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Class 8 Maths Notes Linear Equations in One Variable

🟰 Chapter 2: Linear Equations in One Variable (Class 8) ✍️

Hi Math Detectives! Pichli classes mein humne algebraic expressions aur simple equations ke baare mein seekha tha. Ab hum thoda aage badhenge aur **Linear Equations in One Variable** ko solve karna seekhenge!

Hi Math Detectives! Pichli classes mein humne algebraic expressions aur simple equations ke baare mein seekha tha. Ab hum thoda aage badhenge aur **Ek Char Wale Rakhik Samikaran (Linear Equations in One Variable)** ko solve karna seekhenge!

🔑 What is an Equation? (Samikaran Kya Hai?)

  • An equation is a statement of equality involving variables and constants.
    • Equation ek equality ka statement hota hai jisme variables (char) aur constants (acharaank) hote hain.
  • It has an Equality sign (=).
  • The expression on the left of ‘=’ is the LHS (Left Hand Side).
  • The expression on the right of ‘=’ is the RHS (Right Hand Side).

Linear Equation in One Variable: An equation where the expression forms a line when plotted (linear) and contains only one type of variable with the highest power of 1.

Ek Char Wala Rakhik Samikaran: Ek aisa equation jo graph par ek seedhi line banata hai (linear) aur jisme kewal ek hi prakaar ka variable ho jiski sabse badi power 1 ho.

General Form: ax + b = c (where a, b, c are numbers, a ≠ 0, and x is the variable).

Examples: 2x - 3 = 7, y + 5 = 10, 6z = 12.

NOT Linear: x² + 1 = 5 (power is 2), x + y = 3 (two variables).

The value of the variable which makes LHS equal to RHS is called the solution of the equation.

Variable ki woh value jo LHS ko RHS ke barabar bana deti hai, use equation ka **hal (solution)** kehte hain.

➡️ Solving Equations (Variable on One Side)

Goal: To find the value of the variable (isolate the variable on one side).

Maqsad: Variable ki value nikalna (variable ko ek taraf akela karna).

Method: Transposing (Pakshantaran)

Steps:
  1. Move the constant terms (numbers without variables) from the side of the variable to the other side of the ‘=’ sign.
    2. Constant terms ko variable wali side se doosri taraf le jao.
  2. When transposing (moving) a term to the other side, change its sign:
    • + becomes -
    • - becomes +
    • × becomes ÷ (division)
    • ÷ becomes × (multiplication)
    3. Jab term ko doosri taraf le jate ho, toh uska sign badal do.
  3. Simplify both sides to find the value of the variable.
    4. Dono sides ko saral karo variable ki value nikalne ke liye.
  4. Check: Substitute the value back into the original equation to see if LHS = RHS.
    5. Check Karo: Value ko original equation mein daal kar dekho ki LHS = RHS hai ya nahi.
Example 1: Solve x - 2 = 7

Transpose -2 to RHS (becomes +2):

x = 7 + 2

x = 9

Check: 9 – 2 = 7 (LHS = RHS). ✅

Example 2: Solve 2y + 9 = 4

Transpose +9 to RHS (becomes -9):

2y = 4 - 9

2y = -5

Transpose 2 (which is multiplying y) to RHS (becomes division):

y = -52

Check: 2 × (-52) + 9 = -5 + 9 = 4 (LHS = RHS). ✅

Example 3: Solve t5 = 10

Transpose 5 (which is dividing t) to RHS (becomes multiplication):

t = 10 × 5

t = 50

Check: 50 / 5 = 10 (LHS = RHS). ✅

🔄 Solving Equations (Variable on Both Sides)

Equations like 2x - 3 = x + 2.

