Pair of Linear Equations in Two Variables

Four ways to solve a pair

A pair of linear equations in two variables, $a_1x+b_1y=c_1$ and $a_2x+b_2y=c_2$, is two straight lines. Their solution is the point where the lines meet, if they meet at all. There are four standard ways to find it:

1. Graphical: draw both lines and read off the point where they cross. Comparing the coefficient ratios tells you in advance whether they cross, are parallel or coincide.
2. Substitution: make one variable the subject of one equation, substitute into the other, solve, then back-substitute.
3. Elimination: multiply one or both equations so one variable has matching coefficients, then add or subtract to remove it.
4. Cross-multiplication: a formula that gives both variables in one line, once both equations are written in the "$= 0$" form.

Forming the equations from a word problem

Aftab tells his daughter: "Seven years ago, I was seven times as old as you were then. Three years from now, I'll be three times as old as you'll be." Let Aftab's present age be $x$ and his daughter's be $y$.

Seven years ago: $x-7 = 7(y-7)$, so $x - 7y = -42$.

Three years from now: $x+3 = 3(y+3)$, so $x - 3y = 6$.

Each condition becomes one equation, and each equation is one line. Subtracting the first from the second gives $4y = 48$, so $y = 12$ and $x = 6 + 3(12) = 42$. Check: seven years ago they were 35 and 5 ($35 = 7 \times 5$), and in three years they will be 45 and 15 ($45 = 3 \times 15$).

The ratio test: reading the lines without drawing them

Intersecting. For $5x-4y+8=0$ and $7x+6y-9=0$: $\dfrac{a_1}{a_2} = \dfrac{5}{7}$ and $\dfrac{b_1}{b_2} = \dfrac{-4}{6} = -\dfrac{2}{3}$. These differ, so the lines cross at exactly one point.

Coincident. $4x+6y-12=0$ is exactly twice $2x+3y-6=0$, so all three ratios equal $\dfrac{1}{2}$: the same line, with infinitely many solutions.

Parallel. For $x+2y=3$ and $2x+4y=7$: $\dfrac{1}{2} = \dfrac{2}{4} \ne \dfrac{3}{7}$, so the lines never meet and there is no solution. The elimination example below reaches the same answer.

Worked example: substitution

Solve $x+y=14$ and $x-y=4$. From the second, $x = y+4$. Substituting: $(y+4)+y=14 \Rightarrow 2y=10 \Rightarrow y=5$, so $x=9$. Check: $9+5=14$ and $9-5=4$.

Worked example: elimination

Solve $x+y=5$ and $2x-3y=4$. Multiply the first by 2: $2x+2y=10$. Subtract the second:

Check: $2\left(\dfrac{19}{5}\right) - 3\left(\dfrac{6}{5}\right) = \dfrac{38-18}{5} = 4$.

Worked example: infinitely many solutions

Solve $3x-y=3$ and $9x-3y=9$. From the first, $y = 3x-3$. Substituting: $9x - 3(3x-3) = 9 \Rightarrow 9 = 9$. Every variable cancels and a true statement is left: the second equation is 3 times the first, the same line, so there are infinitely many solutions.

Worked example: no solution

Solve $x+2y=3$ and $2x+4y=7$. Multiply the first by 2: $2x+4y=6$. Subtracting from the second gives $0 = 1$, which is false. The lines are parallel and the system has no solution.

Cross-multiplication

For $a_1x+b_1y+c_1=0$ and $a_2x+b_2y+c_2=0$ with a unique solution, cross-multiplication gives both variables in one line:

Both equations must be in the "$= 0$" form first, so the constants $c_1$ and $c_2$ carry their signs over. For $2x + y = 5$, that means $c_1 = -5$, not 5.

Write the coefficients in this order, repeating the $b$ column at the end:

Each denominator comes from one neighbouring pair of columns: (top-left × bottom-right) minus (bottom-left × top-right). The first pair ($b$, $c$) goes under $x$, the second ($c$, $a$) under $y$, and the third ($a$, $b$) under 1.

Worked example: cross-multiplication

Solve $2x+y=5$ and $3x+2y=8$ by cross-multiplication.

Solution: Rewrite as $2x+y-5=0$ and $3x+2y-8=0$, so $a_1=2,b_1=1,c_1=-5$ and $a_2=3,b_2=2,c_2=-8$:

Check: $2(2) + 1 = 5$ and $3(2) + 2(1) = 8$.

