# If ax+by=ap+bq, must we have x=p and y=q?

This is part of a series on common misconceptions.

True or False?For all real numbers \({\color{purple}a},{\color{purple}b},{\color{red}x},{\color{green}y},{\color{red}p},\color{green}q\), if \(\color{purple}a{\color{red}x}+\color{purple}{b}{\color{green}y}=\color{purple}{a}{\color{red}p}+\color{purple}{b}\color{green}{q}\), then \( {\color{red}x}=\color{red}{p}\) and \({\color{green}y}={\color{green}q}. \)

**Why some people say it's true:** Observe the symmetry of the equation. If \( {\color{red}x}=\color{red}{p}\) and \({\color{green}y}=\color{green}{q}\), then clearly this equation is true.

**Why some people say it's false:**
We have
\[\begin{align}
\color{purple}{a}{\color{red}x}+\color{purple}{b}{\color{green}y}&=\color{purple}{a}{\color{red}p}+\color{purple}{b}\color{green}{q}\\
({\color{purple}a} + {\color{purple}b})({\color{red}x}+{\color{green}y})&=({\color{purple}a} + {\color{purple}b})({\color{red}p}+{\color{green}q})\\
({\color{red}x}+{\color{green}y})&=({\color{red}p}+{\color{green}q}).
\end{align}\]
Therefore, so long as \(({\color{red}x}+{\color{green}y})=({\color{red}p}+{\color{green}q})\), it's not necessarily the case that \({\color{red}x}=\color{red}{p}\) and \({\color{green}y}=\color{green}{q}\).

The statement is \( \color{red}{\textbf{false}}\). (But the argument above for

whyit is false is also completely incorrect.)

Proof(that the statement is false):The statement claims that this algebraic logic holds

for any real numbers\(a, b, x, y, p,\) and \(q\). Therefore, if we can find even a single counterexample, where \(\color{purple}{a}{\color{red}x}+\color{purple}{b}{\color{green}y}=\color{purple}{a}{\color{red}p}+\color{purple}{b}{\color{green}q}\) is true but it isnotthe case that \({\color{red}x}=\color{red}{p}\) and \({\color{green}y}=\color{green}{q}\), then we have disproven the statement.Here is our counterexample:

If we set \({\color{purple}a}={\color{purple}b}=\color{purple}0,\) then \(x,y,p,q\) can be any real numbers. For example, \({\color{red}x}={\color{red}100},\, {\color{green}y}={\color{green}\pi},\, {\color{orange}p} ={\color{orange}1729},\, {\color{turquoise}q}={\color{turquoise}1234} \) gives

\[{\color{purple}0}\times {\color{red}100} + {\color{purple}0} \times {\color{green}\pi} = {\color{purple}0} \times {\color{orange}1729} + {\color{purple}0} \times {\color{turquoise}1234}.\]

Even though we have chosen \(x,y,p,\) and \(q\) so that \(x\neq p\) and \(y\neq q\), the equation above is true since \(0 + 0 = 0 + 0\).

Additionally, we can deduce that if \(a=0,\) then we definitely have \(y=q\) but not necessarily \(x=p\) since \(x,p\) can be any real numbers. Similarly, if \(b=0,\) then we definitely have \(x=p\) but not necessarily \(y=q\) since \(y,q\) can be any real numbers. \(_\square\)

Consider one last case: if \(a\neq 0\) and \(b \neq 0\) but \(a = b\), then what can you deduce about \(x, y, p,\) and \(q?\)

Studying Equations on Different Domains:What we have been interested in on this page is the the study of how various domain constraints on our variables affect the range of values that the other variables can take on. To study this, we fix different sets of the variables in different ways and then examine what freedom the other variables still have in response.

Here is a follow-up question:

An equation is a condition satisfied by only some sets of values assigned to the variables in question. This set is usually a restriction of the natural domains of the variables. For example, the equation \(y = 3x,\) where \(x\) and \(y\) are real numbers, is satisfied by an infinite number of points that, represented on a coordinate plane, form a line. But if you further know that \(y\) is an

even integer, what can you deduce about \(x\)? Must \(x\) also be even? How has your new knowledge about \(y\) affected what values \(x\) might take on? Questions like this are common in both algebra and number theory.

What's wrong with the "proof" at the top of this pagethat uses cancelling \({\color{purple}a} + \color{purple}b\) from both sides?Although it reaches the correct conclusion, the argument at the top of this page is completely invalid. Consider the first step:

\[ \begin{align} \color{purple}{a}{\color{red}x}+\color{purple}{b}{\color{green}y}&=\color{purple}{a}{\color{red}p}+\color{purple}{b}\color{green}{q}\\ ({\color{purple}a} + {\color{purple}b})({\color{red}x}+{\color{green}y})&=({\color{purple}a} + {\color{purple}b})({\color{red}p}+{\color{green}q}). \end{align} \]

Notice that \( ({\color{purple}a} + {\color{purple}b})({\color{red}x}+{\color{green}y}) = \color{purple}{a} {\color{red}x} + \color{purple}{a} {\color{green}y} + \color{purple}{b} {\color{red}x}+\color{purple}{b} \color{green}{y} \), as opposed to \( \color{purple}{a}{\color{red}x}+\color{purple}{b}\color{green}{y} \). Hence, the left-hand sides of both equations are not the same.

Rebuttal: I don't think the case where \(a=b=0\) should count since it's only one case.

Reply: The claim in question asks if this inference is true forall real numbers\(a, b, x, y, p,\) and \(q.\) The sets of real numbers where both \(a\) and \(b\) equal 0 are part of this domain.

Rebuttal: If it's false, then what's the mistake in the proof below?We can also rewrite the equation using the following steps:

\[\begin{align} \color{purple}{a}{\color{red}x}+\color{purple}{b}\color{green}{y}&=\color{purple}{a}{\color{red}p}+\color{purple}{b}\color{green}{q}\\ \color{purple}{a}{\color{red}x}+\color{purple}{b}{\color{green}y}-(\color{purple}{a}{\color{red}p}+\color{purple}{b}{\color{green}q}) &= 0 \\ \color{purple}{a}{\color{red}x}- \color{purple}{a}{\color{red}p} + \color{purple}{b}{\color{green}y} - \color{purple}{b}\color{green}{q} &= 0 \\ {\color{purple}a}({\color{red}x}-{\color{red}p})+{\color{purple}b}({\color{green}y}-{\color{green}q})&=0. \end{align}\]

Clearly \({\color{purple}a} ({\color{red}x}-{\color{red}p})+{\color{purple}b}({\color{green}y}-{\color{green}q})=0\) when \( {\color{red}x}=\color{red}{p}\) and \({\color{green}y}=\color{green}{q}\).

Reply: This proof is a common example of using the symmetry of the equation to justify incorrect algebraic steps.

\[\large {\color{purple}a}{\color{red}x} + {\color{green}5}{\color{purple}b} = {\color{red}2}{\color{purple}a} + {\color{purple}b}{\color{green}q} \\\\\\ \large \text{implies} \\\\\\ \large {\color{red}x} = {\color{red}2} \text{ and } {\color{green}q}={\color{green}5}.\]

**True or False?**

For all real numbers \({\color{purple}a}, {\color{purple}b}, {\color{red}x}, \text{ and } {\color{green}q}\), the statement above must be true.

**See Also**

**Cite as:**If ax+by=ap+bq, must we have x=p and y=q?.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/not-the-distributive-property/