SOLVING LINEAR EQUATIONS IN ONE UNKNOWN PAGE 1

LESSON 1: INTRODUCING LINEAR EQUATIONS IN ONE UNKNOWN

A linear equations in one unknown consists of two expressions that are linked by an = sign. An equation states that two expressions represent the same quantity. Here are some examples of linear equations in one unknown:

x + 8 = 45 2(x – 5) = 108

10 = x + 3 x + (3x • 2) = 21

(3x + 5) – (2x +3) = 16 42 – x = x + 17

Here are some examples that are not linear equations:

x(2x – 6) = 56 x2 + 35 = 84

16x – 12 (x + 2)(4x – 1) = 100

73 = 100 – x3 x

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An equation is a linear equation in one unknown if it can be written in this form:

0 = kx + b In kx+b = 0 the letters k and b are fixed numbers that are usually given. The letter x stands for an unknown. In a linear equation expressed in terms of x, the power (or exponent) of x is 1. When someone solves a linear equation in one unknown, they find the value of x that makes the right-hand side equal to the left-hand side. In other words, the value(s) of x that makes the equation true, or satisfies it, is the solution.

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LESSON 1 TASKS: WHAT ARE LINEAR EQUATIONS IN ONE UNKNOWN AND WHAT DOES IT MEAN TO SOLVE THEM?

1. Decide if each of the following is or is not a linear equation in one unknown and then complete the sentence to give your reason. The first one has been done for you.

3 + x2 = 9 This is/is not a linear equation in one unknown because

it contains an x2 term.

10 = x – 7 This is/is not a linear equation in one unknown because

4x – 11 This is/is not a linear equation in one unknown because

3(x + 2) = 12 This is/is not a linear equation in one unknown because

3x – 4 = 7 + 4x This is/is not a linear equation in one unknown because

5 • 10 = 2 • 25 This is/is not a linear equation in one unknown because

2. Which of the following equations have 3 as their solution?

(a) 3x + 6 = 12 (b) 7x – 14 = 7

(c) 5(x – 2) = 8 – x (d) 6(x+2) = 36

3. Choose the solutions to the following equations from among -1, 0 and 1:

(a) 5x + 2 = 10x – 3 (b) 12 – 4x = 20 + 4x

4. Express each of the following relations in an equation: (a) 2 times a number x plus 4 is equal to 2 times the sum of x plus 2. (b) Half of a number x plus 3 is equal to x minus 2. (c) Bus A has twice as many passengers as Bus B. The total passengers in both buses is 42. 5. Express in words the relations in the following equations: (a) 2x + 10 = 90 (b) x + (x + 1) = 41 (c) 7x = 6x + 4

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LESSON 2: Using mental math to solve linear equation in two unknowns

When you solve an equation you find the value of the unknown that makes the equation true. Some linear equations in one unknown can be solved just by looking at them. For example:

4x = 16 2x = 6

x + 15 = 20 14 = 5 + x

Others are more complicated. For example:

6(x + 2) = x – 7 -2 = y – 5

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In this lesson you will focus on using mental math to solve equations that are not complicated.

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LESSON 2 TASKS: USING MENTAL MATH TO SOLVE LINEAR EQUATIONS IN ONE UNKNOWN

1. (a) 3x = 9 (b) 12x = 24 (c) 6 = 2x

2. (a) = 3 (b) = 12 (c) = 4

3. (a) x + 9 = 15 (b) 27 – x = 12 (c) 3 + x = 15

4. (a) 21 – x = 7 (b) x + 16 = 25 (c) x + x = 18

5. (a) 3x + 5 = 20 (b) 5x + 15 = 20 (c) 4x + 4 = 16

6. (a) 22 – x = 16 (b) 2x + 6 = 18 (c) 14 = 7 + x

7. (a) 3x + 1 = 10 (b) 24 + x = 29 (c) 2x = 16

8. (a) 23 = 14 + x (b) 4x – 4 = 16 (c) 10 = 2x – 2

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LESSON 3: EQUIVALENT EQUATIONS

