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Pair of Linear Equations in Two Variables - Introduction

Grade 10CBSE

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

🔑Concepts

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A linear equation in two variables is an equation that can be put in the form ax+by+c=0ax + by + c = 0, where a,b,a, b, and cc are real numbers, and aa and bb are not both zero (a2+b2≠0a^2 + b^2 \neq 0).

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Each solution (x,y)(x, y) of a linear equation in two variables corresponds to a point on the line representing the equation.

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A pair of linear equations in two variables xx and yy is called a system of linear equations. Its general form is a1x+b1y+c1=0a_1x + b_1y + c_1 = 0 and a2x+b2y+c2=0a_2x + b_2y + c_2 = 0.

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Geometrically, a pair of linear equations represents two lines in a plane. There are three possibilities: the lines intersect at one point, the lines are parallel, or the lines are coincident.

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A system is Consistent if it has at least one solution. It is Inconsistent if it has no solution.

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A system of coincident lines has infinitely many solutions and is called a Dependent system (which is always consistent).

📐Formulae

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

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

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

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

Parallel Lines (No Solution): a1a2=b1b2≠c1c2\text{Parallel Lines (No Solution): } \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.

Solution:

Given: x+3y−6=0  ⟹  a1=1,b1=3,c1=−6x + 3y - 6 = 0 \implies a_1 = 1, b_1 = 3, c_1 = -6 2x−3y−12=0  ⟹  a2=2,b2=−3,c2=−122x - 3y - 12 = 0 \implies a_2 = 2, b_2 = -3, c_2 = -12 Comparing 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 Since a1a2≠b1b2\frac{a_1}{a_2} \neq \frac{b_1}{b_2}, the pair of equations has a unique solution.

Explanation:

Because the ratios of the coefficients of xx and yy are not equal, the lines intersect at a single point, making the system consistent.

Problem 2:

Find the nature of the lines represented by 2x+4y−12=02x + 4y - 12 = 0 and x+2y−4=0x + 2y - 4 = 0.

Solution:

Identify coefficients: a1=2,b1=4,c1=−12a_1 = 2, b_1 = 4, c_1 = -12 a2=1,b2=2,c2=−4a_2 = 1, b_2 = 2, c_2 = -4 Calculate ratios: a1a2=21=2\frac{a_1}{a_2} = \frac{2}{1} = 2 b1b2=42=2\frac{b_1}{b_2} = \frac{4}{2} = 2 c1c2=−12−4=3\frac{c_1}{c_2} = \frac{-12}{-4} = 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.

Explanation:

When the ratios of xx and yy coefficients are equal but not equal to the constant ratio, the lines never meet, resulting in no solution.

Problem 3:

Determine if the lines 9x+3y+12=09x + 3y + 12 = 0 and 18x+6y+24=018x + 6y + 24 = 0 are coincident.

Solution:

Identify coefficients: a1=9,b1=3,c1=12a_1 = 9, b_1 = 3, c_1 = 12 a2=18,b2=6,c2=24a_2 = 18, b_2 = 6, c_2 = 24 Calculate ratios: a1a2=918=12\frac{a_1}{a_2} = \frac{9}{18} = \frac{1}{2} b1b2=36=12\frac{b_1}{b_2} = \frac{3}{6} = \frac{1}{2} c1c2=1224=12\frac{c_1}{c_2} = \frac{12}{24} = \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:

Since all three ratios are identical, one equation is a multiple of the other, meaning they represent the exact same line.