Algebra 1 Questions 1–10
1
Linear Equations
Solving Multi-Step Equations
📘 Key Concept
When solving an equation, apply the distributive property first, then combine like terms, then isolate the variable using inverse operations.
Example: Solve \(2(x+3)=14\).
Step 1: \(2x + 6 = 14\) | Step 2: \(2x = 8\) | Step 3: \(x = 4\)
Step 1: \(2x + 6 = 14\) | Step 2: \(2x = 8\) | Step 3: \(x = 4\)
Solve for \(x\):
\[ 3(2x - 4) = 2(x + 6) \]
\[ 3(2x - 4) = 2(x + 6) \]
📝 Solution
Distribute: \(6x - 12 = 2x + 12\)
Subtract \(2x\): \(4x - 12 = 12\)
Add 12: \(4x = 24\)
Divide by 4: \(\boxed{x = 6}\) ✅
Common Error: Forgetting to distribute to both terms inside the parentheses.
Subtract \(2x\): \(4x - 12 = 12\)
Add 12: \(4x = 24\)
Divide by 4: \(\boxed{x = 6}\) ✅
Common Error: Forgetting to distribute to both terms inside the parentheses.
2
Systems of Equations
Substitution Method
📘 Key Concept
In the substitution method, isolate one variable in one equation and substitute that expression into the other equation.
Example: \(y = 2x\) and \(x + y = 9\) → \(x + 2x = 9\) → \(x = 3,\ y = 6\)
Solve the system:
\[ y = 3x - 1 \] \[ 2x + y = 9 \] What is the value of \(x + y\)?
\[ y = 3x - 1 \] \[ 2x + y = 9 \] What is the value of \(x + y\)?
📝 Solution
Substitute \(y = 3x-1\) into \(2x + y = 9\):
\(2x + (3x-1) = 9 \Rightarrow 5x - 1 = 9 \Rightarrow 5x = 10 \Rightarrow x = 2\)
Then \(y = 3(2)-1 = 5\). So \(x + y = 2 + 5 = \boxed{7}\) ✅
Common Error: The question asks for \(x+y\), not just \(y\). Always re-read the final question!
\(2x + (3x-1) = 9 \Rightarrow 5x - 1 = 9 \Rightarrow 5x = 10 \Rightarrow x = 2\)
Then \(y = 3(2)-1 = 5\). So \(x + y = 2 + 5 = \boxed{7}\) ✅
Common Error: The question asks for \(x+y\), not just \(y\). Always re-read the final question!
3
Inequalities
Compound Inequalities
📘 Key Concept
When you multiply or divide both sides of an inequality by a negative number, you must flip the inequality sign.
Example: \(-2x > 6 \Rightarrow x < -3\) (sign flips!)
Which value of \(x\) satisfies \(-3x + 7 > 16\)?
📝 Solution
\(-3x + 7 > 16\)
Subtract 7: \(-3x > 9\)
Divide by \(-3\) and flip the sign: \(x < -3\)
Check: Only \(x = -4\) satisfies \(x < -3\). \(\boxed{x = -4}\) ✅
Common Error: Forgetting to flip the inequality sign when dividing by a negative.
Subtract 7: \(-3x > 9\)
Divide by \(-3\) and flip the sign: \(x < -3\)
Check: Only \(x = -4\) satisfies \(x < -3\). \(\boxed{x = -4}\) ✅
Common Error: Forgetting to flip the inequality sign when dividing by a negative.
4
Quadratic Equations
Factoring
📘 Key Concept
To factor \(x^2 + bx + c\), find two numbers that multiply to \(c\) and add to \(b\).
Example: \(x^2 + 5x + 6 = (x+2)(x+3)\) since \(2 \times 3 = 6\) and \(2 + 3 = 5\)
What are the solutions to \(x^2 - 5x + 6 = 0\)?
📝 Solution
Find numbers that multiply to 6 and add to −5: those are \(-2\) and \(-3\).
\((x-2)(x-3) = 0\)
\(\boxed{x = 2 \text{ or } x = 3}\) ✅
Common Error: Students often confuse the signs, writing \(-2\) and \(-3\) instead.
