Class 8 Mathematics - Ch 7: Linear Equations & Inequalities Mastery Guide (FBISE)

• Formulate and solve simultaneous linear equations in two variables using Substitution and Elimination methods.
• Solve real-world word problems (cost analysis, age relations, fractions, geometric perimeters).
• Identify and graph horizontal lines ($y = c$) and vertical lines ($x = c$).
• Interpret slope/gradient $m$ and $y$-intercept $c$ in the standard linear equation $y = mx + c$.
• Solve systems of linear equations graphically and classify systems as having one unique solution, no solution (parallel), or infinitely many solutions (coincident).
• Solve linear inequalities ($<, >, \le, \ge$) and represent solution sets on number lines and coordinate graphs.

• Inequality Sign Flip Omission: Forgetting to reverse the inequality sign ($<$ becomes $>$) when multiplying or dividing by a negative number.
• Incomplete Fractional Elimination: Failing to multiply every term (including standalone constants) by the LCD when clearing fractions.
• Substitution Sign Errors: Dropping parentheses when substituting an expression like $(y - 4)$ into $2x - y = 6$, resulting in $-y - 4$ instead of $-y + 4$.
• Graph Scale Inconsistencies: Using unequal grid spacing on the x-axis and y-axis, causing graphical intersection coordinates to be read incorrectly.

Step 1: Express one variable in terms of the other:
From second equation $x - y = 12$, isolate $x$:
$$x = y + 12 \quad \dots\dots (1)$$

Step 2: Substitute equation (1) into the first equation:
$$-(y + 12) + \frac{1}{2}y = -19$$ $$-y - 12 + \frac{1}{2}y = -19$$ $$-\frac{1}{2}y - 12 = -19 \implies -\frac{1}{2}y = -19 + 12 = -7$$ $$y = -7 \times (-2) = 14$$

Step 3: Substitute $y = 14$ back into equation (1):
$$x = 14 + 12 = 26$$

Verification: $-26 + \frac{1}{2}(14) = -26 + 7 = -19$ (True), $26 - 14 = 12$ (True).

Final Answer: **$$x = 26, \quad y = 14 \quad \text{or Solution Set } S.S = \{(26, 14)\}$$**

Step 1: Observe identical coefficients:
Notice both equations contain the exact term $-\frac{2}{3}y$. Subtract the first equation from the second equation:
$$\left(\frac{1}{2}x - \frac{2}{3}y\right) - \left(\frac{1}{4}x - \frac{2}{3}y\right) = -4 - 7$$ $$\left(\frac{1}{2} - \frac{1}{4}\right)x = -11 \implies \frac{1}{4}x = -11 \implies x = -44$$

Step 2: Substitute $x = -44$ into the first equation:
$$\frac{1}{4}(-44) - \frac{2}{3}y = 7 \implies -11 - \frac{2}{3}y = 7$$ $$-\frac{2}{3}y = 7 + 11 = 18 \implies y = 18 \times \left(-\frac{3}{2}\right) = -27$$

Final Answer: **$$x = -44, \quad y = -27 \quad \text{or } S.S = \{(-44, -27)\}$$**

Step 1: Define variables and form linear equations:
Let the cost of 1 pen = Rs. $p$ and cost of 1 notebook = Rs. $n$.
$$5p + 6n = 900 \quad \dots\dots (1)$$ $$3p + 2n = 500 \quad \dots\dots (2)$$

Step 2: Multiply equation (2) by 3 to eliminate $n$:
$$3 \times (3p + 2n) = 3 \times 500 \implies 9p + 6n = 1500 \quad \dots\dots (3)$$ Subtract equation (1) from (3):
$$(9p + 6n) - (5p + 6n) = 1500 - 900 \implies 4p = 600 \implies p = 150$$

Step 3: Find the cost of a notebook $n$:
Substitute $p = 150$ into equation (2):
$$3(150) + 2n = 500 \implies 450 + 2n = 500 \implies 2n = 50 \implies n = 25$$

