Question 1

Solve the following equations and check your results.

1.  $3x = 2x + 18$
2.  $5x - 3 = 3x - 5$
3.  $5x + 9 = 5 + 3x$
4.  $4x + 3 = 6 + 2x$
5.  $2x - 1 = 14 - x$
6.  $8x + 4 = 3(x - 1) + 7$
7.  $x = \frac{4}{5}(x + 10)$
8.  $\frac{2x}{3} + 1 = \frac{7x}{15} + 3$
9.  $2x + \frac{5}{3} = \frac{26}{3} - x$
10.  $3x = 5x - \frac{8}{5}$

Accepted Answer

1. Solve $3x = 2x + 18$:
Subtract $2x$ from both sides:
$3x - 2x = 18$
$x = 18$
Check:
LHS = $3 	imes 18 = 54$
RHS = $2 	imes 18 + 18 = 36 + 18 = 54$
LHS = RHS

2. Solve $5x - 3 = 3x - 5$:
Bring variables to one side and constants to other:
$5x - 3x = -5 + 3$
$2x = -2$
$x = -1$
Check:
LHS = $5(-1) - 3 = -5 - 3 = -8$
RHS = $3(-1) - 5 = -3 - 5 = -8$
LHS = RHS

3. Solve $5x + 9 = 5 + 3x$:
$5x - 3x = 5 - 9$
$2x = -4$
$x = -2$
Check:
LHS = $5(-2) + 9 = -10 + 9 = -1$
RHS = $5 + 3(-2) = 5 - 6 = -1$
LHS = RHS

4. Solve $4x + 3 = 6 + 2x$:
$4x - 2x = 6 - 3$
$2x = 3$
$x = \frac{3}{2}$
Check:
LHS = $4 	imes \frac{3}{2} + 3 = 6 + 3 = 9$
RHS = $6 + 2 	imes \frac{3}{2} = 6 + 3 = 9$
LHS = RHS

5. Solve $2x - 1 = 14 - x$:
$2x + x = 14 + 1$
$3x = 15$
$x = 5$
Check:
LHS = $2 	imes 5 - 1 = 10 - 1 = 9$
RHS = $14 - 5 = 9$
LHS = RHS

6. Solve $8x + 4 = 3(x - 1) + 7$:
Expand RHS: $3x - 3 + 7 = 3x + 4$
Equation: $8x + 4 = 3x + 4$
$8x - 3x = 4 - 4$
$5x = 0$
$x = 0$
Check:
LHS = $8 	imes 0 + 4 = 4$
RHS = $3(0 - 1) + 7 = 3(-1) + 7 = -3 + 7 = 4$
LHS = RHS

7. Solve $x = \frac{4}{5}(x + 10)$:
Multiply both sides by 5:
$5x = 4(x + 10)$
$5x = 4x + 40$
$5x - 4x = 40$
$x = 40$
Check:
LHS = $40$
RHS = $\frac{4}{5}(40 + 10) = \frac{4}{5} 	imes 50 = 40$
LHS = RHS

8. Solve $\frac{2x}{3} + 1 = \frac{7x}{15} + 3$:
Multiply both sides by 15 (LCM of denominators):
$15 	imes \frac{2x}{3} + 15 	imes 1 = 15 	imes \frac{7x}{15} + 15 	imes 3$
$5 	imes 2x + 15 = 7x + 45$
$10x + 15 = 7x + 45$
$10x - 7x = 45 - 15$
$3x = 30$
$x = 10$
Check:
LHS = $\frac{2 	imes 10}{3} + 1 = \frac{20}{3} + 1 = \frac{20}{3} + \frac{3}{3} = \frac{23}{3} \approx 7.67$
RHS = $\frac{7 	imes 10}{15} + 3 = \frac{70}{15} + 3 = \frac{14}{3} + 3 = \frac{14}{3} + \frac{9}{3} = \frac{23}{3} \approx 7.67$
LHS = RHS

