Model Textbook of Mathematics Grade 7 (FBISE / NBF)
Class 7 Mathematics Federal Board of Intermediate and Secondary Education (FBISE) (FBISE) 📚 Model Textbook of Mathematics Grade 7 (FBISE / NBF)

Class 7 Mathematics Ch 8 Mastery Guide: Linear Equations in One & Two Variables, Literal Equations, Graphing & Simultaneous Equations (FBISE)

📖 Chapter 8: Linear Equations 📅 Updated: Sep 12, 2026
Teacher & Parent Roadmap Grade 7 Mathematics • FBISE / National Curriculum (NBF)

Instructional Guide: Unit 08 Linear Equations

Target Learning Outcomes (SLOs)
  • Define and differentiate between algebraic expressions and linear algebraic equations.
  • Solve linear equations in one variable using balance properties of equality, brackets expansion, and cross-multiplication.
  • Rearrange and solve literal equations and scientific/geometric formulas for specified target variables.
  • Translate real-world word problems into linear algebraic equations and solve them.
  • Understand standard form of linear equations in two variables ($ax + by = c$) and test/find ordered pairs $(x, y)$.
  • Graph linear equations on the Cartesian plane and classify lines as horizontal ($y=c$), vertical ($x=c$), or slanted.
  • Solve systems of simultaneous linear equations using the Elimination Method and the Substitution Method.
Prerequisites & Bridge Concepts
  • Integers & Arithmetic: Rules for adding, subtracting, multiplying, and dividing positive and negative numbers.
  • Algebraic Expressions: Combining like terms, distributing constants across brackets ($a(b+c) = ab + ac$).
  • Cartesian Plane: Plotting ordered pairs $(x, y)$ across quadrants and along axes.
Common Misconceptions & Pitfalls
  • Sign Inversion: Forgetting that moving a term across the equals sign changes its sign ($+ \to -$ and $- \to +$).
  • Bracket Distribution: Overlooking the negative sign in expressions like $16 - (2x - 3) = 16 - 2x + 3$.
  • Horizontal vs Vertical Lines: Confusing $x = c$ (vertical line parallel to $y$-axis) with $y = c$ (horizontal line parallel to $x$-axis).
  • Simultaneous Elimination Signs: Subtracting entire equations without distributing the minus sign to every single term.
Pedagogical Tips & Real-World Connections

Illustrate equations as a two-pan balance scale: whatever operation you perform on one pan must be identically performed on the other to preserve perfect balance. Connect simultaneous equations to practical real-world scenarios such as ticket pricing (adult vs child tickets) and finding the break-even point for small business projects.

1. Algebraic Equations & Linear Equations in One Variable

An algebraic expression is a combination of variables, constants, and arithmetic operations (e.g., $4x + 7$). When two algebraic expressions are joined by an equals sign ($=$), it forms an algebraic equation (e.g., $4x + 7 = 19$).

Key Definitions & Properties

  • Linear Equation in One Variable: An algebraic equation where the highest exponent of the variable is strictly $1$. Standard form: $ax + b = 0$ (where $a \ne 0$).
  • Solution (Root): The specific numerical value of the variable that makes the equality statement true.
  • Properties of Equality (Balance Scale Rules):
    • Addition Property: If $a = b$, then $a + c = b + c$.
    • Subtraction Property: If $a = b$, then $a - c = b - c$.
    • Multiplication Property: If $a = b$, then $a \cdot c = b \cdot c$.
    • Division Property: If $a = b$ and $c \ne 0$, then $\frac{a}{c} = \frac{b}{c}$.
  • Literal Equation: An equation consisting primarily of multiple letters/variables representing geometric or physical formulas (such as $P = 2l + 2w$, $V = lwh$, $C = \frac{5}{9}(F - 32)$). Solving a literal equation means isolating one chosen variable in terms of all other variables.

