Algebra II

Everything here builds on Algebra I. The new ideas are factorising a quadratic, solving two equations at once, and the one rule about inequalities that changes the answer if you forget it.

Simplifying algebraic fractions

An algebraic fraction simplifies when the top and bottom share a factor. Factorise both, then cancel what is common.

$$\frac{12x^3}{4x} = \frac{4x \times 3x^2}{4x} = 3x^2$$

With quadratics, factorise first — you cannot see the common factor until you do.

$$\frac{x^2 - 9}{x^2 + 7x + 12} = \frac{(x-3)(x+3)}{(x+3)(x+4)} = \frac{x-3}{x+4}$$

Multiplying and dividing follow the ordinary fraction rules. To divide, multiply by the reciprocal:

$$\frac{x}{4} \div \frac{x^2}{8} = \frac{x}{4} \times \frac{8}{x^2} = \frac{8x}{4x^2} = \frac{2}{x}$$

Adding and subtracting fractions

You cannot add fractions until the denominators match. Find a common denominator, rewrite both, then add the numerators.

$$\frac{x}{3} + \frac{x}{4} = \frac{4x}{12} + \frac{3x}{12} = \frac{7x}{12}$$

With algebraic denominators the common denominator is usually their product:

$$\frac{2}{x} + \frac{3}{x+1} = \frac{2(x+1)}{x(x+1)} + \frac{3x}{x(x+1)} = \frac{2x + 2 + 3x}{x(x+1)} = \frac{5x+2}{x(x+1)}$$

Factorising quadratics

To factorise $x^2 + bx + c$, find two numbers that multiply to $c$ and add to $b$.

Factorise $x^2 + 8x + 15$.

Two numbers multiplying to 15: $1 \times 15$ or $3 \times 5$. Which pair adds to 8? $3 + 5 = 8$.

$$x^2 + 8x + 15 = (x+3)(x+5)$$

Signs follow from the two numbers. In $x^2 + 5x - 14$ the product is negative, so the numbers have opposite signs: $7$ and $-2$ multiply to $-14$ and add to $5$. So $x^2 + 5x - 14 = (x+7)(x-2)$.

Difference of two squares

$$a^2 - b^2 = (a-b)(a+b)$$

So $x^2 - 9 = (x-3)(x+3)$ and $4x^2 - 25 = (2x-5)(2x+5)$. Look for this pattern before anything else — two squares with a minus between them and no middle term.

When the coefficient of $x^2$ is not 1

For $2x^2 + 7x + 3$, find two numbers multiplying to $2 \times 3 = 6$ and adding to $7$: those are $6$ and $1$. Split the middle term and group:

$$2x^2 + 6x + x + 3 = 2x(x+3) + 1(x+3) = (2x+1)(x+3)$$

Check by expanding: $2x^2 + 6x + x + 3 = 2x^2 + 7x + 3$ ✓

Solving quadratics by factorising

If two things multiply to zero, at least one of them is zero. That single fact is the whole method.

Solve $x^2 + 5x - 14 = 0$.

$$(x+7)(x-2) = 0$$ $$x + 7 = 0 \quad \text{or} \quad x - 2 = 0$$ $$x = -7 \quad \text{or} \quad x = 2$$

Check both: $(-7)^2 + 5(-7) - 14 = 49 - 35 - 14 = 0$ ✓ and $2^2 + 5(2) - 14 = 4 + 10 - 14 = 0$ ✓

Simultaneous linear equations

Two equations, two unknowns. The solution is the single pair $(x, y)$ that satisfies both.

Elimination

Make the coefficients of one letter match, then add or subtract to remove it.

Solve $3x + 2y = 16$ and $x + y = 6$.

Multiply the second equation by 2 so the $y$ terms match:

$$2x + 2y = 12$$

Subtract this from the first:

$$(3x + 2y) - (2x + 2y) = 16 - 12$$ $$x = 4$$

Substitute back into $x + y = 6$: $\;4 + y = 6$, so $y = 2$.

Check in the equation you did not use for the substitution: $3(4) + 2(2) = 16$ ✓

Substitution

When one equation already gives a letter on its own, substitution is quicker. From $y = 6 - x$, put that into $3x + 2y = 16$:

$$3x + 2(6-x) = 16$$ $$3x + 12 - 2x = 16$$ $$x = 4$$

Same answer, as it must be.

Linear inequalities

Inequalities are solved exactly like equations, with one exception.

$$3x - 4 < 11$$ $$3x < 15$$ $$x < 5$$

The solution is every value less than 5, not a single number.

The rule everyone forgets

Multiplying or dividing both sides by a negative number reverses the inequality sign.

$$-2x > 8$$ $$x < -4$$

Check with a value: $x = -5$ satisfies $x < -4$, and $-2(-5) = 10 > 8$ ✓. Had you left the sign alone and written $x > -4$, then $x = 0$ would qualify — but $-2(0) = 0$, which is not greater than 8.

Showing the solution

On a number line, use an open circle for $<$ or $>$ (the endpoint is excluded) and a filled circle for $\le$ or $\ge$ (the endpoint is included), with an arrow in the direction of the solution.

When the answer must be a whole number, say so: the integers satisfying $-3 < x \le 2$ are $-2, -1, 0, 1, 2$ — note that $-3$ is out and $2$ is in.