LESSON 6: EQUATIONS WITH DECIMAL NUMBERS, CONSECUTIVE INTEGER WORD PROBLEMS

When an equation contains decimal numbers, it is sometimes helpful to multiply every term in the equation by a power of 10 that will turn all the numbers into integers.

Example: Solve 0.003x = 2.05

Answer: Begin by multiplying every term by x = 2050 x = 550

Many people use the words decimal fraction to describe a number that has an internal decimal point. This is because numbers such as can be written in fractional form. 20,413 10,000 Thus, the general equation for a fractional part of a number can also be used for problems that involve a decimal part of a number.

Example: The students found that of the teachers were either brave or completely fearless. If 300 teachers fell into one of these categories, how many teachers were there in all?

Answer: WD x of = is 0.015T = 300 T = 20,000

Example: An analysis of the old woman’s utterances showed that were vaticinal. If she spoke 2000 times during the period in questions, how many utterances were not vaticinal?

Answer: 1 – = were not vaticinal WD x of = is (0.068)(2000) = NV 136 = NV

In algebra we study problems whose mastery will provide the skills necessary to solve problems that will encountered in higher mathematics and in mathematically based disciplines such as chemistry or physics. Problems about consecutive integers are of this type. They help us remember which numbers are integers and allow us to practice our word problem skills.

We remember that we designate an unspecified integer with the letter N and greater consecutive integers with N + 1, N + 2 and so on. Consecutive integers: N, N + 1, N + 2, etc.

Consecutive odd integers are two units apart, and consecutive even integers are also two units apart. Thus, we can designate both of them with the same notation. Consecutive odd integers: N, N + 2, N + 4, etc. Consecutive even integers: N, N + 2, N + 4, etc.

Example: Find three consecutive even integers such that 5 times the sum of the first and the third is 16 greater than 9 times the second.

Answer: N N + 2 N + 4 5(N + N + 4) – 16 = 9(N + 2) N = 14, 16, 18

Example: Find four consecutive integers such that 5 times the sum of the first and the fourth is 1 greater than 8 times the third.

Answer: NN + 1N + 2N + 3 5(N + N + 3) – 1 = 8(N + 2) N = 1, 2, 3, 4

HW: Lesson 6 #1-30