Ismein variable LHS aur RHS dono taraf hota hai.
Steps:
  1. Transpose all terms containing the variable to one side (usually LHS). Remember to change signs.
    2. Sabhi variable wale terms ko ek taraf (jaise LHS) le aao. Sign badalna yaad rakho.
  2. Transpose all the constant terms (numbers) to the other side (usually RHS). Change signs.
    3. Sabhi constant terms (numbers) ko doosri taraf (jaise RHS) le jao. Sign badal do.
  3. Simplify both sides (combine like terms).
  4. Solve for the variable by transposing the coefficient if necessary.
  5. Check the solution by substituting back into the original equation.
Example 1: Solve 2x - 3 = x + 2

Transpose ‘x’ from RHS to LHS (becomes -x):

2x - x - 3 = 2

Transpose ‘-3’ from LHS to RHS (becomes +3):

2x - x = 2 + 3

Simplify:

x = 5

Check: LHS=2(5)-3=10-3=7. RHS=5+2=7. LHS=RHS. ✅

Example 2: Solve 5t - 3 = 3t - 5

Bring ‘t’ terms to LHS, constants to RHS:

5t - 3t = -5 + 3

Simplify:

2t = -2

Solve for t:

t = -22 = -1

Check: LHS=5(-1)-3=-5-3=-8. RHS=3(-1)-5=-3-5=-8. LHS=RHS. ✅

Example 3: Solve 8x + 4 = 3(x - 1) + 7

First, simplify RHS by opening bracket:

8x + 4 = 3x - 3 + 7

8x + 4 = 3x + 4

Bring ‘x’ terms to LHS, constants to RHS:

8x - 3x = 4 - 4

Simplify:

5x = 0

Solve for x:

x = 05 = 0

Check: LHS=8(0)+4=4. RHS=3(0-1)+7 = 3(-1)+7 = -3+7=4. LHS=RHS. ✅

💡 Applications (Word Problems / Upyog)

Linear equations help solve real-life problems involving unknown quantities.

Linear equations asal zindagi ki samasyao ko hal karne mein madad karte hain jinmein anjaan quantities shamil hoti hain.
Steps to Solve Word Problems:
  1. Read the problem carefully. Identify what is given and what needs to be found.
    2. Problem ko dhyaan se padho. Kya diya hai aur kya nikalna hai, pehchano.
  2. Represent the unknown quantity by a variable (like x, y, t etc.).
    3. Anjaan quantity ko ek variable (jaise x, y, t) se represent karo.
  3. Translate the statements of the problem into mathematical statements (form an equation).
    4. Problem ke statements ko mathematical statements (equation) mein badlo.
  4. Solve the linear equation for the variable.
    5. Variable ke liye linear equation ko solve karo.
  5. Check if the solution satisfies the conditions of the problem. Write the answer clearly.
    6. Check karo ki solution problem ki conditions ko poora karta hai ya nahi. Answer saaf-saaf likho.
Example 1 (Number Problem): Sum of two numbers is 74. One number is 10 more than the other. Find the numbers.

Solution:

Let the smaller number be x.

The other number is 10 more, so it is x + 10.

Their sum is 74: x + (x + 10) = 74 (Equation formed)

Solve: 2x + 10 = 74

2x = 74 - 10

2x = 64

x = 64 / 2 = 32

So, smaller number = 32.

Other number = x + 10 = 32 + 10 = 42.

Check: 32 + 42 = 74. Correct. ✅

Answer: The numbers are 32 and 42.

Example 2 (Age Problem): Amina thinks of a number, subtracts 5/2 from it, multiplies the result by 8. The result now obtained is 3 times the original number. Find the number.

Solution:

Let the number Amina thought of be x.

Subtracts 5/2: x - 52

Multiplies result by 8: 8 × (x - 52)

Result is 3 times original number: 8 (x - 52) = 3x (Equation formed)

Solve:

8x - 8×(52) = 3x (Distributive Property)

8x - 402 = 3x

8x - 20 = 3x

Bring variables to LHS, constants to RHS:

8x - 3x = 20

5x = 20

x = 20 / 5 = 4

Check: Original=4. Subtract 5/2: 4 – 2.5 = 1.5. Multiply by 8: 1.5 * 8 = 12. Is this 3 times original? 3 * 4 = 12. Yes. ✅

Answer: The number is 4.

Example 3 (Perimeter Problem): The perimeter of a rectangle is 13 cm and its width is 234 cm. Find its length.

Solution:

Let the length of the rectangle be L cm.

Given width = 234 = 114 cm.

Perimeter = 13 cm.

Formula for perimeter of rectangle = 2 × (Length + Width).