9. Solve $2x + \frac{5}{3} = \frac{26}{3} - x$:
Bring $x$ terms to one side and constants to other:
$2x + x = \frac{26}{3} - \frac{5}{3}$
$3x = \frac{21}{3} = 7$
$x = \frac{7}{3}$
Check:
LHS = $2 	imes \frac{7}{3} + \frac{5}{3} = \frac{14}{3} + \frac{5}{3} = \frac{19}{3}$
RHS = $\frac{26}{3} - \frac{7}{3} = \frac{19}{3}$
LHS = RHS

10. Solve $3x = 5x - \frac{8}{5}$:
$3x - 5x = - \frac{8}{5}$
$-2x = - \frac{8}{5}$
$x = \frac{8}{10} = \frac{4}{5}$
Check:
LHS = $3 	imes \frac{4}{5} = \frac{12}{5}$
RHS = $5 	imes \frac{4}{5} - \frac{8}{5} = 4 - \frac{8}{5} = \frac{20}{5} - \frac{8}{5} = \frac{12}{5}$
LHS = RHS — Each equation is solved by isolating the variable x on one side, performing algebraic operations such as addition, subtraction, multiplication, or division, and then checking the solution by substituting back into the original equation to verify equality of LHS and RHS.

Question 2

Solve the following linear equations.

1.  $\frac{x}{2} - \frac{1}{5} = \frac{x}{3} + \frac{1}{4}$
2.  $\frac{n}{2} - \frac{3n}{4} + \frac{5n}{6} = 21$
3.  $x + 7 - \frac{8x}{3} = \frac{17}{6} - \frac{5x}{2}$
4.  $\frac{x - 5}{3} = \frac{x - 3}{5}$
5.  $\frac{3t - 2}{4} - \frac{2t + 3}{3} = \frac{2}{3} - t$
6.  $m - \frac{m - 1}{2} = 1 - \frac{m - 2}{3}$

Simplify and solve the following linear equations.

7.  $3(t - 3) = 5(2t + 1)$
8.  $15(y - 4) - 2(y - 9) + 5(y + 6) = 0$
9.  $3(5z - 7) - 2(9z - 11) = 4(8z - 13) - 17$
10.  $0.25(4f - 3) = 0.05(10f - 9)$

Accepted Answer

1. Solve $\frac{x}{2} - \frac{1}{5} = \frac{x}{3} + \frac{1}{4}$
Multiply both sides by 60 (LCM of 2,5,3,4):
$30x - 12 = 20x + 15$
$30x - 20x = 15 + 12$
$10x = 27$
$x = \frac{27}{10} = 2.7$

2. Solve $\frac{n}{2} - \frac{3n}{4} + \frac{5n}{6} = 21$
LCM of 2,4,6 is 12:
$6n - 9n + 10n = 252$
$7n = 252$
$n = 36$

3. Solve $x + 7 - \frac{8x}{3} = \frac{17}{6} - \frac{5x}{2}$
Multiply both sides by 6:
$6x + 42 - 16x = 17 - 15x$
$6x - 16x + 15x = 17 - 42$
$5x = -25$
$x = -5$

4. Solve $\frac{x - 5}{3} = \frac{x - 3}{5}$
Cross multiply:
$5(x - 5) = 3(x - 3)$
$5x - 25 = 3x - 9$
$5x - 3x = -9 + 25$
$2x = 16$
$x = 8$

5. Solve $\frac{3t - 2}{4} - \frac{2t + 3}{3} = \frac{2}{3} - t$
Multiply both sides by 12:
$3(3t - 2) - 4(2t + 3) = 8 - 12t$
$9t - 6 - 8t - 12 = 8 - 12t$
$t - 18 = 8 - 12t$
$t + 12t = 8 + 18$
$13t = 26$
$t = 2$

6. Solve $m - \frac{m - 1}{2} = 1 - \frac{m - 2}{3}$
Multiply both sides by 6:
$6m - 3(m - 1) = 6 - 2(m - 2)$
$6m - 3m + 3 = 6 - 2m + 4$
$3m + 3 = 10 - 2m$
$3m + 2m = 10 - 3$
$5m = 7$
$m = \frac{7}{5} = 1.4$

