Algebraic methods of solving linear pairs

Two straight lines meet at exactly one point, unless they are parallel or the same line — and finding that meeting point by drawing careful graphs is slow, and hopeless if the coordinates are fractions like $\left(\frac12, \frac13\right)$ . Algebra sidesteps the ruler entirely. Given a pair of linear equations in $x$ and $y$ , two systematic techniques — substitution and elimination — turn the geometric question "where do these lines cross" into pure calculation that lands on the exact answer every time. The same toolkit extends further: equations that look non-linear because they contain $\frac1x$ and $\frac1y$ collapse into an ordinary linear pair after one substitution, and real-world word problems (ages, digits, speeds, costs) become solvable the moment their conditions are translated into two such equations. This article builds both methods from the ground up, derives the general formula that guarantees they work, and works through the exact reducible-equations and age problems a CBSE Class 10 exam expects.

The core insight

A pair of linear equations in two variables can be solved exactly, without graphing, by substitution (replace one variable using the other equation) or elimination (scale and combine the equations to cancel a variable), and both methods rest on the same underlying algebra that guarantees a unique solution whenever the equations are not parallel.

The core idea

A pair of linear equations in $x, y$ ,

a_1x + b_1y + c_1 = 0, \qquad a_2x + b_2y + c_2 = 0,

where: $a_1, b_1, c_1$ are the coefficients of the first equation and $a_2, b_2, c_2$ of the second, can be solved by two algebraic routes instead of plotting graphs.

Substitution method. From one equation, express one variable in terms of the other — say $y$ in terms of $x$ — and substitute that expression into the second equation. This leaves a single linear equation in one unknown, which is solved directly, and the result is fed back to find the second variable.
Elimination method. Multiply each equation by a constant chosen so that the coefficient of one variable becomes equal (or equal and opposite) in both equations. Adding or subtracting the two scaled equations then cancels that variable, leaving one equation in the other unknown.

Both methods always agree on the final answer — the choice is about convenience, not correctness. Substitution is fastest when a variable already has coefficient $1$ somewhere; elimination is fastest when the coefficients are already close to lining up for cancellation.

Reducible equations. Some pairs are not linear in $x, y$ directly but become linear after a substitution. If an equation involves only $\frac1x$ and $\frac1y$ (never $x$ or $y$ alone), set $u = \frac1x,\ v = \frac1y$ . The pair is now linear in $u, v$ , is solved by substitution or elimination as usual, and $x = \frac1u,\ y = \frac1v$ are recovered at the very end.

Word problems. Every condition in a word problem becomes one equation: assign variables to the unknown quantities, write one equation per independent piece of information, and solve the resulting pair.

Figure

Two straight lines on a coordinate grid crossing at a single point marked as the solution, with dashed guide lines to the x and y axes showing its coordinates

Derivation AGeneral solution of a linear pair by elimination

The goal is to eliminate the systematic guesswork from elimination itself by deriving one algebraic formula, built purely from the coefficients $a_1, b_1, c_1, a_2, b_2, c_2$ , that hands over the exact values of $x$ and $y$ for any pair $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$ whose lines are not parallel, so that a student can see why the elimination method is guaranteed to work rather than trusting it as a recipe.

Step 1. Eliminate $y$ first. Multiply the first equation by $b_2$ and the second by $b_1$ , so the coefficient of $y$ becomes $b_1b_2$ in both: $a_1b_2x + b_1b_2y + c_1b_2 = 0$ and $a_2b_1x + b_1b_2y + c_2b_1 = 0$ .

Step 2. Subtract the second new equation from the first. The $b_1b_2y$ terms cancel exactly, leaving $(a_1b_2 - a_2b_1)x + (c_1b_2 - c_2b_1) = 0$ , a single equation in $x$ alone.

Step 3. Solve for $x$ , provided $a_1b_2 - a_2b_1 \neq 0$ : $x = \dfrac{b_1c_2 - b_2c_1}{a_1b_2 - a_2b_1}$ . The quantity $a_1b_2 - a_2b_1$ is nonzero exactly when the two lines are not parallel, so this is the condition under which a single, unique solution exists.

Step 4. Repeat the same elimination on $x$ instead of $y$ (multiply by $a_2$ and $a_1$ , subtract) to get the matching expression for $y$ : $y = \dfrac{c_1a_2 - c_2a_1}{a_1b_2 - a_2b_1}$ . Together these two formulas are exactly what the elimination method computes step by step for any specific numbers — the derivation shows the method always terminates in a unique answer whenever $a_1b_2 \neq a_2b_1$ .