Equation: 2 × (L + 114) = 13

Solve:

Divide both sides by 2: L + 114 = 132

Transpose 11/4 to RHS:

L = 132 - 114

Find LCM(2,4)=4:

L = 264 - 114 = 154

Convert to mixed fraction: L = 334 cm.

Check: 2 * (15/4 + 11/4) = 2 * (26/4) = 2 * (13/2) = 13. Correct. ✅

Answer: The length is 334 cm.

✨ Reducing Equations to Simpler Form

Sometimes equations have brackets or fractions that make them look complex. We simplify them first before solving.

Kabhi kabhi equations mein brackets ya fractions hote hain jo unhe mushkil dikhate hain. Hum solve karne se pehle unhe saral banate hain.
Steps to Simplify & Solve:
  1. Remove Brackets: Use distributive property to open brackets.
    2. Brackets Hatayein: Distributive property use karke brackets kholein.
  2. Combine Like Terms: Combine variable terms and constant terms on each side (LHS and RHS).
    3. Ek Jaise Terms Ko Milayein: Dono sides par variable wale terms aur constant wale terms ko milayein.
  3. Solve Normally: Now solve the simpler linear equation using transposing method (bring variables to one side, constants to other).
    4. Normal Solve Karein: Ab simpler linear equation ko transposing method se solve karein.
  4. Check the solution.
Example 1: Solve x/2 - 1/5 = x/3 + 1/4

Solution:

Bring variable terms to LHS, constants to RHS:

x2 - x3 = 14 + 15

Take LCM on LHS (LCM(2,3)=6) and RHS (LCM(4,5)=20):

3x - 2x6 = 5 + 420

x6 = 920

Transpose 6 to RHS:

x = (920) × 6

x = 5420

Simplify:

x = 2710

(Check is tedious but possible). ✅

Answer: x = 2710

Example 2: Solve 3(t - 3) = 5(2t + 1)

Solution:

Open brackets (distributive property):

3t - 9 = 10t + 5

Bring variable terms to LHS, constants to RHS:

3t - 10t = 5 + 9

Simplify:

-7t = 14

Solve for t:

t = 14-7 = -2

Check: LHS=3(-2-3)=3(-5)=-15. RHS=5(2(-2)+1)=5(-4+1)=5(-3)=-15. ✅

Answer: t = -2

Example 3: Solve 15(y - 4) - 2(y - 9) + 5(y + 6) = 0

Solution:

Open brackets:

(15y - 60) - (2y - 18) + (5y + 30) = 0

15y - 60 - 2y + 18 + 5y + 30 = 0 (Careful with signs)

Combine like terms (y terms together, constants together):

(15y - 2y + 5y) + (-60 + 18 + 30) = 0

18y + (-12) = 0

18y - 12 = 0

Transpose constant:

18y = 12

Solve for y:

y = 1218

Simplify:

y = 23

(Checking recommended). ✅

Answer: y = 23

✂️ Equations Reducible to the Linear Form

Some equations are not linear initially (e.g., have variables in the denominator), but can be simplified into a linear equation.

Kuch equations shuru mein linear nahi dikhte (jaise variable denominator mein ho), lekin unhe simplify karke linear banaya ja sakta hai.

Method: Cross-Multiplication

If the equation is in the form AB = CD (where A, B, C, D are expressions, B≠0, D≠0), we can cross-multiply.

A × D = B × C

Steps:
  1. If the equation involves fractions with linear expressions in numerator and/or denominator, try to bring it to the form Linear Expr 1Linear Expr 2 = NumDen or similar.
  2. Use **cross-multiplication** to eliminate denominators.
    3. Denominators hatane ke liye **cross-multiplication** (tirchi guna) ka istemal karein.
  3. This will usually result in a linear equation.
  4. Solve the resulting linear equation using previous methods (opening brackets, transposing).
  5. **Important:** Check if the solution makes any original denominator equal to zero. If it does, that solution is not valid (extraneous solution).
    6. **Zaroori:** Check karein ki solution se koi original denominator zero toh nahi ban raha. Agar ban raha hai, toh woh solution valid nahi hai.
Example 1: Solve x + 12x + 3 = 38

