Multiply the first equation by 2: \(6x-4y=32\). Add this to the second equation \(5x+4y=6\): \(11x = 38\), so \(x=\dfrac{38}{11}\). Substitute back: \(2y = 3x-16 = \dfrac{114}{11}-\dfrac{176}{11}=-\dfrac{62}{11}\), so \(y=-\dfrac{31}{11}\). Then \(x+y = \dfrac{38}{11}-\dfrac{31}{11}=\dfrac{7}{11}\). Common mistake: forgetting to scale the first equation before eliminating \(y\).

By Vieta's formulas, \(r_1+r_2 = \dfrac{7}{2}\) and \(r_1 r_2 = \dfrac{-4}{2} = -2\). Use the identity \(r_1^2+r_2^2 = (r_1+r_2)^2 - 2r_1r_2 = \left(\dfrac{7}{2}\right)^2 - 2(-2) = \dfrac{49}{4}+4 = \dfrac{65}{4}\). Do not just square each root separately — use the identity to avoid the messy radicals.

Rewrite \(27 = 3^3\), so the equation becomes \(3^{2x+1}=3^{3(x-2)}\). Since the bases match, set exponents equal: \(2x+1 = 3x-6\), giving \(x=7\). A common error is forgetting to distribute the 3 across \((x-2)\).

Isolate and square: \(\sqrt{2x+5} = 2+\sqrt{x-1}\). Squaring gives \(2x+5 = 4+4\sqrt{x-1}+(x-1)\), so \(x+2 = 4\sqrt{x-1}\). Squaring again: \(x^2-12x+4=0\)... solving carefully gives \(x=2\) and \(x=10\), and both check in the original equation (no extraneous roots here), so the sum is \(2+10=12\). Always verify each candidate in the original equation.

Multiply numerator and denominator by \(xy\): \(\dfrac{y+x}{y-x}\). With \(x=4,y=9\): \(\dfrac{13}{5}\). Trying to find common denominators separately for top and bottom before combining is the step where students most often make a sign error.

Complete the square: \((x^2-6x+9)+(y^2+10y+25) = -18+9+25\), so \((x-3)^2+(y+5)^2 = 16\). Radius \(=\sqrt{16}=4\). A frequent slip is mismanaging the constant term when moving it to the other side.

By the Remainder Theorem, the remainder equals \(f(3) = 2(3)^2-3(3)-5 = 18-9-5=4\). Long division confirms quotient \(2x+3\) with remainder \(4\).

Use the double angle identity \(\cos2\theta = 1-2\sin^2\theta = 1-2\left(\dfrac{9}{25}\right)=1-\dfrac{18}{25}=\dfrac{7}{25}\). Note the quadrant of \(\theta\) is not actually needed since the identity only uses \(\sin\theta\) — a common trap is wasting time finding \(\cos\theta\) first.

Set \(500(1.08)^t=1000\), so \((1.08)^t=2\). Taking logs: \(t = \dfrac{\ln 2}{\ln 1.08} \approx 9.0\) years. Students often forget to divide by \(\ln(1.08)\) and instead divide by \(1.08\) directly.

The inequality splits into \(-9 \le 2x-5 \le 9\), so \(-4 \le 2x \le 14\), giving \(-2 \le x \le 7\). The sum of endpoints is \(-2+7=5\).

Let \(v\) be the mL of 50% solution and \(300-v\) the mL of 20% solution. Then \(0.5v + 0.2(300-v) = 0.35(300)\). Solving: \(0.5v+60-0.2v=105\), so \(0.3v=45\), \(v=150\) mL.

First find \(g(2)=2(2)+1=5\). Then \(f(5)=5^2-3(5)+2=25-15+2=12\). A common error is plugging \(x=2\) directly into \(f\) instead of computing \(g(2)\) first.

Area ratio equals the square of the side ratio: \(\left(\dfrac{3}{5}\right)^2=\dfrac{9}{25}\). Area of B \(= 75 \times \dfrac{9}{25}=27\). Forgetting to square the ratio is the most common mistake with similar-figure area problems.

\(a_{20}=a_1+(n-1)d = 7+(19)(4)=7+76=83\). A frequent slip is using \(20d\) instead of \((20-1)d\).

The feasible region's vertices are \((0,0)\), \((0,7)\), and the intersection of \(x+2y=14\) with \(3x-y=0\), which is \((2,6)\). Evaluating the objective: \((0,0)\to0\), \((0,7)\to35\), \((2,6)\to 6+30=36\). The maximum is \(36\) at \((2,6)\).

Since \(7 \equiv 2 \pmod5\), we need \(2^{100} \bmod 5\). Powers of 2 mod 5 cycle as \(2,4,3,1\) with period 4. Since \(100\) is a multiple of 4, \(2^{100}\equiv 1 \pmod5\).

In 2 hours, Pipe B completes \(\dfrac{2}{10}=\dfrac{1}{5}\) of the tank. The remaining \(\dfrac{4}{5}\) is filled by Pipe A alone at rate \(\dfrac{1}{6}\) per hour, taking \(\dfrac{4}{5}\times 6 = 4.8\) hours. Total time \(=2+4.8=6.8\) hours. This differs from simply adding both rates together the whole time — a common shortcut error.

Multiply numerator and denominator by the conjugate \(1+2i\): numerator \((3+4i)(1+2i)=3+6i+4i+8i^2=3+10i-8=-5+10i\); denominator \((1-2i)(1+2i)=1+4=5\). Result: \(\dfrac{-5+10i}{5}=-1+2i\).

The maximum occurs at \(t=-\dfrac{64}{2(-16)}=2\). Then \(h(2)=-16(4)+64(2)+5=-64+128+5=69\) feet.

Successive changes multiply: \(1.20 \times 0.75 = 0.90\), representing 90% of the original — a net decrease of 10%. Simply adding \(+20\%\) and \(-25\%\) to get \(-5\%\) is the classic (incorrect) shortcut.

By the Law of Cosines: \(c^2 = 7^2+9^2-2(7)(9)\cos60^\circ = 49+81-126(0.5)=130-63=67\). Forgetting that \(\cos60^\circ=0.5\) exactly (not an approximation) leads to rounding errors.

Adding equations 1 and 2 eliminates nothing directly useful, so instead subtract eq.1 from eq.2: \(x-2y=-3\). Add eq.1 and eq.3: \(2x+3y=11\). Solving these two: from the first, \(x=2y-3\); substitute: \(2(2y-3)+3y=11 \Rightarrow 7y=17 \Rightarrow y=\dfrac{17}{7}\), then \(x=2\left(\dfrac{17}{7}\right)-3=\dfrac{34}{7}-\dfrac{21}{7}=\dfrac{13}{7}\).