Prepared by Muhammad Tayyab, Subject Specialist Mathematics, Govt Christian High School Daska
📌 Based on National Curriculum 2023 / PECTAA 2026 Syllabus
📖 What's Inside: This exercise covers finding intercepts of linear equations graphically, solving systems of linear equations by graphical method, and solving linear-quadratic systems graphically. Perfect for Punjab Boards exam preparation.
📚 Related Resources – Unit 2: Quadratic Equations & Inequalities
1 Points of Intersection with Axes (Graphical)
(i) \(x + y = 8\)
For \(x\)-intercept: Put \(y = 0\)
\[ x + 0 = 8 \quad\Rightarrow\quad x = 8 \]Point: \((8,0)\)
For \(y\)-intercept: Put \(x = 0\)
\[ 0 + y = 8 \quad\Rightarrow\quad y = 8 \]Point: \((0,8)\)
(ii) \(x - y = 1\)
For \(x\)-intercept: Put \(y = 0\)
\[ x - 0 = 1 \quad\Rightarrow\quad x = 1 \]Point: \((1,0)\)
For \(y\)-intercept: Put \(x = 0\)
\[ 0 - y = 1 \quad\Rightarrow\quad -y = 1 \quad\Rightarrow\quad y = -1 \]Point: \((0,-1)\)
(iii) \(x - 2y = 1\)
For \(x\)-intercept: \(y = 0\)
\[ x - 2(0) = 1 \quad\Rightarrow\quad x = 1 \]Point: \((1,0)\)
For \(y\)-intercept: \(x = 0\)
\[ 0 - 2y = 1 \quad\Rightarrow\quad -2y = 1 \quad\Rightarrow\quad y = -\frac{1}{2} = -0.5 \]Point: \((0, -0.5)\)
(iv) \(x - 2y + 2 = 0\)
For \(x\)-intercept: \(y = 0\)
\[ x - 2(0) + 2 = 0 \quad\Rightarrow\quad x + 2 = 0 \quad\Rightarrow\quad x = -2 \]Point: \((-2,0)\)
For \(y\)-intercept: \(x = 0\)
\[ 0 - 2y + 2 = 0 \quad\Rightarrow\quad -2y = -2 \quad\Rightarrow\quad y = 1 \]Point: \((0,1)\)
(v) \(5x - 5y = 1\)
For \(x\)-intercept: \(y = 0\)
\[ 5x - 5(0) = 1 \quad\Rightarrow\quad 5x = 1 \quad\Rightarrow\quad x = \frac{1}{5} = 0.2 \]Point: \((0.2, 0)\)
For \(y\)-intercept: \(x = 0\)
\[ 5(0) - 5y = 1 \quad\Rightarrow\quad -5y = 1 \quad\Rightarrow\quad y = -\frac{1}{5} = -0.2 \]Point: \((0, -0.2)\)
2 Solve Systems of Linear Equations Graphically
(i) \(x + y = 8\) ; \(3x - y = 4\)
For \(x + y = 8\):
\(x\)-intercept (\(y=0\)): \(x = 8\) → \((8,0)\)
\(y\)-intercept (\(x=0\)): \(y = 8\) → \((0,8)\)
For \(3x - y = 4\):
\(x\)-intercept: \(3x = 4\) → \(x = \frac{4}{3} \approx 1.33\) → \((1.33, 0)\)
\(y\)-intercept: \(-y = 4\) → \(y = -4\) → \((0, -4)\)
Point of intersection (Algebraic verification):
\[ \begin{aligned} x + y &= 8 \quad \text{(i)}\\ 3x - y &= 4 \quad \text{(ii)}\\ \text{From (i): } x &= 8 - y \\ \text{Substitute into (ii): } 3(8-y) - y &= 4 \\ 24 - 3y - y &= 4 \\ 24 - 4y &= 4 \\ 20 &= 4y \\ y &= 5 \\ x &= 8 - 5 = 3 \end{aligned} \]Hence intersection point \(M(3,5)\).
(ii) \(x - y = 1\) ; \(x + 2y = 7\)
Line 1: \(x - y = 1\): intercepts \((1,0)\) and \((0,-1)\).
Line 2: \(x + 2y = 7\): \(x\)-intercept: \((7,0)\), \(y\)-intercept: \(y = \frac{7}{2}=3.5\) → \((0,3.5)\).
\[ \begin{aligned} x - y &= 1 \quad \text{(i)}\\ x + 2y &= 7 \quad \text{(ii)}\\ \text{From (i): } x &= 1 + y \\ 1 + y + 2y &= 7 \\ 1 + 3y &= 7 \\ 3y &= 6 \quad\Rightarrow\quad y = 2 \\ x &= 1 + 2 = 3 \end{aligned} \]Intersection: \(M(3,2)\).
(iii) \(x - 2y = 1\) ; \(2x + y = 2\)
\(x-2y=1\): intercepts \((1,0)\), \((0,-0.5)\). \(2x+y=2\): \((1,0)\), \((0,2)\).
\[ \begin{aligned} x - 2y &= 1 \quad\Rightarrow\quad x = 1+2y \\ 2(1+2y) + y &= 2 \\ 2 + 4y + y &= 2 \\ 5y &= 0 \quad\Rightarrow\quad y = 0 \\ x &= 1 + 0 = 1 \end{aligned} \]Intersection: \(M(1,0)\).