\((x-2)(x-3) = 0\)
\(\boxed{x = 2 \text{ or } x = 3}\) ✅
Common Error: Students often confuse the signs, writing \(-2\) and \(-3\) instead.
5
Slope & Linear Equations
Slope-Intercept Form
📘 Key Concept
Slope formula: \(\displaystyle m = \frac{y_2 - y_1}{x_2 - x_1}\). A line parallel to \(y = mx + b\) has the same slope \(m\).
Example: Line through \((1,3)\) and \((3,7)\): \(m = \frac{7-3}{3-1} = 2\)
A line passes through the points \((2, 5)\) and \((6, 13)\). Which equation represents a line parallel to this line that passes through \((0, 3)\)?
📝 Solution
Slope of original line: \(m = \dfrac{13-5}{6-2} = \dfrac{8}{4} = 2\)
Parallel line has slope 2, y-intercept 3 (passes through origin at (0,3)):
\(\boxed{y = 2x + 3}\) ✅
Parallel line has slope 2, y-intercept 3 (passes through origin at (0,3)):
\(\boxed{y = 2x + 3}\) ✅
6
Functions
Domain, Range & Function Notation
📘 Key Concept
\(f(a)\) means "substitute \(a\) for \(x\) in the function." The domain is all allowed inputs; the range is all possible outputs.
Example: If \(f(x) = x^2 - 1\), then \(f(3) = 9 - 1 = 8\)
Given \(f(x) = 2x^2 - 3x + 1\), what is \(f(-2)\)?
📝 Solution
\(f(-2) = 2(-2)^2 - 3(-2) + 1\)
\(= 2(4) + 6 + 1\)
\(= 8 + 6 + 1 = \boxed{15}\) ✅
Common Error: Calculating \(2 \times (-2)^2\) as \(-8\) instead of \(+8\). Remember \((-2)^2 = +4\).
\(= 2(4) + 6 + 1\)
\(= 8 + 6 + 1 = \boxed{15}\) ✅
Common Error: Calculating \(2 \times (-2)^2\) as \(-8\) instead of \(+8\). Remember \((-2)^2 = +4\).
7
Exponents & Polynomials
Rules of Exponents
📘 Key Concept
Key exponent rules: \(x^a \cdot x^b = x^{a+b}\), \(\dfrac{x^a}{x^b} = x^{a-b}\), \((x^a)^b = x^{ab}\), \(x^0 = 1\), \(x^{-n} = \dfrac{1}{x^n}\)
Example: \((2^3)^2 = 2^6 = 64\)
Simplify: \(\dfrac{(3x^2y)^3}{9x^4y^2}\)
📝 Solution
Numerator: \((3x^2y)^3 = 27x^6y^3\)
\(\dfrac{27x^6y^3}{9x^4y^2} = 3 \cdot x^{6-4} \cdot y^{3-2} = \boxed{3x^2y}\) ✅
\(\dfrac{27x^6y^3}{9x^4y^2} = 3 \cdot x^{6-4} \cdot y^{3-2} = \boxed{3x^2y}\) ✅
8
Quadratic Formula
Discriminant & Nature of Roots
📘 Key Concept
\(x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). The discriminant \(\Delta = b^2 - 4ac\): if \(\Delta > 0\): 2 real roots; \(\Delta = 0\): 1 real root; \(\Delta < 0\): no real roots.
Use the quadratic formula to solve \(2x^2 + 3x - 2 = 0\). Which answer is correct?
📝 Solution
\(a=2,\ b=3,\ c=-2\)
\(\Delta = 9 + 16 = 25,\ \sqrt{\Delta} = 5\)
\(x = \dfrac{-3 \pm 5}{4}\)
\(x_1 = \dfrac{2}{4} = \dfrac{1}{2},\quad x_2 = \dfrac{-8}{4} = -2\)
\(\boxed{x = \tfrac{1}{2} \text{ or } x = -2}\) ✅
\(\Delta = 9 + 16 = 25,\ \sqrt{\Delta} = 5\)
\(x = \dfrac{-3 \pm 5}{4}\)
\(x_1 = \dfrac{2}{4} = \dfrac{1}{2},\quad x_2 = \dfrac{-8}{4} = -2\)
\(\boxed{x = \tfrac{1}{2} \text{ or } x = -2}\) ✅
9
Absolute Value
Absolute Value Equations
📘 Key Concept
\(|ax + b| = c\) (where \(c \geq 0\)) splits into two equations: \(ax+b = c\) OR \(ax+b = -c\). Always check both solutions.
