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ACT · Domain 2

Algebra

12–15% of the ACT math section

Linear, polynomial, radical and exponential relationships, and the systems they make. The ACT’s algebra is broader than the SAT’s and each question is shorter — one move, done fast, then the next question.

Algebra · Topic 1 of 3

Linear, Polynomial & Radical Expressions

Solving an equation is undoing what was done to the variable, in reverse order. A radical is undone by squaring both sides, and squaring can create a solution that was not there — so a radical equation is not finished until the answer has been put back in.

The problem
If 2x+7=5\sqrt{2x + 7} = 5, what is the value of xx?
  1. A99
  2. B33
  3. C1818
  4. D−1-1
Working1 / 3
  1. Square both sides to remove the root2x+7=252x + 7 = 25
Algebra · Topic 2 of 3

Systems & Inequalities

A system of two lines has one solution, none, or infinitely many, and the ACT likes the last two: it gives one equation with an unknown coefficient and asks for the value that makes the lines coincide or run parallel. Coincident lines are the same equation twice, scaled.

The problem
For what value of kk does the system have infinitely many solutions?
3x+ky=66x+4y=12\begin{aligned} 3x + ky &= 6 \\ 6x + 4y &= 12 \end{aligned}
  1. F22
  2. G44
  3. H88
  4. J−2-2
Working1 / 3
  1. Infinitely many means the second equation is a multiple of the first6x+4y=12   is   2×(3x+ky=6)6x + 4y = 12 \;\text{ is }\; 2 \times (3x + ky = 6)
Algebra · Topic 3 of 3

Quadratics & Factoring

Factor when the numbers allow it, complete the square when a vertex is wanted, and use the formula when nothing factors. The choices tell you which: radicals in the choices mean the formula; whole numbers and simple fractions mean it factors. A leading coefficient other than 1 means splitting the middle term.

The problem
What are the solutions to 2x2−5x−3=02x^2 - 5x - 3 = 0?
  1. A33 and −12-\tfrac{1}{2}
  2. B−3-3 and 12\tfrac{1}{2}
  3. C33 and 12\tfrac{1}{2}
  4. D−3-3 and −12-\tfrac{1}{2}
Working1 / 4
  1. Two numbers with product ac=−6ac = -6 and sum b=−5b = -5−6 and 1-6 \text{ and } 1