Equivalent equations are equations that have the same solution. Imagine starting with this equation—2x = 8—for short we will call it equation (1). Show that its solution is 4. Now imagine adding 3 to both sides of equation (1): 2x + 3 = 8 + 3 or, 2x + 3 = 11 (2) Now, verify that solution to equation (2) is also 4. Now imagine taking equation (2) and multiplying each term by 4: 4 • 2x + 4 • 3 = 4 • 11 or 8x + 12 = 44 (3) Again, verify that the solution to equation (3) is still 4. Equations (1), (2), and (3) are some examples of equivalent equations You can generate equivalent equations by doing the following: ⇒ add or subtract a number or term to each side of the equation ⇒ multiply or divide each term by a non-zero number ⇒ apply properties such as the distributive property Now start with an equation that is too complicated to solve using mental math alone: 5x – 2(x – 1) = 14 (1) Use the distributive property to generate an equivalent equation: 5x – 2x + 2 = 14 (2) Generate another equivalent equation by subtracting the monomial 2x from the monomial 5x: 5x – 2x + 2 = 14 (2) 3x + 2 = 14 (3) If you can “see” the solution to equation (3) you can stop generating equivalent equations and just write the solution. But if you cannot “see” the solution you can generate another equivalent equation by subtracting the number 2 from both sides of the equation: 3x + 2 – 2 = 14 – 2 to give 3x = 12 (4) Now, use mental math to solve equation 4. Or continue on your own to generate another equivalent equation that will allow you to “see” the solution No matter how complex a linear equation in one unknown may seem, it is important to realize that it can be solved by writing as many equivalent equations as needed to reveal the solution.

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LESSON 3 TASKS: USING EQUIVALENT EQUATIONS TO SOLVE EQUATIONS

Solve each of the following equations by generating equivalent equations:

1. (a) 3x + 4 = 25 (b) 7x + 4 = 39

2. (a) 28 = 2x – 14 (b) 100 = 4x + 20

3. (a) 2(x + 4) = 12 (b) 3(4 + x) = 21

4. (a) 52 = 16 + 9x (b) 10 + 8x = 90

5. (a) 3x + x = 2 + 3x (b) 6x = 4(3 + x)

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LESSON 4 TASKS USING EQUIVALENT EQUATIONS TO SOLVE EQUATIONS

Solve each of the following equations by generating equivalent equations:

1. (a) 5x – 16 = 14 (b) -20 = x – 7

2. (a) 26 = 5x – 14 – x (b) 80 – 5x = 5

3. (a) 2x = 10 – 3x (b) 4x = 45 – 5x

(c) 42 – x = 45 – 2x (d) 4x + 1 = 7 – 7x

4. (a) 16(x + 2) = 64 (b) 15 = x + (7x – 1)

5. (a) 6x – (7x + 12) = -7x + 60 (b) 10x – (x + 14) = 56 – x

(c) 32 + 2x = 2(8 – x) (d) 3(2x + 1) – 3x = 3(13 – x)

(e) 8(x – 1) = -2(14 – 2x) (f) 3(2x – 1) – (4 – x) = 14

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LESSON 5: HOW MANY SOLUTIONS?

How many values of x will satisfy this equation? 3x + 9 = 27 Now, how many values of x will satisfy this equation? 3(x + 2) = 3x + 6 How many values of x will satisfy this equation? x = x + 6 An equation that has an infinite number of solutions is called an identity. Give one example of an identity.

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LESSON 5 TASKS: FINDING THE NUMBER OF SOLUTIONS 1. Identify the number of solutions for each of these equations:

Equation Number of solutions

3x + 10 = 2

2x + 4 = 2(x – 5) + 14

10x – 14 = (50 + 7x) – 4

12x = 6(2x – 8)

9x + 4 = -21 + 2(2 + 3x)

2. Give an example of a linear equation in one unknown that has exactly one solution: 3. Give an example of a linear equation in one unknown that has no solutions: 4. Give an example of a linear equation in one unknown that has an infinite number of

solutions:

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LESSON 8 TASKS: WRAPPING UP!

Solve each of the following equations:

1. (a) 8x = 64 (b) 3x + 6 = 18

2. (a) 17 – x = 7 (b) 9x + 13 = 49

3. (a) 2(x + 9) = 30 (b) 2(x + 12) = 48

4. (a) 2(x + 6) = 4 + (12 + x) (b) 17 – x = 23

5. (a) 4(x + 10) = 3(15 + x) (b) 42 = 4x + 51 – 7x

6. (a) 6x – 7 = 2x + 17 (b) 10x – 12 = 2(25 – x) – 2