How many solutions does the equation \(|2x - 3| = 7\) have, and what are they?
📝 Solution
Case 1: \(2x - 3 = 7 \Rightarrow 2x = 10 \Rightarrow x = 5\)
Case 2: \(2x - 3 = -7 \Rightarrow 2x = -4 \Rightarrow x = -2\)
\(\boxed{x = 5 \text{ and } x = -2}\) ✅
Case 2: \(2x - 3 = -7 \Rightarrow 2x = -4 \Rightarrow x = -2\)
\(\boxed{x = 5 \text{ and } x = -2}\) ✅
10
Rational Expressions
Simplifying Rational Expressions
📘 Key Concept
Factor both numerator and denominator completely, then cancel common factors (not terms!). State any values that make the denominator zero.
Example: \(\dfrac{x^2-4}{x-2} = \dfrac{(x+2)(x-2)}{x-2} = x+2,\quad x \neq 2\)
Simplify: \(\dfrac{x^2 + 2x - 15}{x^2 - 9}\)
📝 Solution
Numerator: \(x^2+2x-15 = (x+5)(x-3)\)
Denominator: \(x^2-9 = (x+3)(x-3)\)
Cancel \((x-3)\): \(\dfrac{(x+5)\cancel{(x-3)}}{(x+3)\cancel{(x-3)}} = \boxed{\dfrac{x+5}{x+3}},\quad x \neq 3,\ x \neq -3\) ✅
Common Error: Canceling terms instead of factors, or forgetting to state the restrictions on \(x\).
Denominator: \(x^2-9 = (x+3)(x-3)\)
Cancel \((x-3)\): \(\dfrac{(x+5)\cancel{(x-3)}}{(x+3)\cancel{(x-3)}} = \boxed{\dfrac{x+5}{x+3}},\quad x \neq 3,\ x \neq -3\) ✅
Common Error: Canceling terms instead of factors, or forgetting to state the restrictions on \(x\).
Geometry Questions 11–20
11
Triangles
Pythagorean Theorem
📘 Key Concept
In a right triangle: \(a^2 + b^2 = c^2\) where \(c\) is the hypotenuse (side opposite the right angle).
Example: Legs 3 and 4 → \(c = \sqrt{9+16} = 5\)
In a right triangle, one leg measures 8 and the hypotenuse measures 17. What is the length of the other leg?
📝 Solution
\(a^2 + 8^2 = 17^2\)
\(a^2 + 64 = 289\)
\(a^2 = 225\)
\(a = \boxed{15}\) ✅ (This is the famous 8-15-17 Pythagorean triple)
\(a^2 + 64 = 289\)
\(a^2 = 225\)
\(a = \boxed{15}\) ✅ (This is the famous 8-15-17 Pythagorean triple)
12
Circles
Arc Length & Sector Area
📘 Key Concept
Arc length: \(L = \dfrac{\theta}{360°} \cdot 2\pi r\) | Sector area: \(A = \dfrac{\theta}{360°} \cdot \pi r^2\)
Example: 90° sector of circle with \(r=4\): Area \(= \frac{90}{360}\pi(16) = 4\pi\)
A circle has radius 6. What is the area of a sector with a central angle of 120°? (Leave your answer in terms of \(\pi\).)
📝 Solution
\(A = \dfrac{120}{360} \cdot \pi(6)^2 = \dfrac{1}{3} \cdot 36\pi = \boxed{12\pi}\) ✅
Common Error: Using diameter instead of radius, or forgetting to reduce the fraction first.
Common Error: Using diameter instead of radius, or forgetting to reduce the fraction first.
13
Angle Relationships
Parallel Lines & Transversals
📘 Key Concept
When a transversal crosses parallel lines: Alternate interior angles are equal; Co-interior (same-side interior) angles are supplementary (sum to 180°); Corresponding angles are equal.
