If the system below has no solution, what is the value of \(k\)?
\[ \begin{cases} 2x + 4y = 8 \\ kx + 6y = 3 \end{cases} \]
No solution = parallel lines. Parallel means the ratio of x-coefficients equals the ratio of y-coefficients, but NOT the ratio of constants.
Equation 1: \(2x+4y=8\) → coefficients \((2,4,8)\)
Equation 2: \(kx+6y=3\) → coefficients \((k,6,3)\)
Set \(\dfrac{k}{2} = \dfrac{6}{4}\) → \(k = 3\).
Check constant ratio: \(\dfrac{3}{8} \ne \dfrac{6}{4}\). ✓ Not equal, so no solution confirmed.
How many solutions does \(|2x - 3| = x + 1\) have?
Case 1: \(2x-3 = x+1 \Rightarrow x=4\). Check: \(|5|=5=4+1\) ✓
Case 2: \(2x-3 = -(x+1) \Rightarrow 2x-3=-x-1 \Rightarrow 3x=2 \Rightarrow x=\tfrac{2}{3}\). Check: \(|{-5/3}|=\tfrac{5}{3}=\tfrac{2}{3}+1\) ✓
Both valid → 2 solutions. Many students forget Case 2!
Which value of \(x\) satisfies \(-3x + 5 > 14\)?
\(-3x+5 > 14 \Rightarrow -3x > 9 \Rightarrow x < -3\) (sign flips when dividing by −3).
Only \(x=-4\) satisfies \(x < -3\).
Trap: students often forget to flip the inequality sign → pick \(x=0\) or \(x=2\) incorrectly.
For what value of \(c\) does \(x^2 - 6x + c = 0\) have exactly one real solution?
One solution → discriminant = 0: \(b^2 - 4ac = 0\).
Here \(a=1, b=-6\): \((-6)^2 - 4(1)(c) = 0 \Rightarrow 36 = 4c \Rightarrow c=9\).
Check: \(x^2-6x+9=(x-3)^2=0 \Rightarrow x=3\) (one solution ✓).
If \(f(x) = x^2 + 1\), what is \(f(x+2)\)?
Replace \(x\) with \((x+2)\):
\(f(x+2) = (x+2)^2 + 1 = x^2+4x+4+1 = x^2+4x+5\).
Common mistake: Writing \(f(x)+f(2) = x^2+1+5 = x^2+6\). Wrong! You must substitute the entire \((x+2)\).
Which expression is equal to \(x^{-\frac{2}{3}}\) for all positive \(x\)?
\(x^{-2/3} = \dfrac{1}{x^{2/3}} = \dfrac{1}{\sqrt[3]{x^2}}\).
The negative exponent makes it a reciprocal; the fractional exponent \(\tfrac{2}{3}\) means "cube root of x squared".
When \(p(x) = x^3 - 2x^2 + 4x - 1\) is divided by \((x - 2)\), the remainder is:
By the Remainder Theorem, plug in \(x=2\):
\(p(2) = 8 - 8 + 8 - 1 = 7\).
No long division needed! The answer is 7.
What is the center of the circle \((x+3)^2 + (y-5)^2 = 16\)?
Standard form: \((x-h)^2+(y-k)^2=r^2\), center \(=(h,k)\).
\((x+3)^2 = (x-(-3))^2\) → \(h=-3\)
\((y-5)^2\) → \(k=5\)
Center \(=(-3,5)\). Trap: students read \((x+3)\) and write \(h=3\). The sign flips!
In a right triangle, \(\sin(\theta) = \dfrac{3}{5}\). What is \(\cos(\theta)\)?
Use \(\sin^2\theta + \cos^2\theta = 1\):
\(\left(\tfrac{3}{5}\right)^2 + \cos^2\theta = 1 \Rightarrow \tfrac{9}{25} + \cos^2\theta = 1 \Rightarrow \cos^2\theta = \tfrac{16}{25} \Rightarrow \cos\theta = \tfrac{4}{5}\).
Or use 3-4-5 right triangle: if opp=3, hyp=5, then adj=4.
Two similar triangles have corresponding sides in ratio \(2:3\). If the area of the smaller triangle is \(12\), what is the area of the larger?