2. Exercise 8.1 — 100% Step-by-Step Solved

Question 1: Solve the following linear equations.
(i) $2x + 5 = 19$
Subtract 5 from both sides:
$2x = 19 - 5 = 14$
Divide by 2:
$x = \frac{14}{2} = \mathbf{7}$
✓ Check: $2(7) + 5 = 14 + 5 = 19$.
(ii) $5x - 7 = 2x + 3$
Group $x$-terms on LHS, constants on RHS:
$5x - 2x = 3 + 7$
$3x = 10 \implies x = \mathbf{\frac{10}{3}}$ (or $3\frac{1}{3}$)
✓ Check: $5(\frac{10}{3}) - 7 = \frac{29}{3} = 2(\frac{10}{3}) + 3$.
(iii) $19x - 13 = 11x + 27$
Subtract $11x$ and add 13:
$19x - 11x = 27 + 13$
$8x = 40 \implies x = \frac{40}{8} = \mathbf{5}$
✓ Check: $19(5) - 13 = 82 = 11(5) + 27$.
(iv) $9y + 17 = 6y + 23$
Subtract $6y$ and subtract 17:
$9y - 6y = 23 - 17$
$3y = 6 \implies y = \frac{6}{3} = \mathbf{2}$
✓ Check: $9(2) + 17 = 35 = 6(2) + 23$.
(v) $2x + 3 = 16 - (2x - 3)$
Expand brackets on RHS:
$2x + 3 = 16 - 2x + 3 = 19 - 2x$
Add $2x$ and subtract 3:
$4x = 19 - 3 = 16 \implies x = \frac{16}{4} = \mathbf{4}$
✓ Check: LHS $= 2(4)+3=11$; RHS $= 16-(5)=11$.
(vi) $\frac{x}{2} + 3 = \frac{1}{2}$
Subtract 3 from both sides:
$\frac{x}{2} = \frac{1}{2} - 3 = \frac{1 - 6}{2} = -\frac{5}{2}$
Multiply both sides by 2:
$x = \mathbf{-5}$
✓ Check: $-\frac{5}{2} + 3 = -2.5 + 3 = 0.5 = \frac{1}{2}$.
(vii) $8\left(\frac{x}{2} - 3\right) = 7x + 2$
Distribute 8 inside parentheses:
$4x - 24 = 7x + 2$
Subtract $7x$ and add 24:
$-3x = 26 \implies x = \mathbf{-\frac{26}{3}}$ (or $-8\frac{2}{3}$)
✓ Check: LHS $= 8(-\frac{22}{3}) = -\frac{176}{3} = \text{RHS}$.
(viii) $x + \frac{4}{5} = 2x - \frac{1}{3}$
Rearrange terms:
$\frac{4}{5} + \frac{1}{3} = 2x - x$
$x = \frac{12 + 5}{15} = \mathbf{\frac{17}{15}}$ (or $1\frac{2}{15}$)
✓ Check: LHS $= \frac{29}{15} = \text{RHS}$.
(ix) $\frac{x}{2} - 3 = 4 - \frac{2x}{3}$
Group $x$-terms on LHS:
$\frac{x}{2} + \frac{2x}{3} = 4 + 3 = 7$
$\frac{3x + 4x}{6} = 7 \implies \frac{7x}{6} = 7 \implies x = \mathbf{6}$
✓ Check: $\frac{6}{2} - 3 = 0 = 4 - \frac{2(6)}{3}$.
(x) $\frac{2x + 5}{x} = \frac{7}{3}$
Cross-multiply denominators:
$3(2x + 5) = 7x$
$6x + 15 = 7x \implies x = \mathbf{15}$
✓ Check: $\frac{2(15)+5}{15} = \frac{35}{15} = \frac{7}{3}$.
(xi) $\frac{y}{9y - 2} = \frac{1}{8}$
Cross-multiply:
$8y = 9y - 2 \implies -y = -2 \implies y = \mathbf{2}$
✓ Check: $\frac{2}{9(2)-2} = \frac{2}{16} = \frac{1}{8}$.
(xii) $\frac{x}{x - 12} = \frac{9}{5}$
Cross-multiply:
$5x = 9(x - 12) \implies 5x = 9x - 108$
$-4x = -108 \implies x = \mathbf{27}$
✓ Check: $\frac{27}{27-12} = \frac{27}{15} = \frac{9}{5}$.
(xiii) $\frac{24}{5z + 4} = \frac{4}{z - 1}$
Divide numerators by 4: $\frac{6}{5z + 4} = \frac{1}{z - 1}$
Cross-multiply: $6(z - 1) = 5z + 4$
$6z - 6 = 5z + 4 \implies z = \mathbf{10}$
✓ Check: $\frac{24}{54} = \frac{4}{9} = \frac{4}{10-1}$.
(xiv) $\frac{2}{-3} = \frac{4x + 4}{2x + 14}$
Cross-multiply:
$2(2x + 14) = -3(4x + 4)$