Solution:

Cross-multiply:

8 × (x + 1) = 3 × (2x + 3)

Open brackets:

8x + 8 = 6x + 9

Transpose:

8x - 6x = 9 - 8

Simplify:

2x = 1

Solve:

x = 12

Check denominator: 2x + 3 with x = 1/2 becomes 2(1/2) + 3 = 1 + 3 = 4 (Not zero, so solution is valid). ✅

Answer: x = 12

Example 2: Solve nn + 15 = 49

Solution:

Cross-multiply:

9 × n = 4 × (n + 15)

Open bracket:

9n = 4n + 60

Transpose:

9n - 4n = 60

Simplify:

5n = 60

Solve:

n = 60 / 5 = 12

Check denominator: n+15 = 12+15=27 (Not zero). ✅

Answer: n = 12

Example 3: Solve y + 64 + y - 35 = 5y - 48

Solution:

Find LCM of denominators (4, 5, 8) = 40.

Multiply the entire equation by LCM (40):

40×(y + 64) + 40×(y - 35) = 40×(5y - 48)

Simplify:

10(y + 6) + 8(y - 3) = 5(5y - 4)

Open brackets:

(10y + 60) + (8y - 24) = (25y - 20)

Combine like terms on LHS:

18y + 36 = 25y - 20

Transpose variables to RHS, constants to LHS:

36 + 20 = 25y - 18y

Simplify:

56 = 7y

Solve for y:

y = 56 / 7 = 8

(Checking recommended). ✅

Answer: y = 8

❓ Sawal Jawab (Questions & Answers)

🤏 Very Short Answer Questions

1. What is a linear equation in one variable?

An equation with one variable where the highest power of the variable is 1.

2. What is the value that satisfies an equation called?

Solution.

3. Solve: x + 5 = 12.

x = 12 – 5 = 7.

4. Solve: y – 3 = -5.

y = -5 + 3 = -2.

5. Solve: 6z = 24.

z = 24 / 6 = 4.

6. Solve: p7 = 4.

p = 4 × 7 = 28.

7. Solve: 2x – 1 = 5.

2x = 6, x = 3.

8. What is transposing?

Moving a term from one side of ‘=’ to the other by changing its sign/operation.

9. Solve: 3x = 2x + 18.

3x – 2x = 18, x = 18.

10. Solve: 5y + 9 = 5 + 3y.

5y – 3y = 5 – 9, 2y = -4, y = -2.

11. Find the solution of 2x/3 = 18.

2x = 18 × 3 = 54, x = 54 / 2 = 27.

12. If 7x + 4 = 25, what is x?

7x = 21, x = 3.

13. Is x=5 a solution of 3x + 2 = 17?

Yes (3(5)+2 = 15+2 = 17).

14. When cross-multiplying ab = cd, what equation do you get?

a × d = b × c.

15. What is the first step usually taken when solving 2(x+3) = 10?

Open the bracket (Distribute 2).

16. What is the variable in the equation 5z - 1 = 9?

z.

17. Find the solution: m3 + 1 = 715?

m/3 = 7/15 – 1 = (7-15)/15 = -8/15. m = -8/15 * 3 = -8/5.

📝 Short Answer Questions

1. Solve the equation: 3x = 2x + 18.

  • Transpose 2x to LHS: 3x - 2x = 18.
  • Simplify: x = 18.

2. Solve: 5t - 3 = 3t - 5.

  • Transpose: 5t - 3t = -5 + 3.
  • Simplify: 2t = -2.
  • Solve: t = -2 / 2 = -1.

3. Solve: x2 + x3 = 10.

  • Find LCM(2,3) = 6.
  • Equation becomes: 3x + 2x6 = 10.
  • Simplify: 5x6 = 10.
  • Transpose 6: 5x = 10 × 6 = 60.
  • Solve: x = 60 / 5 = 12.

4. Solve: 0.25(4f - 3) = 0.05(10f - 9).

  • Open brackets: 1.00f - 0.75 = 0.50f - 0.45.
  • Transpose: 1f - 0.5f = -0.45 + 0.75.
  • Simplify: 0.5f = 0.30.
  • Solve: f = 0.30 / 0.50 = 30 / 50 = 3/5 or 0.6.