Areas scale by the square of the side ratio: \(\left(\tfrac{3}{2}\right)^2 = \tfrac{9}{4}\).
Larger area \(= 12 \times \tfrac{9}{4} = 27\).
Trap: multiplying by \(\tfrac{3}{2}\) gives 18 — wrong! Area needs the ratio squared.
Data set: \(\{2, 4, 4, 5, 6, 100\}\). Which is true?
Mean \(= (2+4+4+5+6+100)/6 = 121/6 \approx 20.2\)
Median: middle two values are 4 and 5 → median \(= 4.5\)
Mean (20.2) \(>\) Median (4.5). The outlier 100 drags the mean up while barely moving the median.
A price increases from \(\$80\) to \(\$100\). What is the percent increase?
\(\dfrac{100-80}{80} \times 100 = \dfrac{20}{80} \times 100 = 25\%\)
Trap: dividing by 100 (the new value) gives 20%. Always divide by the original.
A study shows a strong positive correlation between shoe size and reading level in children ages 5–12. The best explanation is:
Both shoe size and reading level increase with age. Age is a lurking (confounding) variable causing both to rise simultaneously. Neither causes the other.
This is a classic SAT reasoning question: correlation doesn't imply causation.
Pipe A fills a tank in 4 hours; Pipe B fills it in 6 hours. Together, how many hours to fill the tank?
Rate A \(= \tfrac{1}{4}\), Rate B \(= \tfrac{1}{6}\).
Combined: \(\tfrac{1}{4}+\tfrac{1}{6} = \tfrac{3}{12}+\tfrac{2}{12}=\tfrac{5}{12}\) tanks/hour.
Time \(= \tfrac{12}{5} = 2.4\) hours \(= 2\tfrac{2}{5}\) hours.
Trap: averaging the times \((4+6)/2=5\) is always wrong for rate problems.
\(y\) varies inversely with \(x\). When \(x=4\), \(y=9\). What is \(y\) when \(x=12\)?
Inverse variation: \(xy = k\). Find k: \(4 \times 9 = 36\).
When \(x=12\): \(12 \cdot y = 36 \Rightarrow y=3\).
Trap: direct variation \(y=kx\) would give \(y=27\). Read "inversely" carefully!
A bacteria population doubles every 3 hours. Starting at 500, what is the population after 9 hours?
In 9 hours, there are \(9 \div 3 = 3\) doubling periods.
Population \(= 500 \times 2^3 = 500 \times 8 = 4000\).
Trap: multiplying by 3 instead of using \(2^3\).
A parabola has equation \(y = 2(x-1)^2 - 3\). What is the minimum value of \(y\)?
Vertex form: \(y=a(x-h)^2+k\). Vertex is at \((h,k)=(1,-3)\).
Since \(a=2>0\), parabola opens upward → minimum at vertex.
Minimum value of \(y = -3\).
Trap: students confuse vertex as \((-1,-3)\) or pick \(k=1\).
How many valid solutions does \(\dfrac{x}{x-2} = \dfrac{2}{x-2}\) have?
Multiply both sides by \((x-2)\): \(x=2\).
But check: when \(x=2\), denominator \(= 2-2=0\). Undefined!
\(x=2\) is an extraneous solution. There are no valid solutions.
Always check solutions in rational equations!
In a class of 30 students, 18 play soccer, 14 play basketball, and 6 play both. How many play neither?
Using inclusion-exclusion: \(|S \cup B| = 18+14-6 = 26\) play at least one sport.
Neither: \(30-26=4\).
Trap: forgetting to subtract the 6 who play both, giving \(18+14=32 > 30\). That's impossible!
How many points of intersection does the system \(y = x^2\) and \(y = 2x - 1\) have?
Set equal: \(x^2 = 2x-1 \Rightarrow x^2-2x+1=0 \Rightarrow (x-1)^2=0\).
Discriminant \(= 4-4=0\) → exactly one solution: \(x=1, y=1\).
The line is tangent to the parabola at \((1,1)\). One intersection point.
Well done!
Review the explanations above to strengthen your weak spots.