EQN-06-05

Comparison with Algebra & Variables

Compare two algebraic quantities by simplifying both sides and respecting any given constraint.

~5 min read•Difficulty 3/5

Simplify both quantities first

Treat the comparison like an inequality you are trying to settle. Subtract the common part from both sides — what remains decides it.

A : 3 x + 7 B : 3 x + 2

subtract 3 x and 2 from both \to A : 5 B : 0 \to A is greater (for every x)

Use the given constraint

A line of shared information above the quantities usually pins things down. Without it you would often be at "D".

Given x + y = 12. A : x B : y

No constraint links their sizes individually \to D (x could be 10, y 2, or vice versa) .

Never multiply by an unknown sign

If a variable could be positive, negative, or zero, you cannot multiply or divide the comparison by it — the order might flip. This is the single biggest source of "cannot be determined".

Plug in when stuck

If algebra is messy, test allowed values (0, 1, a negative, a fraction). Consistent result $\to A / B / C$ ; conflicting results $\to D .$

Worked examples

Example 1 (difficulty 2/5)

Given

x > 2.

Quantity

A : 4 x - 1.

Quantity

B : 3 x + 1.

Which is greater?

◎ Predict the answer first:

Example 2 (difficulty 3/5)

Given xy

= 12.

Quantity A: x. Quantity B: 4. Which is greater?

◎ Predict the answer first:

Tips

Compute $A - B$ and check its sign against the constraint — cleaner than juggling both sides.
A single equation in two unknowns rarely fixes one of them $\to$ suspect D.
Never multiply the comparison by a variable of unknown sign.

Common mistakes

Assuming x is positive when the problem never says so

Why: Positive values are the default mental picture.

Fix: Unless restricted, test a negative and a fraction too.

Solving one constraint as if it fixed each variable

Why: It looks like enough information.

Fix: Two unknowns generally need two facts; otherwise the answer is often D.

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