Two parallel lines are cut by a transversal. One co-interior (same-side interior) angle measures \((4x + 20)°\) and the other measures \((2x + 40)°\). Find the value of \(x\).
📝 Solution
Co-interior angles are supplementary:
\((4x+20) + (2x+40) = 180\)
\(6x + 60 = 180\)
\(6x = 120 \Rightarrow \boxed{x = 20}\) ✅
Common Error: Setting angles equal (treating them as alternate interior) instead of supplementary.
\((4x+20) + (2x+40) = 180\)
\(6x + 60 = 180\)
\(6x = 120 \Rightarrow \boxed{x = 20}\) ✅
Common Error: Setting angles equal (treating them as alternate interior) instead of supplementary.
14
Similar Triangles
Proportions in Similar Figures
📘 Key Concept
If two triangles are similar, their corresponding sides are proportional: \(\dfrac{a}{A} = \dfrac{b}{B} = \dfrac{c}{C}\)
Triangle \(ABC\) is similar to triangle \(DEF\). If \(AB = 8\), \(BC = 12\), and \(DE = 6\), what is the length of \(EF\)?
📝 Solution
Scale factor: \(\dfrac{DE}{AB} = \dfrac{6}{8} = \dfrac{3}{4}\)
\(EF = BC \cdot \dfrac{3}{4} = 12 \cdot \dfrac{3}{4} = \boxed{9}\) ✅
\(EF = BC \cdot \dfrac{3}{4} = 12 \cdot \dfrac{3}{4} = \boxed{9}\) ✅
15
Volume
Cylinder & Cone Volumes
📘 Key Concept
Volume of cylinder: \(V = \pi r^2 h\) | Volume of cone: \(V = \dfrac{1}{3}\pi r^2 h\)
A cone has exactly \(\dfrac{1}{3}\) the volume of a cylinder with the same base and height.
A cone has exactly \(\dfrac{1}{3}\) the volume of a cylinder with the same base and height.
A cone and a cylinder have the same radius of 3 and the same height of 8. What is the positive difference between their volumes? (Leave in terms of \(\pi\).)
📝 Solution
Cylinder: \(V_c = \pi(3)^2(8) = 72\pi\)
Cone: \(V_k = \dfrac{1}{3}\pi(3)^2(8) = 24\pi\)
Difference: \(72\pi - 24\pi = \boxed{48\pi}\) ✅
Cone: \(V_k = \dfrac{1}{3}\pi(3)^2(8) = 24\pi\)
Difference: \(72\pi - 24\pi = \boxed{48\pi}\) ✅
16
Coordinate Geometry
Midpoint & Distance Formula
📘 Key Concept
Midpoint: \(M = \left(\dfrac{x_1+x_2}{2},\ \dfrac{y_1+y_2}{2}\right)\) | Distance: \(d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\)
Point \(M(3, 7)\) is the midpoint of segment \(\overline{PQ}\). If \(P = (1, 3)\), what are the coordinates of \(Q\)?
📝 Solution
Midpoint formula: \(\dfrac{1+x_2}{2} = 3 \Rightarrow x_2 = 5\)
\(\dfrac{3+y_2}{2} = 7 \Rightarrow y_2 = 11\)
\(\boxed{Q = (5, 11)}\) ✅
\(\dfrac{3+y_2}{2} = 7 \Rightarrow y_2 = 11\)
\(\boxed{Q = (5, 11)}\) ✅
17
Trigonometry
Right Triangle Trig (SOH-CAH-TOA)
📘 Key Concept
\(\sin\theta = \dfrac{\text{opposite}}{\text{hypotenuse}},\quad \cos\theta = \dfrac{\text{adjacent}}{\text{hypotenuse}},\quad \tan\theta = \dfrac{\text{opposite}}{\text{adjacent}}\)
Remember: SOH-CAH-TOA
In right triangle \(ABC\) with the right angle at \(C\), \(AC = 5\) and \(BC = 12\). What is \(\sin A\)?