5. Solve the equation zz+15 = 49.

  • Cross-multiply: 9 × z = 4 × (z + 15).
  • 9z = 4z + 60.
  • Transpose: 9z - 4z = 60.
  • 5z = 60.
  • z = 60 / 5 = 12.

6. The sum of three consecutive multiples of 8 is 888. Find the multiples.

  • Let the multiples be 8x, 8(x+1), 8(x+2).
  • Sum: 8x + 8(x+1) + 8(x+2) = 888.
  • Divide by 8: x + (x+1) + (x+2) = 111.
  • 3x + 3 = 111.
  • 3x = 108.
  • x = 108 / 3 = 36.
  • Multiples are: 8(36)=288, 8(37)=296, 8(38)=304.
  • Answer: 288, 296, 304.

7. Solve: m - m-12 = 1 - m-23.

  • Bring ‘m’ terms to LHS, constants to RHS (1 is constant): m - m-12 + m-23 = 1.
  • LCM(1, 2, 3) = 6.
  • Multiply by 6: 6m - 3(m-1) + 2(m-2) = 6×1.
  • 6m - 3m + 3 + 2m - 4 = 6.
  • Combine terms: (6m-3m+2m) + (3-4) = 6.
  • 5m - 1 = 6.
  • 5m = 7.
  • m = 75.

8. The ages of Rahul and Haroon are in the ratio 5:7. Four years later the sum of their ages will be 56 years. What are their present ages?

  • Let present ages be 5x and 7x years.
  • Four years later, ages will be (5x+4) and (7x+4).
  • Sum after 4 years: (5x+4) + (7x+4) = 56.
  • 12x + 8 = 56.
  • 12x = 48.
  • x = 48 / 12 = 4.
  • Present ages: Rahul=5x=5(4)=20 yrs, Haroon=7x=7(4)=28 yrs.
  • Answer: Rahul 20 years, Haroon 28 years.

9. Solve: 8x - 33x = 2.

  • Assume 2 = 2/1.
  • Cross-multiply: 1 × (8x - 3) = 2 × (3x).
  • 8x - 3 = 6x.
  • Transpose: 8x - 6x = 3.
  • 2x = 3.
  • x = 32.
  • (Check denominator 3x: 3(3/2)=9/2 ≠ 0. Valid.)

10. Solve: 154 - 7x = 9.

  • Transpose 15/4: -7x = 9 - 154.
  • Find LCM on RHS: -7x = 36 - 154 = 214.
  • Solve for x: x = (214) ÷ (-7).
  • x = 214 × 1-7.
  • x = -2128 = -34.

11. Half of a herd of deer are grazing. Three-fourths of the remaining are playing. The rest 9 are drinking water. Find total deer.

  • Let total deer = x.
  • Grazing = x/2.
  • Remaining = x – x/2 = x/2.
  • Playing = (3/4) of remaining = (3/4) × (x/2) = 3x/8.
  • Drinking = 9.
  • Equation: Grazing + Playing + Drinking = Total.
  • x/2 + 3x/8 + 9 = x.
  • Multiply by LCM(2,8)=8: 4x + 3x + 72 = 8x.
  • 7x + 72 = 8x.
  • 72 = 8x - 7x = x.
  • Answer: Total deer = 72.

📜 Long Answer Questions

1. Solve: x + 12x + 3 = 38.

Equation: x + 12x + 3 = 38.

  • Step 1: Cross-multiply. 8 × (x + 1) = 3 × (2x + 3).
  • Step 2: Open brackets. 8x + 8 = 6x + 9.
  • Step 3: Transpose variables to LHS, constants to RHS. 8x - 6x = 9 - 8.
  • Step 4: Simplify. 2x = 1.
  • Step 5: Solve for x. x = 12.
  • Step 6: Check denominator. Original denominator is 2x+3. At x=1/2, it is 2(1/2)+3 = 1+3 = 4, which is not 0. So solution is valid.

Answer: x = 12.

