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Linear Equations in Two Variables - Solve pair of linear equations algebraically using substitution and elimination

Grade 9CBSE

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

🔑Concepts

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A pair of linear equations in two variables xx and yy is represented as a1x+b1y+c1=0a_1x + b_1y + c_1 = 0 and a2x+b2y+c2=0a_2x + b_2y + c_2 = 0, where a1,b1,c1,a2,b2,c2a_1, b_1, c_1, a_2, b_2, c_2 are real numbers.

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Substitution Method: This involves expressing one variable in terms of the other from one equation and substituting this expression into the second equation to get an equation in one variable.

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Elimination Method: This involves multiplying the equations by suitable non-zero constants so that the coefficients of one variable become equal in both equations. The equations are then added or subtracted to eliminate that variable.

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A unique solution exists if the lines represented by the equations intersect at a single point (a1a2≠b1b2)(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}).

📐Formulae

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

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

Substitution Step: y=−c1−a1xb1\text{Substitution Step: } y = \frac{-c_1 - a_1x}{b_1}

💡Examples

Problem 1:

Solve the following pair of linear equations using the Substitution Method: x+y=14x + y = 14 x−y=4x - y = 4

Solution:

From equation (1): x+y=14  ⟹  x=14−yx + y = 14 \implies x = 14 - y. Substitute x=14−yx = 14 - y into equation (2): (14−y)−y=4(14 - y) - y = 4 14−2y=414 - 2y = 4 −2y=4−14-2y = 4 - 14 −2y=−10  ⟹  y=5-2y = -10 \implies y = 5. Substitute y=5y = 5 into x=14−yx = 14 - y: x=14−5=9x = 14 - 5 = 9. Thus, x=9,y=5x = 9, y = 5.

Explanation:

We isolated xx in the first equation and substituted it into the second to find yy, then back-substituted to find xx.

Problem 2:

Solve the following pair of linear equations using the Elimination Method: 3x+4y=103x + 4y = 10 2x−2y=22x - 2y = 2

Solution:

Multiply the second equation by 22 to make the coefficients of yy equal (but opposite in sign): Equation 1: 3x+4y=103x + 4y = 10 Equation 2 (multiplied by 2): 4x−4y=44x - 4y = 4 Now add the two equations: 3x+4y=10+4x−4y=47x=14\begin{array}{r} 3x + 4y = 10 \\ + 4x - 4y = 4 \\ \hline 7x = 14 \end{array} 7x=14  ⟹  x=27x = 14 \implies x = 2. Substitute x=2x = 2 into the first equation: 3(2)+4y=103(2) + 4y = 10 6+4y=106 + 4y = 10 4y=4  ⟹  y=14y = 4 \implies y = 1. Thus, x=2,y=1x = 2, y = 1.

Explanation:

We eliminated yy by making its coefficients 44 and −4-4, then added the equations to solve for xx.