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Pair of Linear Equations in Two Variables

Pair of Linear Equations in Two Variables

Consistency conditions, graphical, substitution and elimination methods, and word problems, weighted towards the consistency and coincident-lines questions and the 5-mark graphical solutions set across the 2026 papers.

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Pair of linear equations in two variables
Two linear equations in the same two variables, written generally as a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0. Each is represented by a straight line, and a solution is a pair of values satisfying both equations.
Consistent pair of linear equations
A pair that has at least one solution. Its lines either intersect at a single point, giving a unique solution, or coincide, giving infinitely many solutions.
Inconsistent pair of linear equations
A pair of linear equations that has no solution at all. Graphically, the two lines are parallel and never meet.
Dependent pair of linear equations
A pair whose equations are equivalent, so they share infinitely many solutions. The two lines coincide, and a dependent pair is always consistent.
Condition for a unique solution
a1/a2 ≠ b1/b2, for the pair a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0. The lines then intersect at exactly one point and the pair is consistent.
Condition for infinitely many solutions
a1/a2 = b1/b2 = c1/c2. The lines coincide, every point on them is a solution, and the pair is dependent and consistent.
Condition for no solution
a1/a2 = b1/b2 ≠ c1/c2. The lines are parallel, so the pair has no solution and is inconsistent.
How do you solve a pair of linear equations graphically?
Find two solutions of each equation, plot the points and draw both lines, then read off any point common to them. In the chapter's example, x + 3y = 6 and 2x − 3y = 12 both pass through (6, 0), so x = 6, y = 0 and the pair is consistent.
Why is the graphical method not always a convenient way to solve a pair of linear equations?
When the solution has non-integral coordinates, such as (−1.75, 3.3), it is easy to misread the point from a graph. Algebraic methods give the exact values instead.
How do you solve a pair of linear equations by the substitution method?
Express one variable in terms of the other from whichever equation is convenient. Substitute this into the other equation to get an equation in one variable, and solve it. Then put that value back into the first expression to find the other variable.
When solving a pair of equations leads to a false statement such as −4 = 0, what does it tell you?
That the equations have no common solution, so the pair is inconsistent. In the chapter's example, the rails x + 2y − 4 = 0 and 2x + 4y − 12 = 0 reduce to −4 = 0, so the rails never cross.
When solving a pair of equations leads to a true statement with no variable, such as 18 = 18, what does it tell you?
That the two equations are really the same, so they have infinitely many common solutions and no unique answer. In the chapter's example, 2x + 3y = 9 and 4x + 6y = 18 give 18 = 18, so a pencil's and an eraser's cost cannot be fixed uniquely.
How do you solve a pair of linear equations by the elimination method?
Multiply the equations by suitable non-zero constants so the coefficients of one variable become numerically equal. Add or subtract to eliminate that variable, solve the resulting equation, then substitute the value into either original equation to find the other variable.
How do you form linear equations from a problem about ages in the past and the future?
Let the present ages be s and t, then shift both by the same number of years for each condition. In the chapter's example, the conditions seven years ago and three years hence give s − 7 = 7(t − 7) and s + 3 = 3(t + 3), so the ages are 42 and 12.
How do you set up equations when incomes and expenditures are given as ratios?
Write the incomes as 9x and 7x and the expenditures as 4y and 3y, then use income − expenditure = saving. In the chapter's example, savings of Rs 2000 each give 9x − 4y = 2000 and 7x − 3y = 2000, so x = 2000, y = 4000 and the incomes are Rs 18,000 and Rs 14,000.
How do you solve a problem about a two-digit number and the number formed by reversing its digits?
Let the tens digit be x and the units digit y, so the number is 10x + y and its reverse 10y + x. Write equations from the conditions, allowing for either digit being larger when only their difference is given. In the chapter's example this gives two answers, 42 and 24.

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