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Pair of Linear Equations in Two Variables - Graphical Method of Solution of a Pair of Linear Equations

Grade 10CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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The general form of a pair of linear equations in two variables is a1x+b1y+c1=0a_1x + b_1y + c_1 = 0 and a2x+b2y+c2=0a_2x + b_2y + c_2 = 0. Geometrically, each equation represents a straight line on a Cartesian plane.

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Intersecting Lines: If a1a2≠b1b2\frac{a_1}{a_2} \neq \frac{b_1}{b_2}, the two lines intersect at exactly one point. This point (x,y)(x, y) is the unique solution to the system, and the pair of equations is called consistent.

Graph showing two lines intersecting at point (3,2) representing a unique solution.
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Coincident Lines: If a1a2=b1b2=c1c2\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}, the two lines lie on top of each other. Every point on the line is a solution, resulting in infinitely many solutions. This system is consistent and dependent.

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Parallel Lines: If a1a2=b1b2≠c1c2\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}, the lines never meet. There is no common point, meaning no solution exists. The pair of equations is called inconsistent.

Graph showing two parallel lines that never intersect, representing no solution.

📐Formulae

a1x+b1y+c1=0a_1x + b_1y + c_1 = 0

a2x+b2y+c2=0a_2x + b_2y + c_2 = 0

Unique Solution (Intersecting): a1a2≠b1b2\text{Unique Solution (Intersecting): } \frac{a_1}{a_2} \neq \frac{b_1}{b_2}

Infinitely Many Solutions (Coincident): a1a2=b1b2=c1c2\text{Infinitely Many Solutions (Coincident): } \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}

No Solution (Parallel): a1a2=b1b2≠c1c2\text{No Solution (Parallel): } \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}

💡Examples

Problem 1:

Solve the following pair of linear equations graphically: x+y=3x + y = 3 and 3x−2y=43x - 2y = 4.

Solution:

  1. For x+y=3x + y = 3: If x=0,y=3x = 0, y = 3; if y=0,x=3y = 0, x = 3. Points are (0,3)(0, 3) and (3,0)(3, 0).
  2. For 3x−2y=43x - 2y = 4: If x=0,y=−2x = 0, y = -2; if x=2,y=1x = 2, y = 1. Points are (0,−2)(0, -2) and (2,1)(2, 1).
  3. Plotting these points on a graph and drawing lines through them, we observe that the two lines intersect at the point (2,1)(2, 1).
  4. Therefore, x=2x = 2 and y=1y = 1 is the unique solution.

Explanation:

To solve graphically, we find at least two solutions (coordinates) for each equation, plot them on a Cartesian plane, and identify the point of intersection. Since 13≠1−2\frac{1}{3} \neq \frac{1}{-2}, the lines must intersect at exactly one point.

Problem 2:

Check whether the pair of equations 2x+3y=92x + 3y = 9 and 4x+6y=184x + 6y = 18 is consistent and dependent.

Solution:

Compare ratios: a1=2,b1=3,c1=−9a_1 = 2, b_1 = 3, c_1 = -9 a2=4,b2=6,c2=−18a_2 = 4, b_2 = 6, c_2 = -18 a1a2=24=12\frac{a_1}{a_2} = \frac{2}{4} = \frac{1}{2} b1b2=36=12\frac{b_1}{b_2} = \frac{3}{6} = \frac{1}{2} c1c2=−9−18=12\frac{c_1}{c_2} = \frac{-9}{-18} = \frac{1}{2} Since a1a2=b1b2=c1c2\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}, the lines are coincident.

Explanation:

When all three ratios of the coefficients are equal, the two equations represent the same line. This means they have infinitely many solutions, and the system is consistent and dependent.

Problem 3:

Determine graphically if the system of equations x+2y=4x + 2y = 4 and 2x+4y=122x + 4y = 12 is consistent or inconsistent.

Graph showing two parallel lines with the same slope, indicating no solution.

Solution:

For x+2y=4x + 2y = 4: If x=0,y=2x=0, y=2; if y=0,x=4y=0, x=4. Points: (0,2),(4,0)(0,2), (4,0)

For 2x+4y=122x + 4y = 12: If x=0,y=3x=0, y=3; if y=0,x=6y=0, x=6. Points: (0,3),(6,0)(0,3), (6,0)

Comparing ratios: a1a2=12\frac{a_1}{a_2} = \frac{1}{2}, b1b2=24=12\frac{b_1}{b_2} = \frac{2}{4} = \frac{1}{2}, c1c2=412=13\frac{c_1}{c_2} = \frac{4}{12} = \frac{1}{3}. Since a1a2=b1b2≠c1c2\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}, the lines are parallel. The system is inconsistent.

Explanation:

Parallel lines have no points in common, hence no solution. The constant slopes but different intercepts confirm they never meet.

Problem 4:

Solve the following pair of equations graphically and find the coordinates of the points where the lines intersect the y-axis: x−y+1=0x - y + 1 = 0 3x+2y−12=03x + 2y - 12 = 0

A coordinate plane showing two intersecting lines. Line 1 passes through (0,1) and (2,3). Line 2 passes through (0,6) and (2,3). The intersection point is labeled P(2,3).

Solution:

Step 1: Find points for x−y+1=0  ⟹  y=x+1x - y + 1 = 0 \implies y = x + 1. If x=0,y=1x = 0, y = 1 (Point A: (0,1)(0, 1)). If x=2,y=3x = 2, y = 3 (Point B: (2,3)(2, 3)). If x=−1,y=0x = -1, y = 0 (Point C: (−1,0)(-1, 0)).

Step 2: Find points for 3x+2y−12=0  ⟹  y=12−3x23x + 2y - 12 = 0 \implies y = \frac{12 - 3x}{2}. If x=0,y=6x = 0, y = 6 (Point D: (0,6)(0, 6)). If x=2,y=3x = 2, y = 3 (Point E: (2,3)(2, 3)). If x=4,y=0x = 4, y = 0 (Point F: (4,0)(4, 0)).

Step 3: Plot the points and draw the lines. The lines intersect at the point (2,3)(2, 3).

Step 4: Identify y-intercepts. For x−y+1=0x - y + 1 = 0, the y-intercept is (0,1)(0, 1). For 3x+2y−12=03x + 2y - 12 = 0, the y-intercept is (0,6)(0, 6).

Explanation:

The graphical solution of a pair of linear equations is the point of intersection of the two lines. By plotting at least two points for each equation, we can draw the straight lines. The common point (2,3)(2, 3) satisfies both equations. The y-intercepts are found by setting x=0x = 0 in each equation.