LINEAR EQUATIONS AND INEQUALITIES IN ONE VARIABLE

Solve the linear equations : 

Example 1 :

x - 3  =  2

Solution : 

Add 3 to each side of the equation.

x - 3 + 3  =  2 + 3

x  =  5

Example 2 : 

3x - 2(x - 1)  =  5

Solution : 

Simplify the left side of the equation.

3x - 2(x - 1)  =  5

3x - 2x + 2  =  5

Combine the like terms.

x + 2  =  5

Subtract 2 from each side of the equation.

x + 2 - 2  =  5 - 2

x  =  3

Example 3 : 

5(x - 3) - 7(6 - x)  =  24 - 3(8 - x) - 3

Solution : 

Simplify each side of the equation. 

5(x - 3) - 7(6 - x)  =  24 - 3(8 - x) - 3

5x - 15 - 42 + 7x  =  24 - 24 + 3x - 3

Combine the like terms. 

12x - 57  =  3x - 3

Subtract 3x from each side of the equation. 

12x - 57 - 3x  =  3x - 3 - 3x

9x - 57  =  - 3

Add 57 to each side of the equation. 

9x - 57 + 57  =  - 3 + 57

9x  =  54

Divide each side by 9. 

9x / 9  =  54 / 9

x  =  6

Solve the linear inequalities :

Example 4 :

2(2x + 3) - 10  ≤  6(x - 2)

Solution :

 2(2x + 3) - 10  ≤  6(x - 2)

4x + 6 - 10  ≤  6x - 12

4x - 4  ≤  6x - 12

Subtract 6x from each side. 

-2x - 4  ≤  - 12

Add 4 to each side.

-2x  ≤  - 8

Divide each side by (-2). 

x  ≥  4

So, the solution set is [4, ∞)

Example 5 :

3x - 7  >  x + 1

Solution :

3x - 7  >  x + 1

Subtract x from each side. 

2x - 7  >  1

Add 7 to each side. 

2x  >  8

Divide each side by 2. 

x  >  4

So, the solution set is (4, ∞)

Example 6 :

- (x - 3) + 4  <  5 - 2x

Solution :

- ( x - 3) + 4  <  5 - 2x

-x + 3 + 4  <  5 - 2x 

-x + 7  <  5 -2x

Add 2x to each side.

x + 7  <  5

Subtract 7 from each side. 

x  <  - 2

So, the solution set is (-∞, -2). 

Example 6 :

Keith has $500 in a savings account at the beginning of the summer. He wants to have at least $200 at the end of the summer. He withdraws $25 per week for food, clothing, and movie tickets. How many weeks can Keith withdraw money from his account?

Solution :

Let x be the number of weeks withdrawn.

Total saving initially = 500

Amount withdrawn each week = $25

500 - 25x ≥ 200

Solving for x, we get 

-25x ≥ 200 - 500

-25x ≥ -300

Dividing by -25 on both sides

x ≤ 300/25

x ≤ 12

She can withdraw for 12 weeks.

Example 7 :

A taxi charges a flat rate of $1.75, plus an additional $0.65 per mile. If Erica has at most $10 to spend on the cab ride, how far could she travel?

Solution :

Charge per mile = $0.65

Let x be the number of hours of cab ride. Savings should be at most $10. Then the condition should be ≤ 10.

1.75 + 0.65x ≤ 10

0.65x ≤ 10 - 1.75

0.65x ≤ 8.25

x ≤ 8.25/0.65

x ≤ 12.6

So, no more than 12.7 miles.

Example 8 :

Find three consecutive odd integers so that the sum of twice the first, the second and three times the third is 152.

Solution :

Let x, x +2 and x + 4 are three odd numbers.

2x + x + 2 + 3(x + 4) = 152

2x + x + 2 + 3x + 12 = 152

6x + 14 = 152

6x = 152 - 14

6x = 138

x = 138/6

x = 23

So, the first odd number is 23, the second is 25 and third is 27.

Related pages

• Linear equations with one unknown
  
• Solving systems of linear equations by graphing
  
• Graphing linear equations using intercepts
• Linear inequalities
  
• Linear inequalities in one variable
  
• Solving linear inequalities in one variable word problems
  
• Graphing linear inequalities in two variables and find common region

Kindly mail your feedback to v4formath@gmail.com

We always appreciate your feedback.