(iv) \(y = 2x + 2\) ; \(3x + 2y = 4\)
Line \(y=2x+2\): intercepts: \(y=0 \Rightarrow 0=2x+2 \Rightarrow x=-1\) → \((-1,0)\), \(x=0 \Rightarrow y=2\) → \((0,2)\).
\(3x+2y=4\): \(x\)-intercept: \(3x=4 \Rightarrow x=4/3\approx1.33\); \(y\)-intercept: \(2y=4 \Rightarrow y=2\) → \((0,2)\).
\[ \begin{aligned} y &= 2x+2 \\ 3x + 2(2x+2) &= 4 \\ 3x + 4x + 4 &= 4 \\ 7x &= 0 \quad\Rightarrow\quad x = 0 \\ y &= 2(0)+2 = 2 \end{aligned} \]Intersection: \(M(0,2)\).
(v) \(3y = 2x + 8\) ; \(x + y = 1\)
\(3y=2x+8\): intercepts: \(y=0 \Rightarrow 0=2x+8 \Rightarrow x=-4\) → \((-4,0)\); \(x=0 \Rightarrow 3y=8 \Rightarrow y=8/3 \approx 2.667\) → \((0,2.667)\).
\(x+y=1\): intercepts \((1,0)\), \((0,1)\).
\[ \begin{aligned} 3y &= 2x+8 \quad\Rightarrow\quad y = \frac{2x+8}{3} \\ x + \frac{2x+8}{3} &= 1 \\ \frac{3x + 2x+8}{3} &= 1 \\ 5x + 8 &= 3 \\ 5x &= -5 \quad\Rightarrow\quad x = -1 \\ y = \frac{2(-1)+8}{3} = \frac{-2+8}{3} = \frac{6}{3}=2 \end{aligned} \]Intersection: \(M(-1,2)\).
3 Solve Graphically: Linear & Quadratic Intersections
(i) \(y = 8x - 32\) ; \(y = x^2 - 6x + 8\)
Linear: \(y = 8x - 32\): \(x\)-intercept: \(0=8x-32 \Rightarrow x=4\) → \((4,0)\); \(y\)-intercept: \(y = -32\) → \((0,-32)\).
Quadratic: \(y = x^2 - 6x + 8\). Vertex: \(x = -\frac{b}{2a} = -\frac{-6}{2}=3\).
| \(x\) | -3 | -2 | -1 | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
|---|---|---|---|---|---|---|---|---|---|---|---|
| \(y = x^2-6x+8\) | 35 | 24 | 15 | 8 | 3 | 0 | -1 | 0 | 3 | 8 | 15 |
Algebraic intersection:
\[ \begin{aligned} x^2 - 6x + 8 &= 8x - 32 \\ x^2 -14x + 40 &= 0 \\ (x-10)(x-4) &= 0 \\ \Rightarrow x = 10 \quad\text{or}\quad x = 4 \end{aligned} \]For \(x=4\): \(y=8(4)-32=0\) → \((4,0)\)
For \(x=10\): \(y=8(10)-32=80-32=48\) → \((10,48)\).
Intersection points: \(M_1(4,0)\) and \(M_2(10,48)\).
(ii) \(y + x = 2\) ; \(y = 2x^2 + x - 10\)
Rewrite linear: \(y = -x + 2\). Intercepts: \((2,0)\) and \((0,2)\).
Quadratic: \(y = 2x^2 + x - 10\). Vertex: \(x = -\frac{b}{2a} = -\frac{1}{4} = -0.25\).
| \(x\) | -4 | -3 | -2 | -1 | 0 | 1 | 2 | 3 |
|---|---|---|---|---|---|---|---|---|
| \(y=2x^2+x-10\) | 18 | 5 | -4 | -9 | -10 | -7 | 0 | 11 |
Algebraic solution:
\[ \begin{aligned} 2x^2 + x - 10 &= -x + 2 \\ 2x^2 + 2x - 12 &= 0 \\ 2x^2 + 6x - 4x -12 &= 0 \\ 2x(x+3) -4(x+3) &= 0 \\ (2x-4)(x+3) &= 0 \\ \Rightarrow x = 2 \quad\text{or}\quad x = -3 \end{aligned} \]When \(x=2\): \(y = -(2)+2 = 0\) → \((2,0)\)
When \(x=-3\): \(y = -(-3)+2 = 3+2=5\) → \((-3,5)\).
Intersection points: \(M_1(-3,5)\) and \(M_2(2,0)\).
Graph reference: Parabola and line intersection clearly marked.
📈 Key Graphical Concepts
- x-intercept: Put \(y=0\) → point where graph meets x-axis.
- y-intercept: Put \(x=0\) → point where graph meets y-axis.
- Solving linear systems graphically: Intersection of two lines gives the solution.
- Linear-Quadratic system: Intersection points satisfy both equations; may have 0,1,2 solutions.