Systems of Equations and Polynomials — ACT Math

Introduction

Systems and polynomials unlock the middle and hard range of ACT Math — students who skip these topics hit a score ceiling around 26.

Systems of equations and polynomial questions appear 5-9 times per ACT, primarily in the medium-to-hard range (questions 25-55). Mastering both can add 2-4 raw score points.

By the end of this lesson you will be able to:

Solve 2×2 systems of linear equations using substitution and elimination
Add, subtract, multiply, and factor polynomials
Identify the number of solutions a system has from its equations

You'll factor a cubic polynomial using grouping and use the factored form to identify all three roots.

The Concept

The Core Rule

A system of two linear equations has: one solution (lines intersect), no solution (parallel lines: same slope, different intercept), or infinitely many solutions (same line). For polynomials: to factor ax² + bx + c, find two numbers that multiply to ac and add to b.

How the ACT tests this

Writing a system from a word problem (number/value, coin, or mixture problems)
Testing whether a system has 0, 1, or infinitely many solutions
Factoring quadratics to find zeros or simplify rational expressions

Solving Systems

Substitution: solve one equation for one variable, substitute into the other. Elimination: add or subtract equations to cancel a variable (multiply first if needed). Always solve for both variables and check in BOTH original equations.

Elimination is fastest when coefficients are already equal or opposites
Substitution is fastest when one equation is already solved for a variable
Inconsistent system: elimination gives 0 = nonzero → no solution

Polynomial Operations

FOIL for (a + b)(c + d) = ac + ad + bc + bd. For higher-degree products, distribute each term of the first polynomial to every term of the second. Combining like terms requires the same variable and same exponent.

(a + b)² = a² + 2ab + b² (memorize this — do not FOIL every time)
(a − b)(a + b) = a² − b² (difference of squares)
Degree of a product = sum of degrees of the factors

Factoring

Step 1: Factor out GCF. Step 2: For ax² + bx + c with a=1, find two numbers that multiply to c and add to b. Step 3: For a≠1, use the ac method or trial-and-error. Step 4: Difference of squares: a² − b² = (a−b)(a+b).

Sum/difference of cubes: a³ + b³ = (a+b)(a²−ab+b²)
Factor by grouping for 4-term polynomials: group in pairs
Zero product property: if ab = 0 then a = 0 or b = 0

Your strategy

For systems in word problems, assign clear variable names and write two equations before solving
Choose elimination when you can line up matching variable terms; choose substitution otherwise
Always factor out the GCF before attempting other factoring techniques
Use the discriminant b²−4ac to predict the number of real roots before factoring

Worked Examples

Easy Example 1 Checking Only One Equation Instead Of Both

What is the solution to the system: 2x + y = 7 and x − y = 2?

A. (1, 5)
B. (2, 3)
C. (3, 1) (Correct answer)
D. (4, −1)
E. (5, −3)

Step 1

Add the two equations: (2x + y) + (x − y) = 7 + 2 → 3x = 9 → x = 3

Step 2

Substitute x = 3 into x − y = 2: 3 − y = 2 → y = 1

Step 3

Check in first equation: 2(3) + 1 = 7 ✓

Correct answer: C

Why C is correct

Correct: (3, 1) satisfies both equations

Why other options are wrong

A: x=1: 2(1)+y=7 → y=5; check: 1−5=−4 ≠ 2

B: x=2: 2(2)+y=7 → y=3; check: 2−3=−1 ≠ 2

D: x=4: 2(4)+y=7 → y=−1; check: 4−(−1)=5 ≠ 2

E: x=5: 2(5)+y=7 → y=−3; check: 5−(−3)=8 ≠ 2

⚠ Trap: Checking only one equation instead of both

Medium Example 2 Using The Wrong Sign: Finding Numbers −2 And −3 But Writing (x+2)(x+3)

Which of the following is a factor of x² − 5x + 6?

A. x + 2
B. x − 1
C. x − 2 (Correct answer)
D. x + 3
E. x + 6

Step 1

Find two numbers that multiply to 6 and add to −5: those are −2 and −3

Step 2

Factor: x² − 5x + 6 = (x − 2)(x − 3)

Step 3

x − 2 is a factor; verify: root x=2 gives 4−10+6=0 ✓

Correct answer: C

Why C is correct

Correct: x − 2 → root x=2: 4−10+6=0 ✓

Why other options are wrong

A: x + 2 → root x=−2: (−2)²−5(−2)+6=4+10+6=20≠0

B: x − 1 → root x=1: 1−5+6=2≠0

D: x + 3 → root x=−3: 9+15+6=30≠0

E: x + 6 → root x=−6: 36+30+6=72≠0

⚠ Trap: Using the wrong sign: finding numbers −2 and −3 but writing (x+2)(x+3)

Hard Example 3 Trying To Find X And Y Individually Instead Of Using The (x+y)² Identity

If x + y = 8 and xy = 15, what is the value of x² + y²?

A. 24
B. 30
C. 34 (Correct answer)
D. 49
E. 64

Step 1

Use the identity: (x + y)² = x² + 2xy + y²

Step 2

Substitute: 8² = x² + 2(15) + y²

Step 3

64 = x² + y² + 30 → x² + y² = 34

Correct answer: C

Why C is correct

Correct: 64 − 30 = 34

Why other options are wrong

A: Subtracted 2xy from 64 incorrectly: 64 − 40 = 24

B: Used only 2xy = 30 as the answer

D: Confused (x+y)² with some other squaring: 7² = 49

E: Used (x+y)² = 64 directly without subtracting 2xy

⚠ Trap: Trying to find x and y individually instead of using the (x+y)² identity

Strategy Tips

Memorize the three special products: (a±b)², (a+b)(a−b), and sum/difference of cubes
For systems with messy fractions, multiply each equation through by its LCD first
When factoring fails, use the quadratic formula: x = (−b ± √(b²−4ac)) / 2a
Look for the GCF first — it simplifies everything downstream
Picking numbers works to verify a factored form: try x = 1 or x = 2

Common pitfalls

Forgetting to solve for BOTH variables in a system (ACT often asks for x + y, not just x)

Sign errors when factoring: (x − 2)(x − 3) gives +6 constant, not −6

Not checking whether a system is inconsistent or dependent before assuming one solution

Systems: 60-90 seconds using elimination. Factoring: 30-60 seconds. For hard polynomial identities, spend up to 2 minutes — they are worth it since they appear near the end.

Summary

Elimination adds equations to cancel a variable; substitution replaces a variable
Factor by always pulling out the GCF first, then checking for special forms
The identity (x+y)² = x²+2xy+y² lets you find x²+y² without solving individually

Factor 2x² + 7x + 3 completely. Verify by expanding. Then solve the system 3x − 2y = 12 and x + y = 1 by elimination.

Next: Function Notation, Domain, and Range All ACT Math lessons