Final Answer: **Cost of 1 Pen = Rs. 150, \quad Cost of 1 Notebook = Rs. 25**

Step 1: Setup system:
Let the numbers be $x$ and $y$ ($x > y$).
$$x + y = 30 \quad \dots\dots (1)$$ $$x - y = 4 \quad \dots\dots (2)$$

Step 2: Add equations (1) and (2):
$$(x + y) + (x - y) = 30 + 4 \implies 2x = 34 \implies x = 17$$

Step 3: Solve for $y$:
$$17 + y = 30 \implies y = 30 - 17 = 13$$

Final Answer: **The numbers are 17 and 13.**

Step 1: Form equations:
Let the fraction be $\frac{x}{y}$.
Case 1: $\frac{x + 1}{y + 1} = \frac{4}{5} \implies 5(x + 1) = 4(y + 1) \implies 5x + 5 = 4y + 4 \implies 5x - 4y = -1 \quad (1)$
Case 2: $\frac{x - 5}{y - 5} = \frac{1}{2} \implies 2(x - 5) = 1(y - 5) \implies 2x - 10 = y - 5 \implies y = 2x - 5 \quad (2)$

Step 2: Substitute (2) into (1):
$$5x - 4(2x - 5) = -1 \implies 5x - 8x + 20 = -1 \implies -3x = -21 \implies x = 7$$

Step 3: Solve for $y$:
$$y = 2(7) - 5 = 14 - 5 = 9$$

Final Answer: **The fraction is $$\frac{7}{9}$$**

Step 1: Translate conditions into algebra:
Let the first number be $x$ and second number be $y$.
Condition 1: $x + 9 = 2y \implies x = 2y - 9 \quad \dots\dots (1)$
Condition 2: $y + 20 = 4x \quad \dots\dots (2)$

Step 2: Substitute (1) into (2):
$$y + 20 = 4(2y - 9) \implies y + 20 = 8y - 36$$ $$20 + 36 = 8y - y \implies 56 = 7y \implies y = 8$$

Step 3: Find $x$:
$$x = 2(8) - 9 = 16 - 9 = 7$$

Final Answer: **The first number is 7 and the second number is 8.**

Step 1: Model ages algebraically:
Let son's present age = $y$ years, Father's present age = $x$ years.
Condition 1: $x = 3y \quad \dots\dots (1)$
After 5 years: Father's age = $x + 5$, Son's age = $y + 5$.
Condition 2: $x + 5 = 2.5(y + 5) = \frac{5}{2}(y + 5) \implies 2(x + 5) = 5(y + 5) \implies 2x + 10 = 5y + 25 \quad (2)$

Step 2: Substitute $x = 3y$ into (2):
$$2(3y) + 10 = 5y + 25 \implies 6y + 10 = 5y + 25 \implies y = 15\text{ years}$$

Step 3: Solve for Father's age $x$:
$$x = 3(15) = 45\text{ years}$$

Final Answer: **Father's age = 45 years, \quad Son's age = 15 years**

Step 1: Set up system:
Let length = $l$ and breadth = $b$.
Condition 1: $l - b = 12 \quad \dots\dots (1)$
Condition 2: Perimeter $= 2(l + b) = 144 \implies l + b = 72 \quad \dots\dots (2)$

Step 2: Add equations (1) and (2):
$$(l - b) + (l + b) = 12 + 72 \implies 2l = 84 \implies l = 42\text{ m}$$

Step 3: Solve for breadth $b$:
$$42 - b = 12 \implies b = 42 - 12 = 30\text{ m}$$

Final Answer: **Length = 42 m, \quad Breadth = 30 m**

(i) $x + 2y = -2 \implies y = \frac{-2 - x}{2}$
- Table: for $x \in \{-4, -2, 0, 2, 4\} \implies y \in \{1, 0, -1, -2, -3\}$.
- $x$-intercept (set $y=0$): $x + 0 = -2 \implies (-2, 0)$.
- $y$-intercept (set $x=0$): $0 + 2y = -2 \implies (0, -1)$.