When $a_1b_2 - a_2b_1 = 0$ the formula can't be used: the lines are parallel or coincident, and the ratio test above tells you which.

Worked example: finding k for no solution

For what value of $k$ does the pair $kx + 3y = k - 3$ and $12x + ky = k$ have no solution?

Solution: No solution needs $\dfrac{k}{12} = \dfrac{3}{k}$, so $k^2 = 36$ and $k = 6$ or $k = -6$. Both must then also satisfy $\dfrac{3}{k} \neq \dfrac{k-3}{k}$:

• $k = 6$: $\dfrac{3}{6} = \dfrac{1}{2}$ and $\dfrac{6-3}{6} = \dfrac{1}{2}$. These are equal, so the lines coincide and there are infinitely many solutions.
• $k = -6$: $\dfrac{3}{-6} = -\dfrac{1}{2}$ but $\dfrac{-9}{-6} = \dfrac{3}{2}$. These differ, so the lines are parallel.

So the pair has no solution only when $k = -6$.

Worked example: equations that reduce to linear ones

Solve $\dfrac{1}{2x}+\dfrac{1}{3y}=2$ and $\dfrac{1}{3x}+\dfrac{1}{2y}=\dfrac{13}{6}$.

Solution: Neither equation is linear in $x$ and $y$, but substituting $p=\tfrac{1}{x}$ and $q=\tfrac{1}{y}$ makes them linear in $p$ and $q$:

Cross-multiplying:

Converting back, $\tfrac{1}{x}=2$ gives $x=\tfrac{1}{2}$, and $\tfrac{1}{y}=3$ gives $y=\tfrac{1}{3}$.

Word problems

Every word problem comes down to naming the two unknowns and writing one equation per condition. Then any method above finishes it.

Two numbers. Two numbers add up to 50 and differ by 12: $x+y=50$ and $x-y=12$. Adding gives $2x=62$, so $x=31$ and $y=19$.

Ages, with two cases. Ani and Biju's ages differ by 3 years. Ani's father Dharam is twice as old as Ani, and Biju is twice as old as his sister Cathy. Cathy and Dharam's ages differ by 30 years. With Ani $x$ and Biju $y$, Dharam is $2x$ and Cathy $\tfrac{y}{2}$, so $2x - \tfrac{y}{2} = 30$, that is $4x - y = 60$. "Differ by 3" doesn't say who is older, so check both cases.

• Ani older: $x - y = 3$. Subtracting from $4x - y = 60$ gives $3x = 57$, so Ani is 19 and Biju 16.
• Biju older: $y - x = 3$. Adding to $4x - y = 60$ gives $3x = 63$, so Ani is 21 and Biju 24.

Both answers satisfy every condition, so both are valid.

Money. One friend says to another: "Give me ₹100 and I'll have twice what you have left." The other replies: "Give me ₹10 instead and I'll have six times what you have left." With the first friend's money $x$ and the second's $y$: $x+100 = 2(y-100)$ gives $x-2y=-300$, and $y+10 = 6(x-10)$ gives $6x-y=70$. Doubling the second and subtracting the first: $11x = 440$, so $x = 40$ and $y = 170$. Check: $40+100 = 140 = 2(170-100)$.

Common mistakes

• Subtracting only some terms. When subtracting equations, change the sign of every term, the constant included.
• Stopping after one variable. Back-substitute to get the other, then check both equations.
• Reading $0 = 0$ as no solution. A true statement means infinitely many solutions; a false one like $0 = 1$ means none.
• Using the constants with the wrong sign. Move everything to one side first. $3x + 2y = 8$ has $c_2 = -8$.
• Mixing up the order of the products. It's always top-left × bottom-right first in each pair of columns. Reversing one of them flips that variable's sign.
• Forgetting to convert back. In reducible equations, $p = 2$ is not the answer. The question asked for $x$, which is $\dfrac{1}{2}$.
• Stopping at $k^2 = 36$. One of the two roots often gives infinitely many solutions instead of none. Test each root against the $c$ ratio.

Where this comes up in exams

The 2025-26 CBSE Class X syllabus lists pairs of linear equations in two variables, their graphical solution and consistency, the algebraic conditions for the number of solutions, and solving them algebraically by substitution and elimination. Cross-multiplication isn't named there, so check whether your board or school still uses it.

Checked against CBSE Mathematics curriculum, Classes IX-X (2025-26). Syllabi can change from year to year, so confirm with the latest official notification for your exam.

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