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Pair of Linear Equations in Two Variables - Classify systems as consistent or inconsistent using graphical and algebraic conditions

Grade 10CBSE

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

🔑Concepts

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A pair of linear equations is called Consistent if it has at least one solution. This occurs when the lines either intersect at a single point (unique solution) or lie exactly on top of each other (infinitely many solutions). Algebraically, consistency is guaranteed if a1a2≠b1b2\frac{a_1}{a_2} \neq \frac{b_1}{b_2} or if a1a2=b1b2=c1c2\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}.

Graph showing two intersecting lines at a single point representing a consistent system with a unique solution.
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A system is Inconsistent if the equations have no common solution. Graphically, this corresponds to Parallel Lines that never meet. This condition occurs when the ratios of the coefficients of xx and yy are equal, but not equal to the ratio of the constant terms: a1a2=b1b2≠c1c2\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}.

Graph showing two parallel lines representing an inconsistent system with no solution.
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A Dependent Consistent system occurs when the two equations represent the same line. Graphically, the lines are Coincident. Every point on the line is a solution, leading to infinitely many solutions. This happens when a1a2=b1b2=c1c2\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}.

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The ratio comparison method allows for quick classification without graphing:

  1. a1a2≠b1b2  ⟹  \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \implies Intersecting lines (Consistent).
  2. a1a2=b1b2=c1c2  ⟹  \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \implies Coincident lines (Consistent/Dependent).
  3. a1a2=b1b2≠c1c2  ⟹  \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \implies Parallel lines (Inconsistent).

📐Formulae

Standard Form: a1x+b1y+c1=0a_1x + b_1y + c_1 = 0 a2x+b2y+c2=0a_2x + b_2y + c_2 = 0

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

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

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

💡Examples

Problem 1:

Check whether the pair of equations x+3y=6x + 3y = 6 and 2x−3y=122x - 3y = 12 is consistent or inconsistent by comparing coefficient ratios.

Solution:

Step 1: Write the equations in standard form: x+3y−6=0x + 3y - 6 = 0 (a1=1,b1=3,c1=−6a_1 = 1, b_1 = 3, c_1 = -6) 2x−3y−12=02x - 3y - 12 = 0 (a2=2,b2=−3,c2=−12a_2 = 2, b_2 = -3, c_2 = -12) Step 2: Calculate the ratios: a1a2=12\frac{a_1}{a_2} = \frac{1}{2} b1b2=3−3=−1\frac{b_1}{b_2} = \frac{3}{-3} = -1 Step 3: Compare ratios: Since 12≠−1\frac{1}{2} \neq -1, we have a1a2≠b1b2\frac{a_1}{a_2} \neq \frac{b_1}{b_2}.

Explanation:

Because the ratios of the coefficients of xx and yy are not equal, the lines intersect at a single point. Therefore, the system is consistent and has a unique solution.

Problem 2:

Find the value of kk for which the system of equations x+2y=3x + 2y = 3 and 5x+ky+7=05x + ky + 7 = 0 has no solution.

Solution:

Step 1: Write in standard form: 1x+2y−3=01x + 2y - 3 = 0 5x+ky+7=05x + ky + 7 = 0 Step 2: Identify coefficients: a1=1,b1=2,c1=−3a_1=1, b_1=2, c_1=-3 and a2=5,b2=k,c2=7a_2=5, b_2=k, c_2=7. Step 3: Apply the condition for no solution: a1a2=b1b2≠c1c2\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}. Step 4: Solve 15=2k\frac{1}{5} = \frac{2}{k}. k=2×5=10k = 2 \times 5 = 10. Step 5: Verify the third ratio: c1c2=−37\frac{c_1}{c_2} = \frac{-3}{7}. Since 15≠−37\frac{1}{5} \neq \frac{-3}{7}, the condition holds.

Explanation:

For a system to have no solution, the lines must be parallel. This requires the xx and yy coefficient ratios to be equal while the constant ratio differs. Solving the proportion gives k=10k=10.

Problem 3:

Show graphically that the system of equations 2x+4y=102x + 4y = 10 and 3x+6y=123x + 6y = 12 is inconsistent.

Coordinate plane showing two parallel lines with negative slopes, illustrating an inconsistent system.

Solution:

  1. Write equations in y=mx+cy = mx + c form: Eq 1: 4y=−2x+10  ⟹  y=−0.5x+2.54y = -2x + 10 \implies y = -0.5x + 2.5 Eq 2: 6y=−3x+12  ⟹  y=−0.5x+26y = -3x + 12 \implies y = -0.5x + 2
  2. Compare coefficients: a1=2,b1=4,c1=−10a_1=2, b_1=4, c_1=-10 and a2=3,b2=6,c2=−12a_2=3, b_2=6, c_2=-12.
  3. Check ratios: a1a2=23\frac{a_1}{a_2} = \frac{2}{3}, b1b2=46=23\frac{b_1}{b_2} = \frac{4}{6} = \frac{2}{3}, c1c2=−10−12=56\frac{c_1}{c_2} = \frac{-10}{-12} = \frac{5}{6}.
  4. Since a1a2=b1b2≠c1c2\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}, the lines are parallel and the system is inconsistent.

Explanation:

Because the slopes are identical (−0.5-0.5) but the y-intercepts are different (2.52.5 vs 22), the lines will never intersect, meaning there is no pair (x,y)(x, y) that satisfies both equations.

Problem 4:

Determine the nature of the system x−2y=0x - 2y = 0 and 3x+4y−20=03x + 4y - 20 = 0 using the graphical method.

Graph of two lines intersecting at the point (4, 2) indicating a consistent system.

Solution:

  1. For x−2y=0x - 2y = 0, when x=0,y=0x=0, y=0; when x=4,y=2x=4, y=2.
  2. For 3x+4y=203x + 4y = 20, when x=0,y=5x=0, y=5; when x=4,y=2x=4, y=2.
  3. Plotting these points, we see both lines pass through (4,2)(4, 2).
  4. Ratio check: a1a2=13\frac{a_1}{a_2} = \frac{1}{3}, b1b2=−24=−12\frac{b_1}{b_2} = \frac{-2}{4} = -\frac{1}{2}.
  5. Since a1a2≠b1b2\frac{a_1}{a_2} \neq \frac{b_1}{b_2}, the system is consistent and has a unique solution.

Explanation:

The lines intersect at the point (4,2)(4, 2). Since there is exactly one point of intersection, the system is consistent with a unique solution.