$4x + 28 = -12x - 12 \implies 16x = -40$
$x = -\frac{40}{16} = \mathbf{-\frac{5}{2}}$ (or $-2.5$)
✓ Check: $\frac{-6}{9} = -\frac{2}{3}$.
Word Problems (Questions 2 – 4)
Question 2: $\left(\frac{x}{2} + 7\right)$ is three times of $\left(\frac{x}{5} + 2\right)$. Find the value of $x$.
Step 1 (Equation): $\frac{x}{2} + 7 = 3\left(\frac{x}{5} + 2\right) = \frac{3x}{5} + 6$
Step 2 (Rearrange): $7 - 6 = \frac{3x}{5} - \frac{x}{2} \implies 1 = \frac{6x - 5x}{10} = \frac{x}{10}$
Step 3 (Solve): $x = 1 \times 10 = \mathbf{10}$.
✓ Verification: LHS $= \frac{10}{2}+7 = 12$; RHS $= 3(\frac{10}{5}+2) = 3(4) = 12$.
Question 3: When a number is added to two-thirds of itself, the result is 45. Find the number.
Step 1: Let the unknown number be $x$.
Step 2: $x + \frac{2}{3}x = 45 \implies \left(1 + \frac{2}{3}\right)x = 45 \implies \frac{5}{3}x = 45$
Step 3: Multiply by $\frac{3}{5}$: $x = 45 \times \frac{3}{5} = 9 \times 3 = \mathbf{27}$.
✓ Verification: $27 + \frac{2}{3}(27) = 27 + 18 = 45$.
Question 4: The numerator of a fraction is 5 less than the denominator. If 1 is added to both the numerator and the denominator, the fraction would become $\frac{2}{3}$. Find the fraction.
Step 1: Let denominator $= d$. Then numerator $= d - 5$. Fraction $= \frac{d - 5}{d}$.
Step 2: Add 1 to both: $\frac{(d - 5) + 1}{d + 1} = \frac{2}{3} \implies \frac{d - 4}{d + 1} = \frac{2}{3}$
Step 3: Cross-multiply: $3(d - 4) = 2(d + 1) \implies 3d - 12 = 2d + 2 \implies d = 14$.
Step 4: Numerator $= 14 - 5 = 9$. Thus, the original fraction is $\mathbf{\frac{9}{14}}$.
✓ Verification: $\frac{9 + 1}{14 + 1} = \frac{10}{15} = \frac{2}{3}$.
Literal Equations & Formula Rearrangements (Questions 5 – 10)
Question 5: $5x + 4y = 20$
• (i) In terms of $y$: $4y = 20 - 5x \implies \mathbf{y = \frac{20 - 5x}{4}}$
• (ii) In terms of $x$: $5x = 20 - 4y \implies \mathbf{x = \frac{20 - 4y}{5}}$
Question 6: Perimeter $P = 2l + 2w$
• (i) Solve for $w$: $2w = P - 2l \implies \mathbf{w = \frac{P - 2l}{2} = \frac{P}{2} - l}$
• (ii) For $P = 100, l = 25$: $w = \frac{100 - 50}{2} = \mathbf{25\text{ units}}$.
Question 7: $C = \frac{5}{9}(F - 32)$
• (i) In terms of $C$: $\frac{9}{5}C = F - 32 \implies \mathbf{F = \frac{9}{5}C + 32}$
• (ii) At $C = 15^\circ\text{C}$: $F = \frac{9}{5}(15) + 32 = 27 + 32 = \mathbf{59^\circ\text{F}}$
• At $C = 10^\circ\text{C}$: $F = \frac{9}{5}(10) + 32 = 18 + 32 = \mathbf{50^\circ\text{F}}$.
Question 8: Cuboid Volume $V = lwh$
Divide both sides by $lh$:
$\mathbf{w = \frac{V}{lh}}$
Question 9: Prism Area $S = 2B + Ph$
Subtract $2B$: $Ph = S - 2B$
Divide by $P$: $\mathbf{h = \frac{S - 2B}{P}}$
Question 10: Solve $x = \frac{a+b+c}{ab}$ for $a$
Multiply by $ab$: $abx = a + b + c$
Collect $a$: $abx - a = b + c \implies a(bx - 1) = b + c$
Divide by $(bx - 1)$: $\mathbf{a = \frac{b + c}{bx - 1}}$