2. Solve: n2 - 3n4 + 5n6 = 21.

  • Step 1: Find LCM of denominators (2, 4, 6). LCM = 12.
  • Step 2: Multiply entire equation by LCM (12). 12×(n2) - 12×(3n4) + 12×(5n6) = 12 × 21.
  • Step 3: Simplify. (6 × n) - (3 × 3n) + (2 × 5n) = 252.
    6n - 9n + 10n = 252.
  • Step 4: Combine like terms on LHS. (6 + 10 - 9)n = 252.
    7n = 252.
  • Step 5: Solve for n. n = 252 / 7.
    (252 ÷ 7: 7×3=21, remainder 4, bring down 2 -> 42. 7×6=42).
  • Result: n = 36.

3. The perimeter of a rectangle is 154 m. Its length is 2 m more than twice its breadth. Find the length and breadth.

  • Let the breadth be b metres.
  • Length = 2 more than twice breadth = 2b + 2 metres.
  • Perimeter = 2 × (Length + Breadth) = 154 m.
  • Equation: 2 × ( (2b + 2) + b ) = 154.
  • Simplify inside bracket: 2 × (3b + 2) = 154.
  • Divide by 2: 3b + 2 = 77.
  • Transpose: 3b = 77 - 2 = 75.
  • Solve for b: b = 75 / 3 = 25 m.
  • Breadth = 25 m.
  • Length = 2b + 2 = 2(25) + 2 = 50 + 2 = 52 m.
  • Answer: Length = 52 m, Breadth = 25 m.

4. Aman’s age is three times his son’s age. Ten years ago he was five times his son’s age. Find their present ages.

  • Let the son’s present age be x years.
  • Aman’s present age is 3x years.
  • Ten years ago: Son’s age = x-10; Aman’s age = 3x-10.
  • Condition 10 years ago: Aman’s age = 5 times Son’s age.
  • Equation: 3x - 10 = 5 × (x - 10).
  • Solve: 3x - 10 = 5x - 50.
  • Transpose: -10 + 50 = 5x - 3x.
  • 40 = 2x.
  • x = 40 / 2 = 20 years (Son’s present age).
  • Aman’s present age = 3x = 3(20) = 60 years.
  • Answer: Son’s age = 20 yrs, Aman’s age = 60 yrs.

5. Solve: 3t - 24 - 2t + 33 = 23 - t.

  • Rewrite: 3t - 24 - 2t + 33 + t = 23 (Bring t to LHS). Note t = t/1.
  • LCM(4, 3, 1) on LHS = 12.
  • Multiply equation by 12: 12×(3t-24) - 12×(2t+33) + 12×(t) = 12×(23).
  • Simplify: 3(3t - 2) - 4(2t + 3) + 12t = 4(2).
  • Open brackets: 9t - 6 - 8t - 12 + 12t = 8 (Careful with -4 multiplied by +3).
  • Combine ‘t’ terms: (9 - 8 + 12)t = 13t.
  • Combine constants: -6 - 12 = -18.
  • Equation: 13t - 18 = 8.
  • Transpose: 13t = 8 + 18 = 26.
  • Solve: t = 26 / 13 = 2.
  • Answer: t = 2.

6. A positive number is 5 times another number. If 21 is added to both numbers, then one of the new numbers becomes twice the other new number. What are the numbers?

  • Let the smaller number be x.
  • The other number is 5x.
  • Add 21 to both: New numbers are x+21 and 5x+21.
  • Condition: One new number is twice the other. Since 5x+21 will be larger than x+21 (as x is positive), we have:
  • Equation: 5x + 21 = 2 × (x + 21).
  • Solve: 5x + 21 = 2x + 42.
  • Transpose: 5x - 2x = 42 - 21.
  • 3x = 21.
  • x = 21 / 3 = 7.
  • Numbers are: x = 7 and 5x = 5(7) = 35.
  • Check: New numbers 7+21=28 and 35+21=56. Is 56 twice 28? Yes (56=2×28).
  • Answer: The numbers are 7 and 35.

