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Pair of Linear Equations in Two Variables - Algebraic Methods of Solution (Substitution and Elimination)

Grade 10CBSE

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

🔑Concepts

A pair of linear equations in two variables is represented by the general form a1x+b1y+c1=0a_1x + b_1y + c_1 = 0 and a2x+b2y+c2=0a_2x + b_2y + c_2 = 0. Visually, these equations describe two straight lines on a Cartesian plane, and the solution corresponds to the point or points where these lines meet.

The Substitution Method is an algebraic technique where you solve one of the equations for one variable (e.g., xx in terms of yy) and substitute that expression into the other equation. This transforms a system of two equations into a single equation with one variable.

The Elimination Method involves multiplying one or both equations by suitable non-zero constants so that the coefficients of one variable (either xx or yy) become numerically equal. Adding or subtracting the equations then 'eliminates' that variable, allowing you to solve for the remaining one.

When comparing ratios of coefficients, if a1a2b1b2\frac{a_1}{a_2} \neq \frac{b_1}{b_2}, the lines are intersecting. Visually, they cross at exactly one specific point, meaning the system is consistent and has a unique solution.

If the ratios satisfy a1a2=b1b2c1c2\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}, the lines are parallel. Visually, these lines maintain a constant distance and never intersect, resulting in no solution (inconsistent system).

If all ratios are equal, a1a2=b1b2=c1c2\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}, the lines are coincident. Visually, one line lies exactly on top of the other, meaning every point on the line is a solution, resulting in infinitely many solutions (consistent and dependent system).

Algebraic methods provide a precise way to find intersection points, especially when the coordinates are fractions or decimals that would be difficult to plot accurately on a graphical grid.

📐Formulae

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

Condition for Unique Solution (Intersecting Lines): a1a2b1b2\frac{a_1}{a_2} \neq \frac{b_1}{b_2}

Condition for No Solution (Parallel Lines): a1a2=b1b2c1c2\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}

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

💡Examples

Problem 1:

Solve the system using Substitution Method: x+y=14x + y = 14 and xy=4x - y = 4.

Solution:

  1. From the second equation xy=4x - y = 4, express xx in terms of yy: x=y+4x = y + 4.
  2. Substitute x=y+4x = y + 4 into the first equation: (y+4)+y=14(y + 4) + y = 14.
  3. Combine like terms: 2y+4=142y + 4 = 14.
  4. Subtract 4 from both sides: 2y=102y = 10.
  5. Divide by 2: y=5y = 5.
  6. Substitute y=5y = 5 back into x=y+4x = y + 4: x=5+4=9x = 5 + 4 = 9. Final Answer: x=9,y=5x = 9, y = 5.

Explanation:

We isolated xx in the simpler second equation and substituted it into the first. This reduced the problem to a basic linear equation in one variable (yy).

Problem 2:

Solve the system using Elimination Method: 3x+4y=103x + 4y = 10 and 2x2y=22x - 2y = 2.

Solution:

  1. To eliminate yy, multiply the second equation by 2: 2(2x2y=2)4x4y=42(2x - 2y = 2) \Rightarrow 4x - 4y = 4.
  2. Now we have: Eq 1: 3x+4y=103x + 4y = 10 Eq 2 (new): 4x4y=44x - 4y = 4
  3. Add the two equations: (3x+4x)+(4y4y)=10+4(3x + 4x) + (4y - 4y) = 10 + 4.
  4. Result: 7x=14x=27x = 14 \Rightarrow x = 2.
  5. Substitute x=2x = 2 into Eq 1: 3(2)+4y=103(2) + 4y = 10.
  6. 6+4y=104y=4y=16 + 4y = 10 \Rightarrow 4y = 4 \Rightarrow y = 1. Final Answer: x=2,y=1x = 2, y = 1.

Explanation:

We identified that multiplying the second equation by 2 would make the yy coefficients additive inverses (44 and 4-4), allowing us to eliminate yy by adding the equations together.