3. Linear Equations in Two Variables & Ordered Pairs

An equation of the form $\mathbf{ax + by = c}$ (where $a, b, c$ are real numbers and $a, b \ne 0$) is called a linear equation in two variables ($x$ and $y$).

  • Solution as an Ordered Pair: A pair of real numbers $(x_0, y_0)$ is a solution if substituting $x = x_0$ and $y = y_0$ into the equation satisfies the equality $\text{LHS} = \text{RHS}$.
  • Infinite Solutions: While a linear equation in one variable has a single unique root, a linear equation in two variables has infinitely many solutions, because for any arbitrary value chosen for $x$, there is a corresponding value of $y$.

4. Exercise 8.2 — 100% Step-by-Step Solved

Question 1: Select the ordered pairs which are solution points to the corresponding equations.
(i) $3x - y = -8$
• $(-1, 10): 3(-1) - 10 = -13 \ne -8$
• $(3, 10): 3(3) - 10 = -1 \ne -8$
• $(-2, 2): 3(-2) - 2 = -8 = \text{RHS}$ ✓
• $(-1, 5): 3(-1) - 5 = -8 = \text{RHS}$ ✓
Solutions: $\mathbf{(-2, 2)}$ and $\mathbf{(-1, 5)}$.
(ii) $5x - 3y = 11$
• $(1, -2): 5(1) - 3(-2) = 11 = \text{RHS}$ ✓
• $(3, 1): 15 - 3 = 12 \ne 11$
• $(-2, -7): -10 + 21 = 11 = \text{RHS}$ ✓
• $(7, 8): 35 - 24 = 11 = \text{RHS}$ ✓
Solutions: $\mathbf{(1, -2)}$, $\mathbf{(-2, -7)}$, and $\mathbf{(7, 8)}$.
(iii) $2x - y = -1$
• $(0, 1): 2(0) - 1 = -1 = \text{RHS}$ ✓
• $(2, 5): 4 - 5 = -1 = \text{RHS}$ ✓
• $(0, 0): 0 \ne -1$
• $(-2, -3): -4 - (-3) = -1 = \text{RHS}$ ✓
Solutions: $\mathbf{(0, 1)}$, $\mathbf{(2, 5)}$, and $\mathbf{(-2, -3)}$.
(iv) $3x - 2y = 6$
• $(4, 0): 12 \ne 6$
• $(2, 0): 3(2) - 0 = 6 = \text{RHS}$ ✓
• $(3, -2): 9 + 4 = 13 \ne 6$
• $(3, -1): 9 + 2 = 11 \ne 6$
Solution: $\mathbf{(2, 0)}$.
(v) $2x + y = 4$
• $(2, 3): 4 + 3 = 7 \ne 4$
• $(1, \frac{1}{2}): 2 + 0.5 = 2.5 \ne 4$
• $(0, 4): 2(0) + 4 = 4 = \text{RHS}$ ✓
• $(-2, 2): -4 + 2 = -2 \ne 4$
Solution: $\mathbf{(0, 4)}$.
(vi) $5x - 2y = -50$
• $(-10, 0): 5(-10) - 0 = -50 = \text{RHS}$ ✓
• $(-5, 5): -25 - 10 = -35 \ne -50$
• $(0, 25): 0 - 50 = -50 = \text{RHS}$ ✓
• $(20, -2): 100 + 4 = 104 \ne -50$
Solutions: $\mathbf{(-10, 0)}$ and $\mathbf{(0, 25)}$.
Question 2: Find any three ordered pairs that are solutions for each equation.
(i) $y - x = 3 \implies y = x + 3$
• For $x = 0 \implies y = 3 \implies \mathbf{(0, 3)}$
• For $x = 1 \implies y = 4 \implies \mathbf{(1, 4)}$
• For $x = -1 \implies y = 2 \implies \mathbf{(-1, 2)}$
(ii) $3x + 2y = -6 \implies y = \frac{-6 - 3x}{2}$
• For $x = 0 \implies y = -3 \implies \mathbf{(0, -3)}$
• For $x = -2 \implies y = 0 \implies \mathbf{(-2, 0)}$
• For $x = 2 \implies y = -6 \implies \mathbf{(2, -6)}$
(iii) $2x + y = 8 \implies y = 8 - 2x$
• For $x = 0 \implies y = 8 \implies \mathbf{(0, 8)}$
• For $x = 2 \implies y = 4 \implies \mathbf{(2, 4)}$
• For $x = 4 \implies y = 0 \implies \mathbf{(4, 0)}$
(iv) $x - 2y = 3 \implies x = 2y + 3$
• For $y = 0 \implies x = 3 \implies \mathbf{(3, 0)}$
• For $y = 1 \implies x = 5 \implies \mathbf{(5, 1)}$
• For $y = -1 \implies x = 1 \implies \mathbf{(1, -1)}$