(ii) $-x + 4y = 12 \implies y = \frac{x + 12}{4}$
- Table: for $x \in \{-4, 0, 4, 8, 12\} \implies y \in \{2, 3, 4, 5, 6\}$.
- $x$-intercept: $-x = 12 \implies x = -12 \implies (-12, 0)$.
- $y$-intercept: $4y = 12 \implies y = 3 \implies (0, 3)$.

(iii) $x + 2y = -10 \implies y = \frac{-10 - x}{2}$
- Table: for $x \in \{-10, -6, -2, 0, 2\} \implies y \in \{0, -2, -4, -5, -6\}$.
- $x$-intercept: $(-10, 0)$; \quad $y$-intercept: $(0, -5)$.

(iv) $3x - 5y = 15 \implies y = \frac{3x - 15}{5}$
- Table: for $x \in \{-5, 0, 5, 10, 15\} \implies y \in \{-6, -3, 0, 3, 6\}$.
- $x$-intercept: $3x = 15 \implies x = 5 \implies (5, 0)$; \quad $y$-intercept: $-5y = 15 \implies y = -3 \implies (0, -3)$.

• (i) $x = 5$: Vertical (parallel to y-axis)
• (ii) $y = -7$: Horizontal (parallel to x-axis)
• (iii) $4x = -16 \implies x = -4$: Vertical
• (iv) $x - 2y = 0 \implies y = \frac{1}{2}x$: Neither (oblique line through origin)
• (v) $-3y = -12 \implies y = 4$: Horizontal
• (vi) $y = 3x + 1$: Neither (sloped line with $m=3, c=1$)
• (vii) $x = 0y + 2 \implies x = 2$: Vertical
• (viii) $x = 3y \implies y = \frac{1}{3}x$: Neither
• (ix) $y = 0$: Horizontal (the x-axis itself)
• (x) $x = 0$: Vertical (the y-axis itself)

Q3 Deductions:
(i) $x = 3, x = -1, x = 7, x = 0, x = 9$ are all vertical lines parallel to the y-axis.
(ii) $y = 6, y = -6, y = 4, y = 0, y = -8$ are all horizontal lines parallel to the x-axis.

Q4 Deductions:
(i) $y = 3x - 2, y = 3x, y = 3x + 2, y = 3x + 3$ have the identical slope $m = 3$. Deduction: Lines with equal slopes are mutually parallel.
(ii) $y = 2x + 1, y = 4x + 1, y = x + 1, y = 5x + 1$ have different slopes but the same $y$-intercept $c = 1$. Deduction: Lines with the same y-intercept intersect concurrently at the single point $(0, 1)$.

(a) $2x + y = 1, \quad 3x - 2y = 12$:
- $(0, 1)$: $2(0)+1=1$ (True), $3(0)-2(1)=-2 \ne 12$ (False).
- $(2, -3)$: $2(2)+(-3)=1$ (True), $3(2)-2(-3)=6+6=12$ (True). Solution: $(2, -3)$.
- $(4, 0)$: $2(4)+0=8 \ne 1$ (False).

(b) $3x - y = 6, \quad 2x + y = 9$:
- $(3, 3)$: $3(3)-3=6$ (True), $2(3)+3=9$ (True). Solution: $(3, 3)$.

(c) $2x + 3y = 6, \quad -2x + 5 = y$:
- Testing $\left(\frac{9}{4}, \frac{1}{2}\right)$: $2(9/4)+3(1/2)=4.5+1.5=6$ (True), $-2(9/4)+5=-4.5+5=0.5=y$ (True). Solution: $\left(\frac{9}{4}, \frac{1}{2}\right)$.

(d) $2x + 7y = 4, \quad 14y + 4x = 8$:
Notice equation 2 is $2 \times$ equation 1 (Coincident lines). Points $\left(-2, \frac{8}{7}\right)$ and $(2, 0)$ satisfy both!

(e) $2x - y = -4, \quad 4x - 2y = -8$:
Coincident system! All three $(0, 4), (3, 10), (-2, 0)$ satisfy both equations.

(f) $3x + 2y = 8, \quad x + y = 1$:
- $(6, -5)$: $3(6)+2(-5)=18-10=8$ (True), $6+(-5)=1$ (True). Solution: $(6, -5)$.