Linear equations in one variable are equations that involve only one variable and the highest power of that variable is 1. They can be expressed in the general form:
ax + b = 0
Where:
- a and b are constants (numbers)
- x is the variable
- a ≠ 0 (if a = 0, it becomes a simple equation, not a linear equation)
Key Concepts:
- Solution: The value of the variable that makes the equation true.
- Solving equations: Involves isolating the variable on one side of the equation using inverse operations (addition/subtraction, multiplication/division).
- Applications: Linear equations are used to solve real-world problems involving age, number, speed, distance, cost, and more.
Example:
- 2x + 5 = 11
- To solve, we subtract 5 from both sides: 2x = 6
- Then, divide both sides by 2: x = 3
Important Properties Used in Solving Equations:
- Commutative property of addition and multiplication
- Associative property of addition and multiplication
- Distributive property of multiplication over addition
By understanding these concepts and properties, you can effectively solve linear equations and apply them to various real-world situations.
Exercise 2.1
Solve the following equations and check your results.
1. 3x = 2x + 18
Ans
Subtract 2x from both sides of the equation: 3x – 2x = 2x + 18 – 2x
Simplifying: x = 18
Therefore, the solution to the equation is x = 18.
Checking the solution:
To verify if x = 18 is correct, we substitute x with 18 in the original equation: Left-hand side (LHS) = 3x = 3 * 18 = 54 Right-hand side (RHS) = 2x + 18 = 2 * 18 + 18 = 54
Since LHS = RHS, the solution x = 18 is correct.
2. 5t – 3 = 3t – 5
Ans :
Step 1: Isolate the variable t To isolate t, we can subtract 3t from both sides of the equation: 5t – 3 – 3t = 3t – 5 – 3t
Simplifying: 2t – 3 = -5
Step 2: Isolate t To isolate t, we can add 3 to both sides of the equation: 2t – 3 + 3 = -5 + 3
Simplifying: 2t = -2
Step 3: Solve for t To find the value of t, we divide both sides by 2: 2t / 2 = -2 / 2
Simplifying: t = -1
Checking the solution:
To verify if t = -1 is correct, we substitute t with -1 in the original equation: Left-hand side (LHS) = 5t – 3 = 5(-1) – 3 = -5 – 3 = -8 Right-hand side (RHS) = 3t – 5 = 3(-1) – 5 = -3 – 5 = -8
Since LHS = RHS, the solution t = -1 is correct.
3. 5x + 9 = 5 + 3x
Ans :
Step 1: Isolate the variable x To isolate x, we can subtract 3x from both sides of the equation: 5x + 9 – 3x = 5 + 3x – 3x
Simplifying: 2x + 9 = 5
Step 2: Isolate x To isolate x, we can subtract 9 from both sides of the equation: 2x + 9 – 9 = 5 – 9
Simplifying: 2x = -4
Step 3: Solve for x To find the value of x, we divide both sides by 2: 2x / 2 = -4 / 2
Simplifying: x = -2
Therefore, the solution to the equation is x = -2.
Checking the solution:
To verify if x = -2 is correct, we substitute x with -2 in the original equation: Left-hand side (LHS) = 5x + 9 = 5(-2) + 9 = -10 + 9 = -1 Right-hand side (RHS) = 5 + 3x = 5 + 3(-2) = 5 – 6 = -1
Since LHS = RHS
The solution x = -2 is correct.
4. 4z + 3 = 6 + 2z
Ans :
Step 1: Combine like terms
To simplify the equation, let’s get all the terms with z on one side and the constants on the other side.
Subtract 2z from both sides: 4z – 2z + 3 = 6 + 2z – 2z
This simplifies to: 2z + 3 = 6
Step 2: Isolate the variable
To isolate z, subtract 3 from both sides: 2z + 3 – 3 = 6 – 3
This simplifies to: 2z = 3
Step 3: Solve for z
To find the value of z, divide both sides by 2: 2z / 2 = 3 / 2
This simplifies to:z = 3/2
Therefore, the solution to the equation is z = 3/2.
5. 2x – 1 = 14 – x
Ans :
Step 1: Combine like terms
To simplify the equation, let’s get all the terms with x on one side and the constants on the other side.