5. Graphing Linear Equations on the Cartesian Plane

The geometric representation of every linear equation is a straight line. We can graph any linear equation by creating a table of values and connecting the plotted points with a straight line.

Horizontal Lines: $y = c$

The $y$-value is constantly $c$ regardless of $x$. The line is parallel to the $x$-axis. (If $c = 0$, it is the $x$-axis itself).

Vertical Lines: $x = c$

The $x$-value is constantly $c$ regardless of $y$. The line is parallel to the $y$-axis. (If $c = 0$, it is the $y$-axis itself).

Slanted Lines: $y = mx + c$

Contains both $x$ and $y$. If $c = 0$ ($y = mx$), the line passes through the origin $(0, 0)$.

6. Exercise 8.3 — 100% Step-by-Step Solved

Question 1: Graph the equations with the given values of $x$.
(i) $y = -x + 1$ with $x \in \{-2, -1, 0, 1, 2\}$
$x$-2-1012
$y$3210-1
Points: $(-2, 3), (-1, 2), (0, 1), (1, 0), (2, -1)$.
(ii) $y = 2x - 5$ with $x \in \{-1, 0, 1, 3, 4\}$
$x$-10134
$y$-7-5-313
Points: $(-1, -7), (0, -5), (1, -3), (3, 1), (4, 3)$.
(iii) $y - 2x = 5 \implies y = 2x + 5$ with $x \in \{-3, -2, -1, 0, 1\}$
$x$-3-2-101
$y$-11357
Points: $(-3, -1), (-2, 1), (-1, 3), (0, 5), (1, 7)$.
(iv) $3x + 2y = 1 \implies y = \frac{1 - 3x}{2}$ with $x \in \{-2, 0, 2, 4\}$
$x$-2024
$y$3.50.5-2.5-5.5
Points: $(-2, 3.5), (0, 0.5), (2, -2.5), (4, -5.5)$.
Question 2: Graph the equations by setting five values of $x$.
(i) $x + 2y = -2 \implies y = \frac{-2 - x}{2}$
Table: $(-4, 1), (-2, 0), (0, -1), (2, -2), (4, -3)$
(ii) $-x + 4y = 12 \implies y = \frac{12 + x}{4}$
Table: $(-4, 2), (0, 3), (4, 4), (8, 5), (-8, 1)$
(iii) $x + 2y = -10 \implies y = \frac{-10 - x}{2}$
Table: $(-4, -3), (-2, -4), (0, -5), (2, -6), (4, -7)$
(iv) $3x - 5y = 15 \implies y = \frac{3x - 15}{5}$
Table: $(-5, -6), (0, -3), (5, 0), (10, 3), (15, 6)$
(v) $2x + 3y = 30 \implies y = \frac{30 - 2x}{3}$
Table: $(-3, 12), (0, 10), (3, 8), (6, 6), (9, 4)$
(vi) $2x + 5y = 6 \implies y = \frac{6 - 2x}{5}$
Table: $(-2, 2), (3, 0), (8, -2), (-7, 4), (13, -4)$
(vii) $6x + 2y = 6 \implies y = 3 - 3x$
Table: $(-1, 6), (0, 3), (1, 0), (2, -3), (3, -6)$
(viii) $2x + 4y = -8 \implies y = -2 - \frac{x}{2}$
Table: $(-4, 0), (-2, -1), (0, -2), (2, -3), (4, -4)$
(ix) $-3x + 4y = -12 \implies y = \frac{3x - 12}{4}$
Table: $(-4, -6), (0, -3), (4, 0), (8, 3), (12, 6)$
(x) $3x - 4y = 36 \implies y = \frac{3x - 36}{4}$
Table: $(0, -9), (4, -6), (8, -3), (12, 0), (16, 3)$
Question 3: Determine whether the graph of each given equation is horizontal, vertical, or neither.
(i) $x = 4$: Vertical (Parallel to $y$-axis)
(ii) $y = -5$: Horizontal (Parallel to $x$-axis)
(iii) $4x = -12 \implies x = -3$: Vertical
(iv) $x - y = 0 \implies y = x$: Neither (Slanted through origin)
(v) $-3y = -15 \implies y = 5$: Horizontal
(vi) $y = 2x - 3$: Neither (Slanted line)
(vii) $x = 0y \implies x = 0$: Vertical (The $y$-axis itself)
(viii) $x = 3y \implies y = \frac{1}{3}x$: Neither (Slanted through origin)

7. Systems of Simultaneous Linear Equations

When two linear equations in two variables are considered together to find a single common ordered pair $(x, y)$ that satisfies both simultaneously, they are called a system of simultaneous linear equations.

Method 1: Elimination Method
  1. Multiply one or both equations by suitable constants so the coefficients of one variable become equal (or opposites).
  2. Add or subtract the two equations to completely eliminate that variable, leaving an equation in one variable.
  3. Solve for the remaining variable, then substitute back to find the other variable.
Method 2: Substitution Method
  1. Pick one equation and isolate one variable (e.g., $x$ in terms of $y$).
  2. Substitute this expression into the other equation, converting it into a single-variable equation.
  3. Solve for that variable, then plug it into the isolated formula to calculate the second variable.

