You've solved hundreds of equations already. Before this course puts vectors and matrices to work, it needs you to notice something you've been doing without naming it: a huge share of those equations belonged to one specific family, and that family is the one this entire course is built around. Linear algebra is the algebra of linear equations — so the first job is pinning down exactly what "linear" does and doesn't allow, because from here on almost everything hinges on staying inside that boundary.
Think of a linear equation as a recipe that allows only two moves on each ingredient. First, scale it by some fixed amount. Then dump all the scaled ingredients into one bowl and add them up. That's it. You're allowed 4x, which scales x by 4, and -x, which scales x by -1. You're allowed to add or subtract any number of those scaled ingredients together. What breaks the recipe is any step that makes one ingredient's contribution depend on how much of it there already is, or on another ingredient. Squaring an amount breaks it. So does multiplying two amounts together, or taking a square root of an amount. Doubling the amount of a linear ingredient always exactly doubles its contribution to the total. That clean, no-surprises proportionality is the entire idea "linear" is naming. It's why straight lines, flat planes, and every higher-dimensional version of them are all called linear too, since nothing curves. Nothing in the recipe can curve.
An equation in the variables x_1, x_2, \ldots, x_n is linear when it can be written in the form a_1 x_1 + a_2 x_2 + \cdots + a_n x_n = b, where a_1, \ldots, a_n and b are fixed real-number constants (the coefficients and the constant term). Every term on the left is a constant multiplied by exactly one variable, and those terms are only ever added or subtracted together. Nothing else is permitted: no variable raised to a power other than 1, no two variables multiplied together, no variable inside a root, a denominator, a trig function, an exponential, or a logarithm. A variable can also be absent — a coefficient of 0 is still a constant — so 3x_1 + 0x_2 = 7 and 3x_1 = 7 describe the same linear equation. The requirement is entirely about what happens to each variable in isolation; the coefficients themselves can be any real numbers, including irrational or negative ones, with no effect on linearity.
A quick way to stress-test that definition: look at the exponent and the company each variable keeps. 5x - 2y = 9 is linear — each variable appears alone, to the first power, multiplied only by a constant. 5x^2 - 2y = 9 is not — the exponent on x broke the rule. 5xy = 9 is not either, even though no exponent is above 1 anywhere — the two variables are multiplied by each other, not by separate constants, and that's exactly the kind of interaction the definition forbids.
A linear equation is a straight line
Predict: is the graph of 5x - 2y = 9 a straight line or a curve? Drag the plane, then trace each line to confirm — every point on it satisfies its equation exactly.
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Check your understanding (counts toward your progress)
Rewrite 4(y - 2) = -3x + 10 in the form ax + by = c (as the left side alone, i.e. give the expression equal to the constant on the right after moving all variable terms to the left — enter it as '3x+4y'). What is that left-hand expression?
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Any equivalent form counts — 2x+3, 3 + 2·x, … Use the keypad below the field for fractions, roots, and other symbols.
Why
Distribute: 4y - 8 = -3x + 10. Move -3x to the left: 3x + 4y - 8 = 10. Move -8 to the right: 3x + 4y = 18. The left side is 3x + 4y.
Correct answer
Also accepted:
Why
Move every variable term to the left side and every constant to the right side, then simplify each side separately.
Which equation is NOT linear?
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
This one IS linear — both variables appear alone, to the first power. Look again for the equation with an unusual exponent or variable placement.
This one IS linear — three variables, each alone to the first power, all added together.
Fractional coefficients like -\frac{1}{3} are still just constants — this equation is linear.
Why
In 4^x + y = 6, the variable x sits in an exponent, not as something multiplied by a constant — exponential terms are never linear.
A student claims \frac{x}{4} + \frac{y}{7} = 1 is not linear because it has fractions. Are they right?
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
The rule restricts what happens to the VARIABLES, not the coefficients — a coefficient can be any real number, fraction included.
The value on the right side (the constant term b) can be any real number, including 1 or 0 — it doesn't affect linearity.
Dividing a variable by a number is the same as multiplying it by a constant (\frac{x}{4} = \frac{1}{4}x) — very different from dividing BY a variable, which is what actually breaks linearity.
Why
\frac{x}{4} + \frac{y}{7} = 1 is really \frac{1}{4}x + \frac{1}{7}y = 1 — constant coefficients times variables to the first power, added together. It's linear.
Ask what the linearity rule actually restricts: the power on each variable, not the type of number multiplying it.
Practice: Is it linear? -- a full pass (5 problems · doesn’t affect your progress)
Which of these equations is linear? (For reference, the choices given are: 7x - 3y = 12; x^2 + y = 5; xy = 10; \sqrt{x} + y = 1.)
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
The exponent 2 on x breaks the rule — every variable must appear to the first power only.
xy multiplies two variables together, which isn't allowed — each term needs a constant times exactly ONE variable.