7. Solve $3(t - 3) = 5(2t + 1)$
$3t - 9 = 10t + 5$
$3t - 10t = 5 + 9$
$-7t = 14$
$t = -2$

8. Solve $15(y - 4) - 2(y - 9) + 5(y + 6) = 0$
$15y - 60 - 2y + 18 + 5y + 30 = 0$
$15y - 2y + 5y - 60 + 18 + 30 = 0$
$18y - 12 = 0$
$18y = 12$
$y = \frac{12}{18} = \frac{2}{3}$

9. Solve $3(5z - 7) - 2(9z - 11) = 4(8z - 13) - 17$
$15z - 21 - 18z + 22 = 32z - 52 - 17$
$-3z + 1 = 32z - 69$
$-3z - 32z = -69 - 1$
$-35z = -70$
$z = 2$

10. Solve $0.25(4f - 3) = 0.05(10f - 9)$
$0.25 \times 4f - 0.25 \times 3 = 0.05 \times 10f - 0.05 \times 9$
$f - 0.75 = 0.5f - 0.45$
$f - 0.5f = -0.45 + 0.75$
$0.5f = 0.3$
$f = 0.6$ — Each equation is solved by first eliminating denominators or brackets by multiplying both sides by the LCM of denominators or expanding brackets. Then, like terms are collected on one side and constants on the other. Finally, the variable is isolated by dividing both sides by the coefficient of the variable.

Question 3

Which of the following is a linear expression in one variable?

Accepted Answer

2x + 3 — A linear expression in one variable has the highest power of the variable as 1. Among the options, only $2x + 3$ has the variable $x$ raised to the power 1. The others have variables raised to powers greater than 1, so they are not linear expressions.

Question 4

Identify the parts labeled in the algebraic expression $3x + 5$ where '3' is the coefficient, 'x' is the variable, and '5' is the constant.

Accepted Answer

The number 3 is the coefficient, 'x' is the variable, and 5 is the constant in the expression $3x + 5$. — In the expression $3x + 5$, the coefficient is the numerical factor multiplying the variable, which is 3. The variable is the letter 'x' representing an unknown quantity. The constant is the fixed number 5 without any variable.

Question 5

Define an algebraic equation and explain the terms LHS and RHS with an example.

Accepted Answer

An algebraic equation is an equality involving variables with an equality sign. The expression on the left side of the equality sign is called the Left Hand Side (LHS), and the expression on the right side is called the Right Hand Side (RHS). For example, in the equation $2x - 3 = 7$, $2x - 3$ is the LHS and 7 is the RHS. — An algebraic equation states that two expressions are equal. The LHS and RHS are the two expressions separated by the equality sign. Understanding these terms helps in solving equations by manipulating both sides equally.

Question 6

Solve the linear equation $2x - 3 = 7$ and verify the solution.

Accepted Answer

x = 5 — Given: 2x - 3 = 7
Find: x
Formula: Solve for x by isolating the variable.
Solution:
Step 1: Add 3 to both sides: 2x = 7 + 3 = 10
Step 2: Divide both sides by 2: x = 10 / 2 = 5
Check:
LHS = 2 × 5 - 3 = 10 - 3 = 7
RHS = 7
LHS = RHS, so x = 5 is the solution.

Question 7

In the equation $2x - 3 = x + 2$, which step correctly isolates the variable $x$?

Accepted Answer

2x = x + 2 + 3 — To isolate $x$, add 3 to both sides to balance the equation: $2x - 3 + 3 = x + 2 + 3$ which simplifies to $2x = x + 5$. This is the correct step to start isolating $x$.

Question 8

Solve the equation $5x + \frac{7}{2} = \frac{3}{2} x - 14$.

Accepted Answer

x = -5 — Given: 5x + 7/2 = 3/2 x - 14
Find: x
Formula: Multiply both sides by 2 to clear denominators.
Solution:
Step 1: Multiply both sides by 2: 2 × (5x + 7/2) = 2 × (3/2 x - 14)
Step 2: Simplify: 10x + 7 = 3x - 28
Step 3: Transpose 3x to LHS: 10x - 3x + 7 = -28
Step 4: Simplify: 7x + 7 = -28
Step 5: Subtract 7: 7x = -35
Step 6: Divide by 7: x = -35 / 7 = -5
Answer: x = -5

Question 9

Which of the following expressions is NOT a linear expression in one variable?