8. Exercise 8.4 — 100% Step-by-Step Solved

Question 1: Solve each system of equations by the ELIMINATION METHOD.
(i) $2x + y = 3$ • $3x - 2y = -6$
• Multiply (1) by 2: $4x + 2y = 6$
• Add to (2): $7x = 0 \implies \mathbf{x = 0}$
• Substitute: $2(0) + y = 3 \implies \mathbf{y = 3}$
Solution: $\mathbf{(0, 3)}$.
(ii) $7x - 2y = 1$ • $3x + 4y = 15$
• Multiply (1) by 2: $14x - 4y = 2$
• Add to (2): $17x = 17 \implies \mathbf{x = 1}$
• Substitute: $7(1) - 2y = 1 \implies 2y = 6 \implies \mathbf{y = 3}$
Solution: $\mathbf{(1, 3)}$.
(iii) $9x + 4y = 5$ • $3x - 5y = 8$
• Multiply (2) by 3: $9x - 15y = 24$
• Subtract from (1): $19y = -19 \implies \mathbf{y = -1}$
• Substitute: $3x - 5(-1) = 8 \implies 3x = 3 \implies \mathbf{x = 1}$
Solution: $\mathbf{(1, -1)}$.
(iv) $2x + 3y = 5$ • $3x + 2y = 10$
• Multiply (1) by 3: $6x + 9y = 15$
• Multiply (2) by 2: $6x + 4y = 20$
• Subtract: $5y = -5 \implies \mathbf{y = -1}$
• Substitute: $2x - 3 = 5 \implies 2x = 8 \implies \mathbf{x = 4}$
Solution: $\mathbf{(4, -1)}$.
Question 2: Solve each system of equations by the SUBSTITUTION METHOD.
(i) $x + 4y = 2$ • $5x - 4y = 10$
• From (1): $x = 2 - 4y$
• In (2): $5(2 - 4y) - 4y = 10 \implies 10 - 24y = 10 \implies \mathbf{y = 0}$
• $x = 2 - 4(0) = \mathbf{2}$
Solution: $\mathbf{(2, 0)}$.
(ii) $x - y = 5$ • $x + y = 19$
• From (1): $x = y + 5$
• In (2): $(y + 5) + y = 19 \implies 2y = 14 \implies \mathbf{y = 7}$
• $x = 7 + 5 = \mathbf{12}$
Solution: $\mathbf{(12, 7)}$.
(iii) $3x - 5y = -1$ • $x - y = -1$
• From (2): $x = y - 1$
• In (1): $3(y - 1) - 5y = -1 \implies -2y = 2 \implies \mathbf{y = -1}$