\sqrt{x} is x raised to the power \frac{1}{2}, not 1, so this fails the exponent requirement.
Why
7x - 3y = 12 has every variable appearing alone, to the first power, multiplied only by a constant coefficient, with those terms added together — exactly the linear recipe.
7x - 3y = 12
In the linear equation 6x - 9y = 15, what is the coefficient of y (include the sign)? (Work through it specifically for 6x - 9y = 15 and y.)
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Why
Written in the standard form a_1x_1 + a_2x_2 = b, the coefficient attached to y is -9.
Correct answer
Also accepted:
Why
Is 2(x - 5) + 3y = x + 1 linear once it's fully expanded and simplified? (For reference, the choices given are: Yes — it simplifies to x + 3y = 11, which fits the linear form; No — the parentheses mean x is being multiplied by another variable; No — expanding introduces an x^2 term; Yes, but only because both sides happen to already match.)
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
The 2 outside the parentheses is a constant, not a variable — distributing 2(x-5) gives 2x - 10, ordinary linear terms.
Distributing a constant across a sum never creates a new exponent — check the expansion again.
The two sides don't already match; you have to distribute and combine like terms first, which is exactly what makes it linear.
Why
Distribute: 2x - 10 + 3y = x + 1. Move terms: 2x - x + 3y = 1 + 10, so x + 3y = 11. Every term is a constant times one variable to the first power — linear.
Distribute the constant 2 across (x-5) first, then combine like terms on both sides before checking the exponents.
Yes — it simplifies to x + 3y = 11, which fits the linear form
Rewrite 4(y - 2) = -3x + 10 in the form ax + by = c (as the left side alone, i.e. give the expression equal to the constant on the right after moving all variable terms to the left — enter it as '3x+4y'). What is that left-hand expression? (Work through it specifically for 4(y - 2) = -3x + 10 and ax + by = c.)
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Any equivalent form counts — 2x+3, 3 + 2·x, … Use the keypad below the field for fractions, roots, and other symbols.
Why
Distribute: 4y - 8 = -3x + 10. Move -3x to the left: 3x + 4y - 8 = 10. Move -8 to the right: 3x + 4y = 18. The left side is 3x + 4y.
Correct answer
Also accepted:
Why
Move every variable term to the left side and every constant to the right side, then simplify each side separately.
Which statement correctly explains why x_1 - 2x_2 + 0x_3 = 7 is linear even though x_3 doesn't visibly appear? (For reference, the choices given are: A missing variable is the same as that variable having coefficient 0, which is a perfectly valid constant; It isn't linear — every variable named in the system must actually appear with a nonzero coefficient; It's linear only because there happen to be exactly three variables; The equation is linear only in x_1 and x_2, and undefined for x_3.)
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
A coefficient of exactly 0 is still a real-number constant — the definition never requires coefficients to be nonzero.
The number of variables mentioned doesn't matter; what matters is that every term present fits the constant-times-first-power pattern.
The equation is a single, fully defined linear equation in all three variables — x_3 is included, just scaled by 0.
Why
The definition only requires each variable to be multiplied by a fixed real constant and added — 0 is a fixed real constant, so an absent variable is just one with coefficient 0, and the equation is linear.
A coefficient of 0 still means x_3 is multiplied by a constant (just a constant equal to 0) -- that's exactly what the linear form allows.
A missing variable is the same as that variable having coefficient 0, which is a perfectly valid constant
The one exponent that's allowed, and the two moves that break it
The fastest way to misclassify an equation is to trust its appearance instead of checking the actual structure once it's fully expanded. An equation can look complicated and still be linear, or look simple and not be. Two classic traps:
Worked example
Is 3(x + 2) - y = 4x - 1 linear?
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1
Distribute the left side first — don't judge linearity from an unexpanded form. 3(x+2) - y = 3x + 6 - y.
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2
Move every term to one side so the equation reads (variables) = (constant): 3x + 6 - y = 4x - 1 \;\Longrightarrow\; 3x - 4x - y = -1 - 6.
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3
Combine the like terms in x: \textcolor{#4f46e5}{3x} - \textcolor{#4f46e5}{4x} = \overset{-1}{\cancel{-x}}, giving -x - y = -7.
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4
Every remaining term is a constant times a single variable to the first power, added together: \boxed{-x - y = -7 \text{ is linear}} — even though the original form had a product-looking 3(x+2), that product was a constant times a sum, which distributes into ordinary linear terms, not a variable-times-variable product.
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Worked example
Is \frac{2}{x} + y = 6 linear?
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1
Check each variable's role separately. The y term is fine: constant 1 times y to the first power.
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2
The x term is \frac{2}{x}, which is 2x^{-1} — x sits in the denominator, so its exponent is -1, not 1.
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3
No amount of rearranging turns x^{-1} into a plain constant-times-x term; the variable is being divided into, not multiplied by a constant.
