What a Contradiction Tells Us
A system of equations asks for one choice of values that makes every equation true at once. If the equations force an impossible statement, such as \(0=5\), then no choice of values can satisfy them all. The important point is not simply that one equation looks unusual: the contradiction must follow from the original equations by steps that preserve what any shared solution would have to satisfy.
Throughout, take the variables and coefficients to be real numbers. A solution of a system is a list of values for its variables that makes every equation in the system true. A system is inconsistent if it has no solution, and consistent if it has at least one solution. A contradiction is a false equality, such as \(0=5\) or \(3=4\). In contrast, \(0=0\) is true and does not itself rule out any solution.
When an equation is true, multiplying both sides by the same number or adding an equation that is also true gives another true equality. This elementary fact lets us combine equations without needing to know the variable values. If the resulting equality is impossible, the original equations could not all have been true for a shared choice of values.
A General Test from Linear Combinations
Write a system of \(m\) linear equations in \(n\) variables as
Here \(a_{ij}\) is the coefficient of \(x_j\) in equation \(i\), and \(b_i\) is its right-hand side. A linear combination of the equations means multiplying equation \(i\) by a chosen real number \(c_i\), for each \(i\), and adding the resulting equalities. The same multiplier must be used on the left-hand side and the right-hand side of each equation.
Proof. Suppose \((x_1,\ldots,x_n)\) is a solution. Then for each equation \(i\), \(a_{i1}x_1+\cdots+a_{in}x_n=b_i\). Multiply this equality by \(c_i\), and add the resulting \(m\) equalities. Because addition and multiplication distribute over finite sums, the result is
Thus every solution of the original system must satisfy the combined equation. \(\square\)
The theorem gives a direct test for inconsistency: choose multipliers so that every variable’s coefficient cancels, while the combined right-hand side does not. Then the combined equation has the form \(0=d\), where \(d\ne0\).
Proof. Suppose, to the contrary, that the system has a solution. By the theorem, that solution must satisfy the linear combination using the multipliers \(c_1,\ldots,c_m\). The coefficient of every variable in this equation is zero, so its left-hand side is zero. Its right-hand side is \(\sum_{i=1}^{m}c_i b_i\), which is nonzero by assumption. The solution would therefore have to satisfy \(0=\sum_{i=1}^{m}c_i b_i\) with a nonzero right-hand side. This is false, so the system cannot have a solution. \(\square\)
This test proves that a system is inconsistent when a suitable combination is found. It does not say that a system is consistent whenever no such combination has yet been found. A search that has not produced a contradiction is not, by itself, a proof that a solution exists.
Finding Contradictions in Small Systems
Worked Example: Same Left-Hand Side, Different Right-Hand Sides
Consider the system \(x+y=4\) and \(x+y=9\). Decide whether one pair \((x,y)\) can satisfy both equations.
Subtract the second equation from the first. If a pair satisfied both equations, subtracting their equal left- and right-hand sides would give another true equality:
The left side is zero because each variable term cancels, while the right side is \(4-9=-5\). Since \(0\ne-5\), the combined equation is false. No pair can satisfy both equations.
Check: A supposed solution would have to make the same number \(x+y\) equal to both \(4\) and \(9\). That would require \(4=9\), which is false. The subtraction records that conflict as \(0=-5\).
Worked Example: A Multiple of One Equation Conflicts with Another
Test the system \(2x-3y=1\) and \(6x-9y=4\). The left side of the second equation is three times the left side of the first, but its right side is not three times as large.
Multiply the first equation by \(3\), multiplying both sides:
The original second equation says \(6x-9y=4\). Subtract it from the multiplied first equation. Any shared solution would therefore have to satisfy
Since the right side is \(-1\), the equality is impossible. Thus the system has no solution.
Check: If a value of \(6x-9y\) existed that satisfied both equations, it would have to equal \(3\) and \(4\) at the same time. Those are unequal, so the contradiction is genuine.
Worked Example: Combining Three Equations
Consider the four equations
Add the first three equations. Every term is included, so the left-hand side is \(2x+2y+2z\), and the right-hand side is \(2+3+5=10\):
Now subtract twice the fourth equation, \(2x+2y+2z=12\). A shared solution would have to satisfy the result:
This is false, so there is no triple \((x,y,z)\) satisfying all four equations.
Check: The first three equations together imply \(2(x+y+z)=10\), hence \(x+y+z=5\). The fourth equation requires \(x+y+z=6\). A single number cannot equal both \(5\) and \(6\), in agreement with the derived contradiction.
What Counts as a Contradiction?
A zero-variable equation can be either false or true. The equation \(0=5\) is false and rules out a shared solution if it follows from the system. The equation \(0=0\) is true and supplies no such evidence. It merely says that a particular combination of the equations has canceled on both sides.
For example, the equation \(x+y=3\) is not contradictory. Subtracting it from itself gives \(0=0\), which is true. It would be an error to treat every equation with zero coefficients as proof of inconsistency. The right-hand side matters: the contradiction test applies only when the variables cancel and the remaining right-hand side is nonzero.
There is also an important difference between finding a false statement in a system and deriving one from a system. An equation in the original system might itself be false, such as \(0=5\), and then the system has no solution immediately. More often, each original equation is individually possible, but together they imply a contradiction. In either case, the conclusion depends on showing that any shared solution would have to satisfy the false equality.
Common Mistake: Changing Only One Side
A frequent error is to multiply or subtract equations on one side only. For instance, starting with \(2x-3y=1\), a student might multiply the left side by \(3\) but leave the right side unchanged, and write \(6x-9y=1\). That does not follow from the original equation. Multiplying an equality by \(3\) requires multiplying both sides: \(3(2x-3y)=3(1)\), so the correct result is \(6x-9y=3\).
A related sign error can hide a real contradiction. In the first worked example, subtracting the second equation from the first gives \(4-9=-5\), not \(5\). Either \(0=-5\) or \(0=5\) would be false, but the written arithmetic should still be accurate: a proof must preserve the actual equality at every step.
Use this short check whenever you combine equations:
- Keep the same variable terms on both sides of each equation.
- Apply every multiplier to the whole equation, including its right-hand side.
- When subtracting, subtract each term, including the constants, in the same order.
- After cancellation, check whether the remaining equality is false, rather than merely unusual.
Why This Reasoning Is Useful
A contradiction is a certificate of inconsistency: a short chain of valid equalities shows why no shared solution can exist. This can be more informative than simply reporting that a system has no solution, because it identifies which equations conflict and how the conflict arises. In the examples, the issue was not an inability to find values; the equations demanded incompatible values for the same expression.
The central logic is one-way and precise. If the equations imply \(0=d\) for \(d\ne0\), there is no solution. If a combination gives \(0=0\), that combination has not ruled out solutions. And if no contradiction has been found, more reasoning may still be needed to determine whether solutions exist. Keeping those conclusions separate prevents a common jump from “I have not found a contradiction” to “the system must be consistent.”
Check Your Understanding
For each question, justify your conclusion by stating what any shared solution would have to satisfy.
- Do \(x-y=2\) and \(x-y=7\) have a common solution? Write the equality obtained by subtracting the first equation from the second.
- Multiply \(3x+2y=4\) by \(2\). What must happen to both sides, and what equation results?
- Explain why \(0=0\) does not prove that a system is inconsistent, while \(0=-3\) does.
- Suppose a combination of equations cancels every variable and leaves \(0=8\). What conclusion follows, and why?
- Is “no contradiction has been found” enough to prove that a system has a solution? Explain the difference between this statement and the contradiction test.