• $x = -1 - 1 = \mathbf{-2}$
Solution: $\mathbf{(-2, -1)}$.
(iv) $3x - 7y = -10$ • $y - 2x = 3$
• From (2): $y = 2x + 3$
• In (1): $3x - 7(2x + 3) = -10 \implies -11x = 11 \implies \mathbf{x = -1}$
• $y = 2(-1) + 3 = \mathbf{1}$
Solution: $\mathbf{(-1, 1)}$.
Question 3: Solve the following simultaneous equations.
(i) $3x - y = 12$ • $2x + y = 13$
Add equations: $5x = 25 \implies \mathbf{x = 5}, \mathbf{y = 3}$. Solution: $\mathbf{(5, 3)}$.
(ii) $2x + y = 3$ • $3x - 2y = 1$
Multiply 1st by 2, add: $7x = 7 \implies \mathbf{x = 1}, \mathbf{y = 1}$. Solution: $\mathbf{(1, 1)}$.
(iii) $13x - 6y = 20$ • $7x + 4y = 18$
$26x - 12y = 40$ & $21x + 12y = 54 \implies 47x = 94 \implies \mathbf{x = 2}, \mathbf{y = 1}$. Solution: $\mathbf{(2, 1)}$.
(iv) $x + y = 3$ • $x - y = 7$
Add: $2x = 10 \implies \mathbf{x = 5}$; Subtract: $2y = -4 \implies \mathbf{y = -2}$. Solution: $\mathbf{(5, -2)}$.
(v) $x + \frac{y}{2} = 4$ • $\frac{x}{3} + 2y = 5$
$2x + y = 8$ & $x + 6y = 15 \implies \mathbf{x = 3}, \mathbf{y = 2}$. Solution: $\mathbf{(3, 2)}$.
(vi) $3x - 2y - 9 = 0$ • $2x - 2y - 7 = 0$
Subtract: $x = 2 \implies -2y = 3 \implies \mathbf{y = -\frac{3}{2}}$. Solution: $\mathbf{(2, -\frac{3}{2})}$.
(vii) $(x+y) = \frac{2}{5}(x-y)$ • $3x + 11y = 4$
$3x + 7y = 0$ & $3x + 11y = 4 \implies 4y = 4 \implies \mathbf{y = 1}, \mathbf{x = -\frac{7}{3}}$. Solution: $\mathbf{(-\frac{7}{3}, 1)}$.
(viii) $\frac{x+1}{y+2} = \frac{2}{3}$ • $\frac{x-2}{y-1} = \frac{1}{3}$
$3x - 2y = 1$ & $3x - y = 5 \implies \mathbf{y = 4}, \mathbf{x = 3}$. Solution: $\mathbf{(3, 4)}$.