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4
\boxed{\frac{2}{x} + y = 6 \text{ is NOT linear}} — one offending term is enough to disqualify the whole equation, no matter how ordinary the rest of it looks.
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Practice: Linear despite appearances -- reading past the surface (4 problems · doesn’t affect your progress)
A student claims \frac{x}{4} + \frac{y}{7} = 1 is not linear because it has fractions. Are they right? (For reference, the choices given are: No — the fractions are constant coefficients (\frac{1}{4} and \frac{1}{7}), which is allowed; Yes — fractions of any kind disqualify an equation from being linear; No, but only because the right side equals 1 instead of 0; Yes — dividing by a number is the same restriction as dividing by a variable.)
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
The rule restricts what happens to the VARIABLES, not the coefficients — a coefficient can be any real number, fraction included.
The value on the right side (the constant term b) can be any real number, including 1 or 0 — it doesn't affect linearity.
Dividing a variable by a number is the same as multiplying it by a constant (\frac{x}{4} = \frac{1}{4}x) — very different from dividing BY a variable, which is what actually breaks linearity.
Why
\frac{x}{4} + \frac{y}{7} = 1 is really \frac{1}{4}x + \frac{1}{7}y = 1 — constant coefficients times variables to the first power, added together. It's linear.
Ask what the linearity rule actually restricts: the power on each variable, not the type of number multiplying it.
No — the fractions are constant coefficients (\frac{1}{4} and \frac{1}{7}), which is allowed
A tutoring app charges a flat \$5 sign-up fee plus \$2 per session, x. Which equation correctly models total cost y, and is it linear? (For reference, the choices given are: y = 2x + 5 — linear; y = 2x^2 + 5 — not linear; y = 2x + 5 — not linear, because of the added constant 5; y = 5x^2 — linear.)
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
There's no reason to square the number of sessions here — cost grows by a fixed \$2 per session, which is the linear model, not a squared one.
An added constant term (+5) doesn't break linearity — the form ax + b = y (equivalently y - ax = b) is exactly the allowed structure.
This squares x and drops the flat fee entirely — it doesn't match the situation described.
Why
Total cost is a constant (2) times the number of sessions plus a fixed fee (5): y = 2x + 5, or rearranged -2x + y = 5 — a constant times x to the first power, plus a constant, is linear.
Write total cost as flat fee plus (price per session) times (number of sessions), then check whether that expression matches the linear form ax+b.
y = 2x + 5 — linear
For which value of k does kx^2 + 3y = 5 become a linear equation? (For reference, the choices given are: k = 0; k = 1; k = -1; No value of k makes it linear.)
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
With k=1 the term is x^2, still degree 2 — the exponent is the problem, and no coefficient value fixes an exponent.
With k=-1 the term is -x^2 — still squared, just negated. The exponent hasn't changed.
One value does work: making the coefficient of the offending term exactly 0 erases it entirely, leaving a linear equation in y alone.
Why
When k=0, the term kx^2 becomes 0 \cdot x^2 = 0, so the equation collapses to 3y = 5 — linear (a constant coefficient of 0 on x is perfectly allowed, it just means x doesn't appear).
Think about what value of k would make the coefficient on x^2 vanish entirely, leaving no x^2 term at all.
k = 0
The equation mx - 4y = 2m - 8 is linear for every real value of m except one value of m where a variable's coefficient becomes 0 in a way that still keeps the equation linear (coefficient 0 is always allowed) — so actually, for how many values of m does this equation FAIL to be linear? (Work through it specifically for mx - 4y = 2m - 8 and m and m.)
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Why
mx - 4y = 2m - 8 is linear for every real m: m and 2m-8 are just fixed numbers once you pick m, and x, y each still appear to the first power only. There's no value of m that introduces a forbidden exponent or a variable product, so it fails for 0 values of m.
Correct answer
Also accepted:
Why
Substitute a few small values of m and see whether the equation ever stops looking like a_1x+a_2y=b -- remember a coefficient of 0 is still allowed.
What's next
That two-move recipe — scale each variable by a constant, then add — is about to scale up hard: instead of one equation in two variables, this course spends its first several chapters on many linear equations in many variables at once, and on the compact notation (vectors, matrices) built specifically to keep that scaling under control. Coordinates and distance are next, to make sure the plane those equations live on is on solid footing too.
Where this goes next
Every later chapter of this course
This definition isn't a warm-up fact you'll leave behind — it's the load-bearing wall. A system of linear equations (Chapter 3) is just a stack of equations that each individually pass this test, and the entire reason matrices are worth inventing is that a linear equation's coefficients are the only information needed to reconstruct it: no exponents to track, no products of variables to worry about, just a row of numbers. Chapter 9's linear transformations — functions that respect addition and scaling exactly the way this section's "scale, then add" recipe describes — are this same idea promoted from equations to functions. Every time a later chapter says "linear," it means precisely the two moves you just learned to check for here, applied to something new.