GMAT · Algebra and constraints · TOPIC LESSON
Original GMAT Problem Solving instruction: reason from constraints, calculate efficiently without a calculator, and check the requested quantity.
BY THE END, YOU CAN
BEFORE YOU START
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01 / UNDERSTAND & APPLY
Choose a variable with a precise meaning and unit. “Five less than twice x” means 2x−5; “five minus twice x” means 5−2x. “A is three times B” means A=3B. Reading the equation back in words catches reversed relationships.
Expressions represent quantities; equations assert equality between them. A word problem may ask for an expression rather than the underlying variable. Keep the requested quantity visible so that a correct intermediate result is not mistaken for the answer.
ORIGINAL WORKED EXAMPLE
A delivery fee is $7 plus $3 per parcel. Write the total cost for p parcels and find the cost of 8 parcels.
Answer 7+3p dollars; $31 for 8 parcels.
ORIGINAL PRACTICE · 1
n/2+4.
Half of n is n/2, then add 4. The expression (n+4)/2 halves the added four as well and means something different.
02 / UNDERSTAND & APPLY
When substituting a negative number or an expression, use parentheses to preserve signs and grouping. Evaluate powers before a leading minus. If a formula has restrictions such as a nonzero denominator, check them before arithmetic.
Substituting a strategically chosen value can test whether an identity is false; one mismatch disproves it. Agreement at a single value does not establish a general identity or determine every possible variable value.
ORIGINAL WORKED EXAMPLE
If a=−2 and b=3, evaluate a²−2ab+b².
Answer 25.
ORIGINAL PRACTICE · 1
−21.
The expression is −(x²)+4x, so −9−12=−21. Squaring the leading minus as though the term were (−x)² changes the given expression.
03 / UNDERSTAND & APPLY
Distribution multiplies every term inside parentheses. A leading minus is multiplication by −1, so it changes every sign. Factoring extracts a common multiplicative structure and is often the fastest way to simplify without large intermediate calculations.
The zero-product property says a product is zero only if at least one factor is zero. It does not apply to a sum, and it does not let you set factors to zero when their product equals a nonzero number.
ORIGINAL WORKED EXAMPLE
Simplify 3(2x−5)−2(x+4).
Answer 4x−23.
ORIGINAL PRACTICE · 1
3x(2x−5).
Both terms contain 3x, which leaves 2x−5. Writing 3x(2x−15) fails because distributing it would produce −45x, not −15x.
04 / UNDERSTAND & APPLY
An equation remains equivalent when the same quantity is added to both sides or both sides are multiplied by the same nonzero constant. Isolate the variable by reversing operations. Clear fractional denominators with a common multiple to reduce bookkeeping.
If variable terms cancel, the resulting statement may be always true or impossible. The equation 2x+3=2x+3 has all real solutions; 2x+3=2x+5 has none. Do not divide by a variable expression without considering when it could be zero.
ORIGINAL WORKED EXAMPLE
Solve (x−2)/3+(x+1)/2=6.
Answer 37/5.
ORIGINAL PRACTICE · 1
None.
Expanding gives 4x−8=4x−7. Subtracting 4x gives −8=−7, an impossibility. Reporting x=0 ignores the fact that x has cancelled completely.
05 / UNDERSTAND & APPLY
A solution to simultaneous equations must satisfy every equation at once. Substitution is efficient when one equation already isolates a variable. Elimination is efficient when coefficients match or can be made to match with small multipliers.
Proportional equations may express the same constraint or contradictory constraints. Two equations do not guarantee a unique solution. Check whether they represent genuinely independent restrictions, and distinguish solving for each variable from determining a requested combination.
ORIGINAL WORKED EXAMPLE
Tickets cost $8 for adults and $5 for students. A group buys 14 tickets for $94. How many are adult tickets?
Answer 8 adult tickets.
ORIGINAL PRACTICE · 1
x=5, y=4.
Adding eliminates y: 3x=15, so x=5. Then y=9−5=4. Checking both equations catches sign errors in elimination.
06 / UNDERSTAND & APPLY
A quadratic equation must be rearranged to zero before applying the zero-product property. For x²+bx+c, search for two numbers with sum b and product c. Their factorization yields candidate roots that must respect the problem’s domain.
Dividing by x can lose the solution x=0. For example, x²=5x becomes x(x−5)=0 and has roots 0 and 5. If a word problem requires a positive length or count, reject inadmissible roots only after identifying the algebraic solutions.
ORIGINAL WORKED EXAMPLE
Solve x²−7x+12=0.
Answer x=3 or x=4.
ORIGINAL PRACTICE · 1
x=0 or x=6.
Move all terms to one side and factor x(x−6)=0. Dividing both sides by x immediately would exclude x=0 without justification.
07 / UNDERSTAND & APPLY
An identity holds for every allowed input. Difference of squares is a²−b²=(a−b)(a+b); a square of a sum is a²+2ab+b². These patterns can turn an awkward numerical calculation into a few small operations.
An equation involving an identity may have no unique variable value. If x+y and xy are known, identities can determine x²+y² without solving separately for x and y. Match the requested expression to a known combination.
ORIGINAL WORKED EXAMPLE
If x+y=11 and xy=24, find x²+y².
Answer 73.
ORIGINAL PRACTICE · 1
1,200.
Use (103−97)(103+97)=6×200=1200. Subtracting bases first and then squaring gives 36, which is not the difference-of-squares identity.
08 / UNDERSTAND & APPLY
A rational expression excludes zeros of its original denominator. Multiplying an equation by a potentially zero expression can introduce candidates that do not satisfy the original equation. Squaring can also erase a sign restriction. Record exclusions before transforming.
At the end, check candidates in the original problem and interpret units and count restrictions. A correct algebraic candidate can still fail a requirement such as an integer number of boxes or a nonnegative elapsed time.
ORIGINAL WORKED EXAMPLE
Solve (x²−4)/(x−2)=5.
Answer x=3.
ORIGINAL PRACTICE · 1
No solution.
The original restriction is x≠3. Cancelling yields x+3=6 and candidate x=3, but that candidate makes the original denominator zero. A simplified expression cannot restore an excluded input.
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