9. Review Exercise 8 — 100% Step-by-Step Solved

Question 1: Multiple Choice Questions (MCQs)
(i) Which is an algebraic equation?
(a) $3x = 8$ (Contains variable and equals sign).
(ii) Product of $y$ and 6 is 60. What is $y$?
(c) 10 ($6y = 60 \implies y = 10$).
(iii) In a linear equation, degree of polynomial is:
(c) 1.
(iv) $4x + 7$ is:
(b) an expression (No equality sign).
(v) Solution of linear equation means finding:
(d) unknown quantity.
(vi) In $4y - 1 = 15$, the value of $y$ is:
(d) 4 ($4y = 16 \implies y = 4$).
(vii) "3 added to 6 times a number gives 9":
(a) $6x + 3 = 9$.
(viii) Solution of $2x + 3 = 9x + 4$:
(b) $-\frac{1}{7}$ ($-7x = 1 \implies x = -\frac{1}{7}$).
(ix) Simple linear equation in ONE variable:
(a) $7x + 2 = 20$.
(x) One solution of $\frac{1}{2}x + y = 7$:
(b) $(2, 6)$ ($\frac{1}{2}(2) + 6 = 1 + 6 = 7$).
(xi) Linear equation in two variables represents:
(a) straight line.
(xii) $(-2, -\frac{1}{2})$ is the solution of:
(b) $-x + 4y = 0$ ($2 + 4(-0.5) = 0$).
(xiii) Intersection of $x = 3$ and $y = 4$:
(a) $(3, 4)$.
Subjective Questions (Q2 – Q7)
Q2: Solve $\frac{x - 8}{3} = \frac{x - 3}{5}$
$5(x - 8) = 3(x - 3) \implies 5x - 40 = 3x - 9$
$2x = 31 \implies \mathbf{x = \frac{31}{2}}$ (or $15.5$).
Q3: $5x - 4 = 41$
$5x = 41 + 4 = 45 \implies \mathbf{x = 9}$.
Q4: Rectangle Perimeter
Let width $= w$, length $= w + 4$.
$2(w + w + 4) = 84 \implies 4w + 8 = 84 \implies 4w = 76$
$\mathbf{\text{Breadth} = 19\text{ m}}$, $\mathbf{\text{Length} = 23\text{ m}}$.
Q5: Graphing Equations
• (i) $5x - 4y = 8 \implies y = \frac{5x - 8}{4}$
• (ii) $x = y - 4 \implies y = x + 4$
• (iii) $x + y = 2 \implies y = 2 - x$
• (iv) $2y + x = -4 \implies y = \frac{-4 - x}{2}$
Q6: Simultaneous Systems
• (i) $6x + 2y = 11$ and $3x - 8y = 1 \implies \mathbf{\left(\frac{5}{3}, \frac{1}{2}\right)}$
• (ii) $4x + 3y = 5$ and $6x - 9y = 0 \implies \mathbf{\left(\frac{5}{6}, \frac{5}{9}\right)}$
Q7: Horizontal / Vertical / Neither
• (i) $x = -6$: Vertical
• (ii) $y = 5$: Horizontal
• (iii) $4x = 12y \implies y = \frac{1}{3}x$: Neither
• (iv) $x = 0, y = 0$: $x=0$ is Vertical ($y$-axis); $y=0$ is Horizontal ($x$-axis).

10. Chapter 8 Master Formula Card & Active Recall

Linear Equation (1 Var):
$ax + b = 0 \quad (a \ne 0)$
Unique root: $x = -\frac{b}{a}$
Linear Equation (2 Vars):
$ax + by = c$
Graph: Straight Line
Solutions: Infinitely Many $(x, y)$
Special Lines:
Horizontal: $y = c$ (parallel to $x$-axis)
Vertical: $x = c$ (parallel to $y$-axis)
Through Origin: $y = mx$
Solving Systems:
1. Elimination (Match coefficients)
2. Substitution (Isolate & plug in)
Intersection point satisfies both.

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