Linear equations in two variables

What "linear" and "two variables" mean together. An equation is called linear when every variable appears only to the first power — no $x^2$, no $y^2$, and no product term like $xy$. When two such variables, conventionally $x$ and $y$, are joined by addition, subtraction, and multiplication by constants, the result is a linear equation in two variables. The standard way to write any such equation is:

where: $x$ and $y$ are the two variables, $a$ and $b$ are their coefficients, $c$ is the constant term, and $a, b, c$ are real numbers with the single restriction that $a$ and $b$ are not both zero — at least one of them must be non-zero, or there would be no variable left in the equation at all.

Any linear relation can be forced into this shape. An equation like $2x = 3y - 7$ does not look like $ax + by + c = 0$ at first glance, but moving every term to one side gives $2x - 3y + 7 = 0$, so $a = 2$, $b = -3$, $c = 7$. This rearranging step — collect everything on one side, keep the right side zero — is the entire skill this topic is built on.

A one-variable equation is a hidden special case. Something like $x + 5 = 0$ can be rewritten as $1\cdot x + 0\cdot y + 5 = 0$: it is a two-variable equation where the coefficient of $y$ happens to be zero. This is exactly why the condition is "$a$ and $b$ not both zero" rather than "both non-zero" — an equation with $b = 0$ (or $a = 0$) is still perfectly valid, as long as the other coefficient survives.

Why one equation cannot fix two unknowns. With a single linear equation and two independent letters, choosing any value for $x$ still leaves an equation in $y$ alone to solve, and a different choice of $x$ gives a different $y$. So instead of one definite answer, the equation is satisfied by an unending list of pairs $(x, y)$ — the idea that the next topic (finding and listing solutions) develops in full.

Problem. The cost of a notebook is twice the cost of a pen. Write this as a linear equation in two variables in standard form, identify $a$, $b$, $c$, and check whether the price pair notebook ₹20, pen ₹10 fits the equation.

Step 1. Choose letters for the two unknown prices. Let the cost of a notebook be $\text{₹}\,x$ and the cost of a pen be $\text{₹}\,y$.

Step 2. Translate the sentence into an equation: "notebook cost is twice pen cost" becomes $x = 2y$.

Step 3. Rearrange into standard form by moving every term to one side: $x - 2y = 0$, i.e. $x - 2y + 0 = 0$. Comparing with $ax + by + c = 0$ gives $a = 1$, $b = -2$, $c = 0$; since $a \ne 0$, the condition on $a, b$ holds.

Step 4. Test the given pair. Substitute $x = 20, y = 10$ into $x - 2y$: $20 - 2(10) = 20 - 20 = 0$, which matches the right-hand side, so $(20, 10)$ satisfies the equation — a sensible price pair, since ₹20 is indeed twice ₹10.

Answer: the standard form is $x - 2y + 0 = 0$ with $a = 1, b = -2, c = 0$, and (notebook ₹20, pen ₹10) is one of its solutions.

Competitive stretch

Would (notebook ₹15, pen ₹5) also work? Substitute: $15 - 2(5) = 15 - 10 = 5$, which is not $0$, so this pair does not satisfy the equation — ₹15 is not twice ₹5. Only pairs where the notebook price is exactly double the pen price will work, and there are infinitely many such pairs, e.g. $(10, 5)$, $(20, 10)$, $(30, 15)$, all lying on the same relationship.

• Calling an equation "linear" just because it looks short: $x^2 + 2y = 7$ has a squared term and is not linear, regardless of how simple it appears.
• Forgetting that a one-variable equation such as $3x - 9 = 0$ is also a two-variable equation in disguise, $3x + 0y - 9 = 0$ — questions sometimes test exactly this connection.
• Misreading the condition on $a$ and $b$: it only forbids both being zero at once, so $0x + 5y - 10 = 0$ (i.e. $5y = 10$) is still perfectly valid.
• Sloppy sign-flips while moving a term across the equals sign, which silently changes the value of $a$, $b$, or $c$ that gets read off.

Ask the tutor: Want to practice converting a few messy equations into standard form, or check whether a particular pair of numbers is a solution — ask the AI tutor to work through more examples with you.