Algebra

Two lines, three outcomes

For $a_1x+b_1y=c_1$ and $a_2x+b_2y=c_2$ compare the ratios: $\frac{a_1}{a_2} \ne \frac{b_1}{b_2} \Rightarrow \text{unique solution} \qquad \frac{a_1}{a_2} = \frac{b_1}{b_2} \ne \frac{c_1}{c_2} \Rightarrow \text{no solution (parallel)} \qquad \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \Rightarrow \text{infinitely many (same line)}$ Find-k questions fix one ratio from the known pair and set it equal to (or apart from) the third: $3x+4y=10$, $6x+ky=20$ → infinitely many needs $\frac36 = \frac4k = \frac{10}{20}$ → $k=8$. No solution needs $\frac36 = \frac{k}{...}$ matched on the first two ratios but NOT on $c$: e.g. $2x+ky=5$, $4x+6y=10$ → $\frac24 = \frac k6 = \frac12$ gives $k=3$ and then $\frac55$-checking $c$: $\frac12 \ne \frac{10}{10}=1$ → indeed no solution at $k=3$ only if the $c$-ratio differs. Always finish the check with the $c$-ratio.

Area with the axes

A line $ax+by=c$ cuts the axes at $\left(\frac ca, 0\right)$ and $\left(0, \frac cb\right)$: $\text{area} = \frac12 \times \frac ca \times \frac cb = \frac{c^2}{2ab}$ $3x+4y=24$ → intercepts 8 and 6 → area 24. With axes and one more line, find the intersection first, then use the base-height read-off. This little formula answers a surprisingly common CGL question in five seconds.

Remainder and factor theorems

Remainder theorem: dividing $p(x)$ by $(x-\alpha)$ leaves $p(\alpha)$ — just substitute. Dividing by $(x+1)$? Substitute $x=-1$. Two remainders give two linear equations in the unknown coefficients: $p(x)=x^4-2x^3+3x^2-ax+b$ with remainders 3 and 9 at $x=1$ and $x=-1$ → $p(1)=2-a+b=3$, $p(-1)=6+a+b=9$ → $a=1$, $b=2$. Factor theorem: $(x-\alpha)$ is a factor $\iff p(\alpha)=0$. "Is a factor" questions are remainder questions with the remainder forced to zero.

Quadratic roots

For $x^2-px+q=0$ (i.e. $ax^2+bx+c$): $\alpha+\beta = -\frac ba$, $\alpha\beta = \frac ca$. The standard follow-ups: $\alpha^2+\beta^2 = (\alpha+\beta)^2-2\alpha\beta \qquad \alpha^3+\beta^3 = (\alpha+\beta)^3-3\alpha\beta(\alpha+\beta) \qquad \frac1\alpha+\frac1\beta = \frac{\alpha+\beta}{\alpha\beta}$ Roots real & distinct $\iff b^2-4ac>0$; equal $\iff =0$; no real roots $\iff <0$. A "for what $k$" question is just that inequality solved.

Common traps

• Quoting the parallel-lines condition without checking the $c$-ratio (many "find k" questions hinge on it).
• Using $\frac cb$ as the $x$-intercept (it is the $y$-intercept).
• Substituting the wrong sign into the remainder theorem ($x+3$ → use $-3$).
• Sign slips in $\alpha+\beta = -b/a$ — it is MINUS $b$ over $a$.

Quick revision

• Ratios decide: unequal → unique; equal-equal-unequal → none; all equal → infinite.
• Area with axes $= \frac{c^2}{2ab}$.
• Remainder $= p(\alpha)$; factor $\iff p(\alpha)=0$.
• $\alpha+\beta = -\frac ba$, $\alpha\beta = \frac ca$; $\alpha^3+\beta^3 = (\alpha+\beta)^3-3\alpha\beta(\alpha+\beta)$.
• Real distinct roots $\iff b^2 > 4ac$.