Class 10 Mathematics · Chapter 3 NotesPair of Linear Equations in Two Variables

A linear equation in two variables, such as 3x + 4y = 20, has infinitely many solutions. When two such equations are considered together, we call them a pair of linear equations in two variables. The general form is a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0, where a₁, b₁ are not both zero and a₂, b₂ are not both zero. A solution of the pair is a pair of values (x, y) that satisfies both equations at the same time. In the opening problem, Akhila's rides and Hoopla games give y = ½x and 3x + 4y = 20. Solving the pair means finding the number of rides and games that fit both conditions.

• General forma₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0
• A solution is a pair of values (x, y) satisfying both equations
• A pair may have one solution, no solution, or infinitely many solutions

Each linear equation in two variables represents a straight line. To solve a pair graphically, find two points on each line, plot them, and draw the lines. The way the lines behave tells you about the solutions. If the lines intersect at a single point, that point is the unique solution and the pair is consistent. If the lines are parallel, they never meet, so there is no solution and the pair is inconsistent. If the lines coincide, every point on the line is a solution, giving infinitely many solutions; such a pair is dependent and also consistent. The graphical method is easy to visualise but less convenient when the solution involves non-integral coordinates.

• Intersecting lines → unique solution → consistent pair
• Parallel lines → no solution → inconsistent pair
• Coincident lines → infinitely many solutions → dependent and consistent pair

Instead of drawing graphs, you can compare the ratios of the coefficients to decide how many solutions a pair has. For the pair a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0, calculate a₁/a₂, b₁/b₂, and c₁/c₂. If a₁/a₂ ≠ b₁/b₂, the lines intersect at exactly one point, so the pair is consistent with a unique solution. If a₁/a₂ = b₁/b₂ ≠ c₁/c₂, the lines are parallel, so the pair is inconsistent and has no solution. If a₁/a₂ = b₁/b₂ = c₁/c₂, the lines coincide, so the pair is dependent and consistent with infinitely many solutions. These conditions work in both directions for any pair of lines.

• a₁/a₂ ≠ b₁/b₂ → intersecting lines, unique solution, consistent
• a₁/a₂ = b₁/b₂ ≠ c₁/c₂ → parallel lines, no solution, inconsistent
• a₁/a₂ = b₁/b₂ = c₁/c₂ → coincident lines, infinitely many solutions, dependent and consistent

The substitution method removes one variable by expressing it in terms of the other. Start with either equation and rearrange it to write one variable, say y, in terms of x. Substitute this expression into the other equation. The result is a single linear equation in one variable, which you solve. Then put that value back into the expression from the first step to find the other variable. If, during substitution, you get a true statement with no variable, such as 18 = 18, the pair has infinitely many solutions. If you get a false statement, such as –4 = 0, the pair has no solution and is inconsistent.

• Step 1Express one variable in terms of the other from one equation
• Step 2Substitute into the other equation and solve for one variable
• Step 3Substitute back to find the second variable
• True statement with no variable → infinitely many solutions; false statement → no solution

The elimination method removes one variable by making its coefficients numerically equal in both equations and then adding or subtracting the equations. Multiply one or both equations by suitable non-zero constants so that the coefficients of x (or y) become equal in magnitude. Then add or subtract to cancel that variable, leaving a linear equation in the other variable. Solve it, then substitute the value into either original equation to find the remaining variable. As with substitution, if the variable disappears and you get a true statement, there are infinitely many solutions; a false statement means no solution. This method is often quicker when coefficients are easy to match.

• Step 1Multiply equations to make coefficients of one variable numerically equal
• Step 2Add or subtract to eliminate that variable
• Step 3Solve the resulting one-variable equation
• Step 4Substitute back to find the other variable

Some situations lead to equations that are not linear at first glance, but can be transformed into a pair of linear equations by a suitable substitution. For example, equations involving terms like 1/(x + y) or 1/(x – y) can be simplified by letting u = 1/(x + y) and v = 1/(x – y). The new equations in u and v are linear and can be solved by substitution or elimination. After finding u and v, substitute back to get x and y. This technique widens the range of problems you can handle with the same algebraic tools.

• Use substitutions like u = 1/(x + y) to convert to linear form
• Solve the linear pair in u and v
• Substitute back to find the original variables

Many word problems can be solved by forming a pair of linear equations. Identify the two unknown quantities and assign them variables. Translate the given conditions into two equations. Solve the pair by substitution, elimination, or graphically, and then interpret the solution in the context of the problem. Common examples include age problems, cost and quantity problems, digit problems, fraction problems, and geometry problems involving perimeter or angles. Always verify that your answer satisfies the original conditions and makes sense in the real situation.

• Define variables for the unknown quantities
• Translate conditions into two linear equations
• Solve and check the solution against the problem context