Add x to both sides: 2x – 1 + x = 14 – x + x
This simplifies to: 3x – 1 = 14
Step 2: Isolate the variable
To isolate x, add 1 to both sides: 3x – 1 + 1 = 14 + 1
This simplifies to: 3x = 15
Step 3: Solve for x
To find the value of x, divide both sides by 3: 3x / 3 = 15 / 3
This simplifies to:x = 5
Checking the solution:
To verify if x = 5 is correct, we substitute x with 5 in the original equation: Left-hand side (LHS) = 2x – 1 = 2(5) – 1 = 10 – 1 = 9 Right-hand side (RHS) = 14 – x = 14 – 5 = 9
Since LHS = RHS, the solution x = 5 is correct.
6. 8x + 4 = 3(x – 1) + 7
Ans :
Step 1: Expand the brackets
First, let’s distribute the 3 on the right side of the equation: 8x + 4 = 3x – 3 + 7
Step 2: Combine like terms
Combine the constant terms on the right side: 8x + 4 = 3x + 4
Step 3: Isolate the variable
Subtract 3x from both sides: 8x – 3x + 4 = 3x – 3x + 4
Simplify: 5x + 4 = 4
Subtract 4 from both sides: 5x + 4 – 4 = 4 – 4
Simplify: 5x = 0
Step 4: Solve for x
Divide both sides by 5: 5x / 5 = 0 / 5
Simplify: x = 0
Checking the solution:
Substitute x = 0 back into the original equation: Left-hand side (LHS): 8(0) + 4 = 0 + 4 = 4 Right-hand side (RHS): 3(0 – 1) + 7 = 3(-1) + 7 = -3 + 7 = 4
Since LHS = RHS, the solution x = 0 is correct.
7. x = 4/5 (x + 10)
Ans :
Step 1: Eliminate the fraction
To simplify the equation, let’s multiply both sides by 5: 5 * x = 5 * (4/5)(x + 10)
This simplifies to: 5x = 4(x + 10)
Step 2: Expand the brackets
Multiply 4 by both terms inside the brackets: 5x = 4x + 40
Step 3: Isolate the variable
Subtract 4x from both sides: 5x – 4x = 4x + 40 – 4x
Simplify: x = 40
Checking the solution:
Substitute x = 40 back into the original equation: Left-hand side (LHS): x = 40 Right-hand side (RHS): (4/5)(x + 10) = (4/5)(40 + 10) = (4/5)(50) = 40
Since LHS = RHS, the solution x = 40 is correct.
Therefore, the solution to the equation is x = 40.
8. 2x/3 + 1 = 7x/15 + 3
Ans :
Step 1: Eliminate the fractions
To simplify the equation, let’s find the least common multiple (LCM) of the denominators 3 and 15, which is 15. Multiply both sides of the equation by 15:
15 * [(2x)/3 + 1] = 15 * [(7x)/15 + 3]
This simplifies to: 10x + 15 = 7x + 45
Step 2: Combine like terms
Subtract 7x from both sides: 10x – 7x + 15 = 7x – 7x + 45
Simplify: 3x + 15 = 45
Step 3: Isolate the variable
Subtract 15 from both sides: 3x + 15 – 15 = 45 – 15
Simplify: 3x = 30
Step 4: Solve for x
Divide both sides by 3: 3x / 3 = 30 / 3
Simplify:x = 10
Checking the solution:
Substitute x = 10 back into the original equation: Left-hand side (LHS): (2 * 10) / 3 + 1 = 20/3 + 1 = (23/3) Right-hand side (RHS): (7 * 10) / 15 + 3 = 70/15 + 3 = 14/3 + 3 = (23/3)
Since LHS = RHS, the solution x = 10 is correct.
9. 2y + 5/3 = 26/3– y
Ans :
Step 1: Eliminate the fractions
To simplify the equation, multiply both sides by 3: 3 * [(2y + 5)/3] = 3 * [26/3 – y]
This simplifies to: 2y + 5 = 26 – 3y
Step 2: Combine like terms
Add 3y to both sides: 2y + 3y + 5 = 26 – 3y + 3y
Simplify: 5y + 5 = 26
Step 3: Isolate the variable
Subtract 5 from both sides: 5y + 5 – 5 = 26 – 5
Simplify: 5y = 21
Step 4: Solve for y
Divide both sides by 5: 5y / 5 = 21 / 5
Simplify:y = 21/5
Therefore, the solution to the equation is y = 21/5.