Worked example

Reducible equations. Solve: $\dfrac{2}{x} + \dfrac{3}{y} = 13$ and $\dfrac{5}{x} - \dfrac{4}{y} = -2$ .

Step 1. Substitute $u = \dfrac1x,\ v = \dfrac1y$ . The pair becomes $2u + 3v = 13$ ...(i) and $5u - 4v = -2$ ...(ii), an ordinary linear pair in $u, v$ .

Step 2. Eliminate $u$ by matching its coefficient. Multiply (i) by $5$ and (ii) by $2$ : $10u + 15v = 65$ and $10u - 8v = -4$ . Subtracting the second from the first cancels $u$ : $23v = 69$ , so $v = 3$ .

Step 3. Back-substitute into (i): $2u + 3(3) = 13 \Rightarrow 2u = 4 \Rightarrow u = 2$ .

Step 4. Recover $x$ and $y$ using $x = \frac1u,\ y = \frac1v$ : since $u = 2$ , $x = \frac12$ ; since $v = 3$ , $y = \frac13$ .

Step 5. Verify against the original equations: $\dfrac{2}{1/2} + \dfrac{3}{1/3} = 4 + 9 = 13$ ✓ and $\dfrac{5}{1/2} - \dfrac{4}{1/3} = 10 - 12 = -2$ ✓. Final answer: $x = \dfrac12,\ y = \dfrac13$ .

Competitive stretch

Apply the derivation's own check before solving. Writing the $u,v$ pair as $2u+3v-13=0$ and $5u-4v+2=0$ gives $a_1=2, b_1=3, c_1=-13$ and $a_2=5, b_2=-4, c_2=2$ , so $a_1b_2-a_2b_1 = 2(-4)-5(3) = -23 \neq 0$ — a unique solution is guaranteed before any elimination arithmetic is done. Plugging these coefficients into the general formula confirms the earlier answer directly: $v=\dfrac{c_1a_2-c_2a_1}{a_1b_2-a_2b_1}=\dfrac{(-13)(5)-(2)(2)}{-23}=\dfrac{-69}{-23}=3$ , matching Step 2 without repeating the multiply-and-subtract process.

Common traps

Using $x$ and $y$ directly in a reducible equation instead of substituting $u = \frac1x, v = \frac1y$ first — this produces a non-linear mess that neither substitution nor elimination can handle cleanly.
Forgetting to convert back from $u, v$ to $x, y$ at the end, or inverting incorrectly (writing $x = u$ instead of $x = \frac1u$ ).
Sign errors while scaling equations for elimination — multiplying only one side of an equation, or forgetting that subtracting a negative flips a sign.
Misreading word problems, especially "ago" and "hence/later" phrases — always write the quantity's value at that specific time explicitly before forming the equation, and verify the final answer against the original story, not just the equations formed from it.

Ask the tutor: bring your own pair of equations or word problem to the AI tutor — it can help you decide the fastest method and check each step of your working.

How do I decide between substitution and elimination on an exam, quickly?

Glance at the coefficients before writing anything. If one equation already has a variable alone or with coefficient $1$ (like $x = 5 - 2y$ or an equation where $y$ 's coefficient is $1$ ), substitution needs almost no extra work. If instead the coefficients of a variable are already equal, or become equal after multiplying by small numbers like $2$ or $3$ , elimination is quicker. Neither choice is wrong — both give the same answer — so pick whichever needs fewer steps for the specific numbers in front of you.

Why does the reducible-equations trick work — why is substituting u = 1/x allowed?

Because $\frac1x$ is just a new quantity in its own right once $x$ is fixed and nonzero. Renaming it $u$ does not change the equation's truth, it only changes which letter you are solving for. Once the pair is written purely in terms of $u$ and $v$ , it is an ordinary linear pair with no fractions involving $x$ or $y$ left, so the usual substitution or elimination method applies without modification. The only extra care needed is the last step: after finding $u$ and $v$ , you must invert them ( $x = \frac1u$ ) to answer the original question, and you must check $x \neq 0, y \neq 0$ since $\frac1x$ was undefined otherwise.

In a word problem, how many equations do I need, and how do I know they are correct?

Exactly as many independent equations as unknowns — two unknowns need two equations, each capturing a separate condition stated in the problem. Assign a variable to each unknown quantity first, then translate each sentence of the problem into one equation using those variables, being precise about the moment in time or the specific quantity each sentence describes. Once solved, the surest check is not to stop at verifying the two equations — substitute the values back into the original story (ages, digits, or costs as described) to make sure the answer actually makes sense, since equations can be satisfied by answers that are mathematically valid but contextually impossible, such as a negative age.

Algebraic methods of solving linear pairs

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads