What Makes an Equation Linear?
An equation in two or three variables is linear when its variable-dependent part has a restricted form: each variable may occur to the first power, but variables may not be multiplied together, raised to higher powers, or placed inside nonlinear functions. This gives a quick recognition test, but the meaning of “linear” needs to be precise. In this tutorial, classification is based on the equation’s algebraic form after valid simplification—not merely on whether some other equation happens to have the same solutions.
We work over \(\mathbb{R}\). For two variables, use the labels \(x\) and \(y\); for three, use \(x\), \(y\), and \(z\). A real constant is a fixed real number, independent of the values of those variables. Thus \(5x\) has a constant coefficient, while \(xy\) does not have a constant coefficient on \(y\): its coefficient \(x\) depends on a variable.
Moving a term across the equals sign changes its sign. Expanding brackets and collecting like terms are valid simplifications because they preserve the two sides’ equality for every real assignment of the variables. Multiplying or dividing both sides by a nonzero constant also preserves the equation’s solutions and does not introduce a variable-dependent coefficient. By contrast, dividing by a variable is not generally valid for all assignments: that variable could be zero.
In the standard form, each variable appears only in a term that is a constant multiple of that variable. A constant term is allowed. After expansion, the variable-dependent terms must therefore be of the form \(ax\), \(by\), and \(dz\). The terms \(x^2\), \(xy\), and \(\sqrt{x}\) do not fit this pattern. The absence of these terms is a test of form, not a claim about whether the equation has a solution or what its solution set looks like.
A Precise Recognition Test
Proof. Suppose first that the equation is linear. By definition, in two variables it can be written \(ax+by=c\), or in three variables \(ax+by+dz=c\), with constant coefficients. Moving the right-hand side to the left gives, respectively, \(ax+by-c=0\) or \(ax+by+dz-c=0\). Each resulting expression is a sum of constant multiples of individual variables and a constant. Conversely, suppose moving terms and simplifying leaves such an expression equal to zero. In two variables it has the form \(ax+by+k=0\), so moving \(k\) to the right gives \(ax+by=-k\). In three variables it has the form \(ax+by+dz+k=0\), so \(ax+by+dz=-k\). These are the required standard forms. \(\square\)
For a quick first check, expand brackets and inspect the terms. A variable raised to a power other than one, two variables multiplied together, a variable in a denominator, or a variable inside a root or other nonlinear function is a warning sign. If the offending term cancels under valid simplification, it is no longer an obstacle. If it remains, the expression is not in standard linear form.
Worked Example: Moving Terms in a Two-Variable Equation
Decide whether \(3x-2y+5=y+7x-1\) is linear, and write it in standard form.
Move the entire right-hand side to the left, changing the sign of each term on that side:
Collect the \(x\)-terms, the \(y\)-terms, and the constants. The \(x\)-terms give \(3x-7x=-4x\), the \(y\)-terms give \(-2y-y=-3y\), and the constants give \(5+1=6\). Thus the equation becomes \(-4x-3y+6=0\), or, after multiplying by \(-1\) and moving the constant,
This is linear: the coefficients \(4\) and \(3\) and the right-hand side \(6\) are constants, and each variable appears only to the first power.
Check: Subtracting the right side of the original equation from its left side gives \(3x-2y+5-(y+7x-1)=-4x-3y+6\). This is the negative of \(4x+3y-6\), so setting either expression equal to zero gives the same equation.
Worked Example: Expanding a Three-Variable Equation
Classify \(2(x-3)+y=4z-(x+2y)+1\) and put it in standard form.
Expand the left side: \(2(x-3)+y=2x-6+y\). Expand the right side, taking care to distribute the minus sign through the parentheses: \(4z-(x+2y)+1=4z-x-2y+1\). Move the right side to the left:
Collect like terms: \(2x+x=3x\), \(y+2y=3y\), the \(z\)-term is \(-4z\), and \(-6-1=-7\). Therefore \(3x+3y-4z-7=0\), so the standard form is
Every variable has a constant coefficient and occurs only to the first power. The equation is linear.
Check: Subtracting the original right side from its left side gives \((2x-6+y)-(4z-x-2y+1)=3x+3y-4z-7\). Thus the derived standard form is exactly the original equation rearranged.
Why Higher Powers and Products Fail the Test
There is a useful way to understand the restriction. In an affine expression such as \(ax+by+c\), changing \(x\) by an amount \(h\) changes the expression by \(ah\), regardless of the starting values of \(x\) and \(y\). In contrast, changing \(x\) in \(x^2\) changes its value by an amount that depends on the starting value. A variable product such as \(xy\) also makes the change in one variable depend on the value of the other.
Proof. First suppose \(f(x_1,\dots,x_n)=a_1x_1+\cdots+a_nx_n+c\). Changing only \(x_i\) to \(x_i+h\) changes the value to \(a_1x_1+\cdots+a_i(x_i+h)+\cdots+a_nx_n+c\). Subtracting the original value leaves \(a_i h\), as required. Conversely, suppose the stated change rule holds for each coordinate. Starting at \((0,\dots,0)\), change the first coordinate to \(x_1\), then the second to \(x_2\), and continue through the \(n\)-th coordinate. At step \(i\), the change in \(f\) is \(a_i x_i\), because the rule holds for every starting point and every change. Adding these changes gives \(f(x_1,\dots,x_n)-f(0,\dots,0)=\sum_{i=1}^n a_i x_i\). Setting \(c=f(0,\dots,0)\) proves the stated form. \(\square\)
This result explains why the first-power rule matters: affine expressions have a fixed change per unit change in each variable. It is not a replacement for the standard-form definition. In particular, the classification of an equation below is about its algebraic form, not solely about the set of points that satisfy it.
Worked Example: A Nonlinear Equation and a Cancellation
Classify \(x^2=0\) in two variables, and compare it with \(x^2-x^2+2x-3y=5\).
For the first equation, the left-minus-right expression is \(x^2\). The square term remains, so it cannot be written in the required form \(ax+by+k\) by collecting or expanding terms. The equation is not linear under the standard-form definition. This is true even though its solutions are exactly the same as those of \(x=0\): for real \(x\), \(x^2=0\) holds exactly when \(x=0\). Equality of solution sets does not make the original equation a linear equation in standard form.
The change in \(x^2\) also shows why the square is not an affine term. Increasing \(x\) from \(0\) to \(1\) changes \(x^2\) by \(1^2-0^2=1\). Increasing \(x\) from \(1\) to \(2\) changes it by \(2^2-1^2=4-1=3\). Those changes are not the same for the same increase \(h=1\), so \(x^2\) does not have the constant-change property.
In the second equation, the two square terms cancel identically: \(x^2-x^2=0\) for every real \(x\). The equation simplifies to \(2x-3y=5\), which is linear.
Check: The simplification of the second equation subtracts equal quantities on the same side, so it leaves the equation unchanged for every \(x,y\). Its remaining coefficients, \(2\) and \(-3\), and its right-hand side, \(5\), are constants. The first equation has no such cancellation.
Common Mistakes and Reliable Checks
Mistake: judging an equation by its solution set alone. A tempting but incorrect argument is: “\(x^2=0\) has the same solutions as \(x=0\), so \(x^2=0\) is linear.” The correction is to distinguish two questions. The equations have the same solutions over \(\mathbb{R}\), but \(x^2=0\) does not have linear standard form. Under the definition used here, it is not a linear equation. That distinction prevents the test from classifying expressions by accidental properties of their solutions.
Mistake: treating a variable coefficient as a constant. The expression \(xy\) can be described as “\(x\) times \(y\),” but neither factor is a fixed real coefficient. The equation \(xy=4\) is not linear in \(x\) and \(y\). Likewise, \(x(y+1)=0\) expands to \(xy+x=0\); the mixed term \(xy\) remains, so the equation is not linear in those variables.
Mistake: overlooking a variable inside a function. The expression \(\sqrt{x}+y=3\) is not linear because \(\sqrt{x}\) is not a constant multiple of \(x\). Similarly, \(1/x+y=3\) is not linear; it also is not defined at \(x=0\), so multiplying by \(x\) would not be a simplification valid for every real value of the variables.
A reliable classification can be made in three checks:
- Move one side of the equation to the other and expand parentheses, distributing negative signs carefully.
- Collect like terms and simplify only by identities valid for all real values of the variables.
- Inspect the remaining expression: every variable term must be a constant multiple of one variable to the first power, with no variable products or nonlinear functions.
Constants are allowed, and some variables may be absent. For example, \(2x=5\) is a linear equation in \(x,y\) with coefficient \(0\) on \(y\). The equation \(0x+0y=0\) is a degenerate linear equation under the convention stated here. These cases do not change the recognition test; the deciding issue is the form of the variable terms.
Check Your Understanding
For each equation, decide whether it is linear in the named variables. If it is linear, write it in standard form. For each non-linear equation, identify the term that fails the test.
- In \(x,y\), classify \(5(x-1)=2y+3\).
- In \(x,y,z\), classify \(x+2(y-z)=4\).
- In \(x,y\), classify \(3x+y^2=8\).
- In \(x,y\), classify \(xy-yx+x=2\), simplifying validly before deciding.
- Explain why \(x^2=0\) is not linear under the standard-form definition even though it has the same solutions as \(x=0\).