Checking the solution:
Substitute y = 21/5 back into the original equation: Left-hand side (LHS): (2 * (21/5)) + 5/3 = 42/5 + 5/3 = (126 + 25)/15 = 151/15 Right-hand side (RHS): 26/3 – (21/5) = (130 – 63)/15 = 67/15
Since LHS = RHS, the solution y = 21/5 is correct.
10. 3m = 5m – 8/5
Ans :
Step 1: Combine like terms
To simplify the equation, let’s get all the terms with m on one side and the constants on the other side.
Subtract 5m from both sides: 3m – 5m = 5m – 8/5 – 5m
Simplify: -2m = -8/5
Step 2: Solve for m
To find the value of m, divide both sides by -2: (-2m) / -2 = (-8/5) / -2
Simplify: m = 4/5
Therefore, the solution to the equation is m = 4/5.
Checking the solution:
Substitute m = 4/5 back into the original equation: Left-hand side (LHS): 3m = 3 * (4/5) = 12/5 Right-hand side (RHS): 5m – 8/5 = 5 * (4/5) – 8/5 = 20/5 – 8/5 = 12/5
Since LHS = RHS, the solution m = 4/5 is correct.
Exercise 2.2
1. Solve the following linear equations.
Ans :
The LCM of 2, 3, 4, and 5 is 60. Multiply both sides of the equation by 60
x/2*60-⅕*60 = x/3*60 + ¼*60
[30x – 12 = 20x + 15]
[30x – 20x = 15 + 12]
[10x = 27]
[x = {27}/{10}]
Therefore, the solution to the equation is x = 27/10.
2.
Ans :
Given equation:
n/2 – 3n/4 + 5n/6 = 21
Step 1: Find LCD
The LCD of 2, 4, and 6 is 12.
Step 2: Convert fractions to equivalent fractions with the LCD
(6n/12) – (9n/12) + (10n/12) = 21
Step 3: Combine like terms
(6n – 9n + 10n)/12 = 21
Simplify the numerator:
7n/12 = 21
Step 4: Solve for n
Multiply both sides by 12:
7n = 21 * 12
7n = 252
Divide both sides by 7:
n = 252 / 7
n = 36
= 36.
3.
Ans :
Given equation:
x + 7 – (8x)/3 = 17/6 – (5x)/2
Step 1: (LCD)
The LCD of 3, 6, and 2 is 6.
Step 2: Convert fractions to equivalent fractions with the LCD
(6x)/6 + 42/6 – (16x)/6 = 17/6 – (15x)/6
Step 3: Combine like terms
(6x – 16x + 42)/6 = (17 – 15x)/6
Simplify the numerator on the left side:
(-10x + 42)/6 = (17 – 15x)/6
Step 4: Eliminate the denominators
-10x + 42 = 17 – 15x
Step 5: Solve for x
Add 15x to both sides:
-10x + 42 + 15x = 17 – 15x + 15x
Simplify:
5x + 42 = 17
Subtract 42 from both sides:
5x + 42 – 42 = 17 – 42
Simplify:
5x = -25
Divide both sides by 5:
x = -25 / 5
x = -5
x = -5.
4.
Ans :
Given equation:
(x-5)/3 = (x-3)/5
Step 1: Cross-multiply
To eliminate the fractions, we can cross-multiply:
5(x – 5) = 3(x – 3)
Step 2: Expand the brackets
5x – 25 = 3x – 9
Step 3: Collect like terms
Subtract 3x from both sides:
5x – 3x – 25 = 3x – 3x – 9
Simplify:
2x – 25 = -9
Step 4: Isolate x
Add 25 to both sides:
2x – 25 + 25 = -9 + 25
Simplify:
2x = 16
Step 5: Solve for x
Divide both sides by 2:
x = 16 / 2
x = 8
5.
Ans :
Given equation:
(3t – 2)/4 – (2t + 3)/3 = 2/3 – t
Step 1: Find (LCD)
The LCD of 4 and 3 is 12.
Step 2: Convert fractions to equivalent fractions with the LCD
(3(3t – 2))/12 – (4(2t + 3))/12 = (8 – 12t)/12
Step 3: Simplify the numerators
(9t – 6 – 8t – 12)/12 = (8 – 12t)/12
Combine like terms in the numerator:
(t – 18)/12 = (8 – 12t)/12
Step 4: Eliminate the denominators
t – 18 = 8 – 12t
Step 5: Solve for t
Add 12t to both sides:
t – 18 + 12t = 8 – 12t + 12t
Simplify:
13t – 18 = 8
Add 18 to both sides:
13t – 18 + 18 = 8 + 18
Simplify:
13t = 26
Divide both sides by 13:
t = 26 / 13
t = 2
Therefore, the solution to the equation is t = 2.
