Pair of Linear Equations in Two Variables

A pair of linear equations in two variables is written in the general form a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0. Each equation is a straight line, so solving the pair means finding the point(s) common to both lines.

Two lines in a plane can behave in only three ways:

• They intersect at one point — exactly one common solution. The pair is consistent with a unique solution. This happens when a₁/a₂ ≠ b₁/b₂.
• They are parallel — no common point, so no solution. The pair is inconsistent. This happens when a₁/a₂ = b₁/b₂ ≠ c₁/c₂.
• They coincide (one line lies exactly on the other) — every point is common, so infinitely many solutions. The pair is consistent (also called dependent). This happens when a₁/a₂ = b₁/b₂ = c₁/c₂.

Worked example (unique solution). Solve x + y = 10 and x − y = 4 graphically. For x + y = 10, take points (0, 10) and (10, 0). For x − y = 4, take points (4, 0) and (0, −4). Plotting both lines, they cross at (7, 3). Check: 7 + 3 = 10 and 7 − 3 = 4, so x = 7, y = 3 is the solution.

Checking the case by ratios. For 2x + 3y − 9 = 0 and 4x + 6y − 18 = 0 we get a₁/a₂ = 2/4 = 1/2, b₁/b₂ = 3/6 = 1/2 and c₁/c₂ = −9/−18 = 1/2. All three ratios are equal, so the lines coincide and there are infinitely many solutions.

The substitution method is an algebraic way to get an exact answer without drawing a graph. It works best when one variable already has coefficient 1 (or is easy to make so).

The steps are:

1. From one equation, express one variable in terms of the other (say, write x in terms of y).
2. Substitute this expression into the other equation. You now have one equation in one variable.
3. Solve for that variable.
4. Put the value back into the expression from step 1 to find the second variable.

Worked example. Solve x + y = 14 and x − y = 4.

From the first equation, x = 14 − y. Substitute into the second equation:

• (14 − y) − y = 4
• 14 − 2y = 4
• −2y = −10, so y = 5.

Now x = 14 − y = 14 − 5 = 9. Check: 9 + 5 = 14 and 9 − 5 = 4. So x = 9, y = 5.

Tip: if while solving you reach a false statement like 0 = 7, the pair has no solution; if you reach a statement that is always true like 0 = 0, the pair has infinitely many solutions.

In the elimination method we remove one variable by adding or subtracting the two equations after matching its coefficients. It is often faster than substitution when no coefficient is 1.

The steps are:

1. Multiply one or both equations by suitable numbers so that the coefficients of one variable become equal in magnitude.
2. Add the equations if those coefficients have opposite signs, or subtract if they have the same sign, to eliminate that variable.
3. Solve the resulting single-variable equation.
4. Substitute back into either original equation to find the other variable.

Worked example. Solve 3x + 4y = 10 and 2x − 2y = 2.

Multiply the second equation by 2 to match the y-terms: 4x − 4y = 4. Now add it to the first equation so that +4y and −4y cancel:

• (3x + 4y) + (4x − 4y) = 10 + 4
• 7x = 14, so x = 2.

Substitute x = 2 into 2x − 2y = 2: 4 − 2y = 2, giving −2y = −2 and y = 1. Check: 3(2) + 4(1) = 10 and 2(2) − 2(1) = 2. So x = 2, y = 1.

For the pair a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0, the solution can be obtained in one shot using:

x / (b₁c₂ − b₂c₁) = y / (c₁a₂ − c₂a₁) = 1 / (a₁b₂ − a₂b₁)

This works provided a₁b₂ − a₂b₁ ≠ 0 (which is exactly the condition a₁/a₂ ≠ b₁/b₂ for a unique solution). The safe way to remember the pattern is to write the coefficients in the order b, c, a, b under a criss-cross and take the diagonal products.

Worked example. Solve 2x + y − 5 = 0 and 3x + 2y − 8 = 0. Here a₁ = 2, b₁ = 1, c₁ = −5 and a₂ = 3, b₂ = 2, c₂ = −8.

• b₁c₂ − b₂c₁ = (1)(−8) − (2)(−5) = −8 + 10 = 2
• c₁a₂ − c₂a₁ = (−5)(3) − (−8)(2) = −15 + 16 = 1
• a₁b₂ − a₂b₁ = (2)(2) − (3)(1) = 4 − 3 = 1

So x/2 = y/1 = 1/1. Therefore x = 2/1 = 2 and y = 1/1 = 1. Check: 2(2) + 1 = 5 and 3(2) + 2(1) = 8. So x = 2, y = 1.

Word problems test whether you can turn a story into equations. The reliable routine is:

1. Read carefully and identify the two unknowns; assign variables such as x and y.
2. Convert each given condition into a linear equation.
3. Solve the pair using substitution, elimination, or cross-multiplication.
4. State the answer in words and check it against the original conditions.

Worked example. The cost of 5 pens and 7 pencils is ₹50, while 7 pens and 5 pencils cost ₹46. Find the cost of one pen and one pencil.

Let the cost of a pen be ₹x and a pencil be ₹y. The conditions give:

• 5x + 7y = 50
• 7x + 5y = 46

Adding the two equations: 12x + 12y = 96, so x + y = 8. Subtracting the first from the second: 2x − 2y = −4, so x − y = −2. Adding these two simpler equations: 2x = 6, giving x = 3; then y = 8 − 3 = 5.

So a pen costs ₹3 and a pencil costs ₹5. Check: 5(3) + 7(5) = 15 + 35 = 50 and 7(3) + 5(5) = 21 + 25 = 46. Both conditions are satisfied.

The same approach handles problems on ages, two-digit numbers, fractions, speed of boat and stream, and geometry — only the way you form the two equations changes.