📝 Solution
First find the hypotenuse: \(AB = \sqrt{5^2+12^2} = \sqrt{169} = 13\)
\(\sin A = \dfrac{\text{opposite}}{\text{hypotenuse}} = \dfrac{BC}{AB} = \boxed{\dfrac{12}{13}}\) ✅
Common Error: Using \(AC\) (adjacent) instead of \(BC\) (opposite) for \(\sin A\).
\(\sin A = \dfrac{\text{opposite}}{\text{hypotenuse}} = \dfrac{BC}{AB} = \boxed{\dfrac{12}{13}}\) ✅
Common Error: Using \(AC\) (adjacent) instead of \(BC\) (opposite) for \(\sin A\).
18
Transformations
Reflections & Rotations
📘 Key Concept
Reflection over x-axis: \((x, y) \to (x, -y)\)
Reflection over y-axis: \((x, y) \to (-x, y)\)
Rotation 90° CCW about origin: \((x, y) \to (-y, x)\)
Rotation 180°: \((x, y) \to (-x, -y)\)
Reflection over y-axis: \((x, y) \to (-x, y)\)
Rotation 90° CCW about origin: \((x, y) \to (-y, x)\)
Rotation 180°: \((x, y) \to (-x, -y)\)
Point \(P(4, -3)\) is rotated 90° counterclockwise about the origin. What are the new coordinates?
📝 Solution
90° CCW: \((x,y) \to (-y, x)\)
\(P(4,-3) \to (-(-3),\ 4) = \boxed{(3, 4)}\) ✅
Common Error: Using the clockwise rule instead of counterclockwise, giving \((-3,-4)\).
\(P(4,-3) \to (-(-3),\ 4) = \boxed{(3, 4)}\) ✅
Common Error: Using the clockwise rule instead of counterclockwise, giving \((-3,-4)\).
19
Proofs & Properties
Triangle Congruence Theorems
📘 Key Concept
Triangle congruence shortcuts: SSS, SAS, ASA, AAS, HL (right triangles only).
⚠ SSA and AAA are NOT valid congruence theorems!
⚠ SSA and AAA are NOT valid congruence theorems!
Two triangles share a common side. In \(\triangle ABC\) and \(\triangle DBC\): \(AB = DB\), \(\angle ABC = \angle DBC\), and \(BC = BC\). Which congruence theorem applies?
📝 Solution
We have: Side \(AB = DB\), included Angle \(\angle ABC = \angle DBC\), Side \(BC = BC\) (reflexive property).
This is Side-Angle-Side = \(\boxed{SAS}\) ✅
Key: The angle must be between (included by) the two sides to use SAS.
This is Side-Angle-Side = \(\boxed{SAS}\) ✅
Key: The angle must be between (included by) the two sides to use SAS.
20
Surface Area
Composite Figures
📘 Key Concept
Total Surface Area of a rectangular prism: \(SA = 2(lw + lh + wh)\)
For composite figures, add or subtract faces that are joined or hidden.
For composite figures, add or subtract faces that are joined or hidden.
A rectangular prism has length 8, width 5, and height 3. A square hole with side length 2 is cut completely through the prism from top to bottom. What is the total surface area of the resulting solid?
📝 Solution
Original SA: \(2(8\cdot5 + 8\cdot3 + 5\cdot3) = 2(40+24+15) = 2(79) = 158\)
The square hole removes two \(2\times2 = 4\) sq unit faces (top and bottom): \(-8\)
The hole adds 4 interior tunnel walls, each \(2 \times 3 = 6\) sq units, total: \(+24\)
Total SA: \(158 - 8 + 24 = \boxed{174}\) ✅
Common Error: Forgetting to add the 4 interior walls of the tunnel, or forgetting to subtract the 2 removed faces on top and bottom.
The square hole removes two \(2\times2 = 4\) sq unit faces (top and bottom): \(-8\)
The hole adds 4 interior tunnel walls, each \(2 \times 3 = 6\) sq units, total: \(+24\)
Total SA: \(158 - 8 + 24 = \boxed{174}\) ✅
Common Error: Forgetting to add the 4 interior walls of the tunnel, or forgetting to subtract the 2 removed faces on top and bottom.
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