Accepted Answer

x^2 + 1 — Linear expressions have variables raised only to the first power. $x^2 + 1$ has the variable $x$ raised to the power 2, so it is not linear. The other options have variables raised to power 1 or are constants.

Question 10

Explain why $x = 10$ is not a solution of the equation $2x - 3 = 7$.

Accepted Answer

For $x = 10$, LHS = $2 	imes 10 - 3 = 20 - 3 = 17$, which is not equal to RHS = 7. Hence, $x = 10$ does not satisfy the equation and is not a solution. — A solution of an equation must make the LHS equal to the RHS. Substituting $x = 10$ results in unequal sides, so it is not a solution.

Question 11

In solving $2x - 3 = x + 2$, what operation is performed when subtracting $x$ from both sides?

Accepted Answer

Transposing $x$ to the left-hand side — Subtracting $x$ from both sides is equivalent to transposing $x$ from the right-hand side to the left-hand side with a change of sign. This helps collect all variable terms on one side.

Question 12

Solve the equation $\frac{6x + 1}{3} + 1 = \frac{x - 3}{6}$ by reducing it to a simpler form.

Accepted Answer

x = -1 — Given: (6x + 1)/3 + 1 = (x - 3)/6
Find: x
Formula: Multiply both sides by LCM of denominators (6).
Solution:
Step 1: Multiply both sides by 6:
6 × ((6x + 1)/3) + 6 × 1 = 6 × ((x - 3)/6)
Step 2: Simplify:
2(6x + 1) + 6 = x - 3
Step 3: Open brackets:
12x + 2 + 6 = x - 3
Step 4: Combine like terms:
12x + 8 = x - 3
Step 5: Transpose x:
12x - x + 8 = -3
11x + 8 = -3
Step 6: Subtract 8:
11x = -11
Step 7: Divide by 11:
x = -11 / 11 = -1
Answer: x = -1

Question 13

Explain the importance of multiplying both sides of an equation by the LCM of denominators when solving equations like $\frac{6x + 1}{3} + 1 = \frac{x - 3}{6}$.

Accepted Answer

Multiplying both sides by the LCM of denominators clears the fractions, converting the equation into one without fractions. This simplifies the equation and makes it easier to solve. — Fractions complicate solving equations. Multiplying by the LCM eliminates denominators, allowing standard algebraic operations on simpler expressions.

Question 14

Solve the equation $5x - 2(2x - 7) = 2(3x - 1) + \frac{7}{2}$ and verify the solution.

Accepted Answer

x = \frac{5}{2} — Given: 5x - 2(2x - 7) = 2(3x - 1) + 7/2
Find: x
Formula: Open brackets and simplify both sides.
Solution:
Step 1: LHS: 5x - 4x + 14 = x + 14
Step 2: RHS: 6x - 2 + 7/2 = 6x - 4/2 + 7/2 = 6x + 3/2
Step 3: Equation: x + 14 = 6x + 3/2
Step 4: Transpose terms: 14 - 3/2 = 6x - x
Step 5: Simplify left: (28/2) - (3/2) = 5x
Step 6: 25/2 = 5x
Step 7: Divide both sides by 5: x = (25/2) × (1/5) = 5/2
Check:
LHS = 5 × 5/2 - 2(2 × 5/2 - 7) = 25/2 - 2(5 - 7) = 25/2 - 2(-2) = 25/2 + 4 = 33/2
RHS = 2(3 × 5/2 - 1) + 7/2 = 2(15/2 - 1) + 7/2 = 2(13/2) + 7/2 = 26/2 + 7/2 = 33/2
LHS = RHS, solution verified.

Question 15

Match the following algebraic expressions with their correct classification as linear or non-linear:

Accepted Answer

— Linear expressions have variables raised to power 1. Non-linear expressions have variables raised to powers greater than 1.

Some examples of expressions we have so far worked with are

Some examples of expressions we have so far worked with are — Study Notes

2.1 Introduction

2.2 Solving Equations having the Variable on both Sides

EXERCISE 2.1

Practice Questions — Some examples of expressions we have so far worked with are

All 13 Chapters in Mathematics