6.
Ans :
Given equation:
m – (m-1)/2 = 1 – (m-2)/3
Step 1: Find (LCD)
The LCD of 2 and 3 is 6.
Step 2: Convert fractions to equivalent fractions with the LCD
(6m)/6 – (3(m-1))/6 = 6/6 – (2(m-2))/6
Step 3: Simplify the numerators
(6m – 3m + 3)/6 = (6 – 2m + 4)/6
Combine like terms in the numerators:
(3m + 3)/6 = (10 – 2m)/6
Step 4: Eliminate the denominators
3m + 3 = 10 – 2m
Step 5: Solve for m
Add 2m to both sides:
3m + 3 + 2m = 10 – 2m + 2m
Simplify:
5m + 3 = 10
Subtract 3 from both sides:
5m + 3 – 3 = 10 – 3
Simplify:
5m = 7
Divide both sides by 5:
m = 7/5
Therefore, the solution to the equation is m = 7/5
Simplify and solve the following linear equations.
7. 3(t – 3) = 5(21 + 1)
Ans :
Given equation: 3(t – 3) = 5(21 + 1)
Step 1: Simplify both sides
- Left side:
- 3(t – 3) = 3t – 9
- Right side:
- 5(21 + 1) = 5 * 22 = 110
So the equation becomes: 3t – 9 = 110
Step 2: Isolate the variable t
- Add 9 to both sides:
- 3t – 9 + 9 = 110 + 9
- 3t = 119
Step 3: Solve for t
- Divide both sides by 3:
- 3t / 3 = 119 / 3
- t = 119/3
Therefore, the solution to the equation is t = 119/3.
8. 15(y – 4) – 2(y – 9) + 5(y + 6) = 0
Ans :
Given equation: 15(y – 4) – 2(y – 9) + 5(y + 6) = 0
Step 1: Expand the brackets
Multiply the terms inside the brackets by the corresponding coefficients: 15y – 60 – 2y + 18 + 5y + 30 = 0
Step 2: Combine like terms
Combine the like terms together: (15y – 2y + 5y) + (-60 + 18 + 30) = 0 18y – 12 = 0
Step 3: Isolate the variable
Add 12 to both sides: 18y – 12 + 12 = 0 + 12 18y = 12
Step 4: Solve for y
Divide both sides by 18: 18y / 18 = 12 / 18 y = 2/3
Therefore, the solution to the equation is y = 2/3.
9. 3(5z – 7) – 2(9z – 11) = 4(8z – 13) – 17
Ans :
Given equation: 3(5z – 7) – 2(9z – 11) = 4(8z – 13) – 17
Step 1: Expand the brackets
- Multiply the terms inside the brackets by the corresponding coefficients:
- 15z – 21 – 18z + 22 = 32z – 52 – 17
Step 2:
Combine the z terms and the constant terms separately on both sides:
- -3z + 1 = 32z – 69
Step 3: Isolate the variable
- Add 3z to both sides:
- -3z + 1 + 3z = 32z – 69 + 3z
- 1 = 35z – 69
- Add 69 to both sides:
- 1 + 69 = 35z – 69 + 69
- 70 = 35z
Step 4: Solve for z
- Divide both sides by 35:
- 70 / 35 = 35z / 35
- 2 = z
z = 2.
10. 0.25(4f – 3) = 0.05(10f – 9)
Ans :
Given equation: 0.25(4f – 3) = 0.05(10f – 9)
Step 1: Simplify the decimals
To make the equation easier to work with, let’s multiply both sides by 100 to eliminate the decimals: 100 * 0.25(4f – 3) = 100 * 0.05(10f – 9)
This simplifies to: 25(4f – 3) = 5(10f – 9)
Step 2: Expand the brackets
Multiply the terms inside the brackets by the corresponding coefficients: 100f – 75 = 50f – 45
Step 3: Combine like terms
Subtract 50f from both sides: 100f – 75 – 50f = 50f – 45 – 50f 50f – 75 = -45
Add 75 to both sides: 50f – 75 + 75 = -45 + 75 50f = 30
Step 4: Solve for f
Divide both sides by 50: 50f / 50 = 30 / 50 f = 3/5
Therefore, the solution to the equation is f = 3/5.
