Exercise 2.1

Solve the following equations and check your results.

Instructions

1. 3x=2x+18

3x = 2x+18 ⟹ 3x−2x = 18 ⟹ x = .

Substituting x in equation: 3x=2x+18 ⟹ 3() = 2() + 18.

3×18 = 2×18 + 18 ⟹ = + 18 ⟹ = .

So, LHS to RHS.

2. 5t−3 = 3t−5

5t = 3t−5+3 ⟹ 5t=3t−2 ⟹ 5t−3t = −2

⟹ 2t = −2 ⟹ t =−22 ⟹ t =

Substituting t in equation: 5t−3 = 3t−5 ⟹ 5() - 3 = 3() - 5

⟹ -5-3 ⟹ -3-5 ⟹ =

So, LHS to RHS.

3. 5x+9 = 5+3x

Substracting 3x in both sides: 5x−3x+9 = 5+3x−3x ⟹ 2x+9 = 5

Subtracting 9 from both sides: 2x+9−9 = 5−9 ⟹ 2x =

Dividing both sides by 2 to solve: 2x2 = −42 ⟹ x =

Substituting x in the equation: 5 () + 9 = 5 + 3()

−10+9 = 5−6 ⟹ =

So, LHS to RHS.

4. 4z+3 = 6+2z

Subtracting 2z from both sides: 4z−2z+3 = 6+2z−2z ⟹ 2z+3=6

Substracting 3 from both sides: 2z+3−3 = 6−3 ⟹ 2z =

Dividing both sides by 2: 2z2 = 32 ⟹ z =

Substituting in equation: 4() + 3 = 6 + 2()

6 + 3 = 6 + 3 ⇒ =

So, LHS to RHS.

5. 2x−1= 14−x

Adding x to both sides: 2x+x−1 = 14−x+x ⇒ 3x - =

Adding 1 to both sides: 3x−1+1 = 14+1 ⇒ 3x =

Dividing both sides by 3: 3x3 = 153 ⇒ x =

Substituting x in equation: 2 × - 1 = 14−5

⇒ 10−1 = 9 ⇒ =

So, LHS to RHS.

6. 8x+4 = 3x−1+7

Distributing the 3 on the right side: 8x+4 = 3x−3+7

Combining like terms on the right side: 8x+4 = 3x+4

Subtracting 3x from both sides: 8x−3x+4 = 4 ⇒ 5x+4 =

Subtracting 4 from both sides: 5x+4−4 = 4−4 ⇒ 5x =

Dividing both sides by 5: 5x5 = 05 ⇒ x =

Substituting in equation: 8 × + 4 = 3 (-1) + 7 ⇒ 4 = 3−1+7 ⇒ 4 = 3−1+7 ⇒ 4 = −3+7 ⇒ = .

So, LHS to RHS.

7. x = 4x+105

Expanding RHS: x = 4x5+8

Subtracting 4x5 from both sides: x−4x5= 8 ⇒ = 8

Multiplying both sides by 5: x = 5 × 8 ⇒ x =

Substituting x in equation: 40 = 45×40+10

40 = 45×50 ⇒ =

So, LHS to RHS.

8. 2x3+1=7x15+3

Subtracting 1 from both sides: 2x3 + 1 - 1 = 7x15 + 3 - 1 ⇒ 2x3 = 7x15 + 2.

Subtracting 7x15 from both sides: 2x3 - 7x15 = 2

Finding common denominator: 10x15 - 7x15 = 2 ⇒ = 2.

Simplify the Equation: 3x15 = 2 ⇒ = 2.

Multiplying both sides by 5: x = 2×5 ⇒ x = .

Substituting x in equation: 2 × /3 + 1 = 7 × /15 + 3

⇒ 203+1= 7015+3

Convert 7015 to a common denominator: 203 + 1 = 143+ 3 ⇒ 203 + 33 = 143 + 93 ⇒ / = /

So, LHS to RHS.

9. 2y+53 = 263−y

Simplifying the equation: 2y+y+53 = 263 ⇒ 3y+53=263 ⇒ 3y = /

3y = 213 ⇒ 3y = 7 ⇒ y =

Substituting y in equation: 273 + 53 = 263 -

⇒ 143 + 53 = 193 ⇒ =

So, LHS to RHS.

10. 3m = 5m−85

Subtracting 5m from both sides: 3m−5m = 5m−5m−85 ⇒ −2m = −85

Dividing both sides by −2: −2m−2 = −85−2 ⇒ m =

Substituting m in equation: 3×45 = 5×45 - 85

⇒ 125 = 205 - 85 ⇒ =

So, LHS to RHS.

Chapter 2: Linear Equations in One Variable

Glossary

Exercise 2.1

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Chapter 2: Linear Equations in One Variable

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Glossary

Exercise 2.1