identity.

Exercise 4: Solve 7 + 2(x + 1) = 2x + 9

Closing Activity: Check for understanding by using this as a quick review before class is over. It should take about the last five to ten minutes. I would use it for my students as their 'ticket out the door'. Click Here.

Homework: The homework to be assigned for tonight would be: 15 - 39 odd, 40, 41, 43 - 49

Alternative Homework: Enriched: 14 - 38 even, 40 - 49

Extra Practice: Students book page 763 Lesson 3-5

Extra Practice Worksheet: Click Here.

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