7. Solve: 7y + 4y + 2 = -43.

  • Condition: y + 2 ≠ 0 (or y ≠ -2).
  • Cross-multiply: 3 × (7y + 4) = -4 × (y + 2).
  • Open brackets: 21y + 12 = -4y - 8.
  • Transpose: 21y + 4y = -8 - 12.
  • Combine: 25y = -20.
  • Solve: y = -20 / 25.
  • Simplify: y = -45.
  • Check: Original denominator y+2 = -4/5 + 2 = (-4+10)/5 = 6/5 ≠ 0. Valid.
  • Answer: y = -4/5.

8. The denominator of a rational number is greater than its numerator by 8. If the numerator is increased by 17 and the denominator is decreased by 1, the number obtained is 3/2. Find the rational number.

  • Let the numerator be x.
  • Denominator = greater than numerator by 8 = x + 8.
  • Original rational number = xx + 8.
  • New numerator = increased by 17 = x + 17.
  • New denominator = decreased by 1 = (x + 8) - 1 = x + 7.
  • New rational number = x + 17x + 7.
  • Given: New number = 3/2.
  • Equation: x + 17x + 7 = 32.
  • Cross-multiply: 2 × (x + 17) = 3 × (x + 7).
  • 2x + 34 = 3x + 21.
  • Transpose: 34 - 21 = 3x - 2x.
  • 13 = x.
  • Numerator = x = 13.
  • Denominator = x + 8 = 13 + 8 = 21.
  • Answer: The rational number is 1321.

9. Simplify and solve the linear equation: 3(5z - 7) - 2(9z - 11) = 4(8z - 13) - 17.

  • Step 1: Open all brackets.
    15z - 21 - 18z + 22 = 32z - 52 - 17.
  • Step 2: Combine like terms on LHS.
    (15z - 18z) + (-21 + 22) = -3z + 1.
  • Step 3: Combine like terms on RHS.
    32z + (-52 - 17) = 32z - 69.
  • Step 4: Write simplified equation.
    -3z + 1 = 32z - 69.
  • Step 5: Transpose variable terms to RHS, constants to LHS.
    1 + 69 = 32z + 3z.
  • Step 6: Simplify.
    70 = 35z.
  • Step 7: Solve for z.
    z = 70 / 35 = 2.
  • Answer: z = 2.

10. Solve the equation and check your result: 5x + 72 = 32x - 14.

  • Step 1: Transpose variable terms to LHS, constants to RHS.
    5x - 32x = -14 - 72.
  • Step 2: Simplify LHS (LCM=2).
    10x - 3x2 = 7x2.
  • Step 3: Simplify RHS (LCM=2).
    -28 - 72 = -352.
  • Step 4: Write simplified equation.
    7x2 = -352.
  • Step 5: Solve for x. Multiply both sides by 2/7.
    x = (-352) × (27) = -357 = -5.
Check Result (x = -5):
  • LHS = 5(-5) + 7/2 = -25 + 7/2 = (-50+7)/2 = -43/2.
  • RHS = (3/2)(-5) – 14 = -15/2 – 14 = (-15 – 28)/2 = -43/2.
  • Since LHS = RHS, the solution is correct. ✅

Answer: x = -5.

🤔 Check Your Understanding! (Quiz Time!)

1. Which is a linear equation in one variable?

2. The solution to x – 5 = 3 is:

3. Solve for y: 3y = 21

4. Solve for p: p4 = 5

5. Solution of 2x + 3 = 9 is:

6. Moving a term to the other side of ‘=’ by changing sign is called:

7. Solve: 7x – 9 = 16

8. Solve: 4z + 3 = 6 + 2z

9. Solution for 3m = 5m – 8/5 is:

10. Simplify: 2(x + 4) = 12

11. Solve: x = 45(x + 10)

12. Sum of two numbers is 95. If one exceeds the other by 15, find the smaller number.

13. Solve: x3 + 1 = 715

14. If you subtract 1/2 from a number and multiply by 1/2, you get 1/8. The number is:

15. Solve: 3y + 42 - 6y = -25

16. Solve: x - 53 = x - 35

17. What should be done first when solving x/2 - 1/5 = x/3 + 1/4?

18. What is the solution to 1.6 = y / 1.5?

19. Simplify t(5t - 5) = 5(t+1)? This is not linear, but what value might seem like a start? [Class 8 often starts with simple ones first]

20. An expression with only one term is called:

21. The expression ax + b is:

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