Core Ideas in Plain Terms

Study Sheet

Core Ideas in Plain Terms

The core of algebra, in everyday language

Lines and Slope

Concept
Slope is steepness: rise over run
xy(1, 2)(2, 4)

The slope of a line is how much it goes up for each step right — "rise over run." In y=mx+by = mx + b, the mm is that slope and bb is where the line crosses the yy-axis. A bigger mm is steeper; a negative mm goes downhill.

Slope is the same between any two points of a line, so it can be measured anywhere: that constant rate of change is what makes the graph straight.

Example
Slope through two points

From (1,2)(1,2) to (4,8)(4,8) the line rises 66 over a run of 33, so the slope is 63=2\tfrac{6}{3} = 2.

Try it
Try it: slope

Find the slope of the line through (−1,7)(-1, 7) and (3,−1)(3, -1), and say whether it rises or falls.

Show the answer

m=−1−73−(−1)=−84=−2m = \tfrac{-1 - 7}{3 - (-1)} = \tfrac{-8}{4} = -2. Negative, so the line falls 22 for each step right.

Factoring and Quadratics

Concept
Factoring is un-multiplying

To factor means to rewrite an expression as a product. For x2+5x+6x^2 + 5x + 6, look for two numbers that multiply to 66 and add to 55 — that is 22 and 33 — so it factors as (x+2)(x+3)(x+2)(x+3). Factoring reveals the roots through the zero-product property: if ab=0ab = 0 then a=0a = 0 or b=0b = 0. And factoring itself is the distributive property run in reverse — expanding back gives (x+2)(x+3)=x2+3x+2x+6(x+2)(x+3) = x^2 + 3x + 2x + 6.

Factoring turns a sum into a product, and a product is zero only when a factor is zero; that single fact is why factoring solves equations.

Example
Solve x² + 5x + 6 = 0

It factors as (x+2)(x+3)=0(x+2)(x+3) = 0, so by the zero-product property x=−2x = -2 or x=−3x = -3.

Try it
Try it: factor and solve

Solve x2−x−20=0x^2 - x - 20 = 0.

Show the answer

Two numbers with product −20-20 and sum −1-1: −5-5 and 44. So (x−5)(x+4)=0(x - 5)(x + 4) = 0, giving x=5x = 5 or x=−4x = -4.

Systems of Equations

Concept
Two equations, two unknowns
xy(3, 2)

A system is two equations that must both be true. Solve by substitution (solve one for a variable and plug in) or elimination (add or subtract the equations to cancel a variable). The solution is the point where the two lines cross.

Each equation alone has a whole line of solutions; asking for both at once keeps only the point the lines share.

Example
Solve the system

x+y=5x + y = 5 and x−y=1x - y = 1: add them — elimination, the yy-terms cancel — to get 2x=62x = 6, so x=3x = 3, and then y=2y = 2.

Try it
Try it: a system

Solve 2x+y=112x + y = 11 and x−y=1x - y = 1.

Show the answer

Add the equations: 3x=123x = 12, so x=4x = 4; then 4−y=14 - y = 1 gives y=3y = 3. Check: 8+3=118 + 3 = 11. Solution (4,3)(4, 3).

Going Deeper: The Quadratic Formula and Absolute Value

Concept
The quadratic formula, demystified

Completing the square once, in general, on ax2+bx+c=0ax^2 + bx + c = 0 gives x=−b±b2−4ac2ax = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} — so the formula is not magic, it is the same move you already know, done once for every equation. The part under the root, b2−4acb^2 - 4ac, tells you how many real solutions there are: positive →\to two, zero →\to one, negative →\to none.

Reminder — The quadratic formula:  x=−b±b2−4ac2a(a≠0)x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\qquad (a\neq 0)
Example
Use it

x2−4x−5=0x^2 - 4x - 5 = 0: x=4±16+202=4±62x = \dfrac{4 \pm \sqrt{16 + 20}}{2} = \dfrac{4 \pm 6}{2}, so x=5x = 5 or x=−1x = -1. (Check: (x−5)(x+1)=x2−4x−5(x-5)(x+1) = x^2 - 4x - 5.)

Concept
Absolute value means distance

∣x−3∣=7|x - 3| = 7 says "xx is 77 away from 33" — so x=10x = 10 or x=−4x = -4. Reading absolute-value equations as distances turns them from a rule to memorize into a picture on the number line.

Example
A multi-step equation

Solve 3(2x−4)+5=2(x+3)−13(2x - 4) + 5 = 2(x + 3) - 1.

Step 1 — distributive property: 6x−12+5=2x+6−16x - 12 + 5 = 2x + 6 - 1.

Step 2 — collect like terms: 6x−7=2x+56x - 7 = 2x + 5, so 4x=124x = 12.

Step 3 — divide: x=3x = 3.

Example
Where two graphs cross
xy(2, 5)

Find where y=2x+1y = 2x + 1 meets y=−x+7y = -x + 7. At an intersection the yy-values agree, so set them equal: 2x+1=−x+72x + 1 = -x + 7, giving 3x=63x = 6 and x=2x = 2; then y=2(2)+1=5y = 2(2) + 1 = 5. The graphs cross at (2,5)(2, 5).

Problem-Solving Playbook

Concept
Get zero on one side

For anything quadratic, move everything to one side first — the zero-product property only works against zero. Then factor (or use the quadratic formula), and check by substituting back.

Example
Worked: rearrange, factor, check

Solve x2=5x−6x^2 = 5x - 6. Step 1: x2−5x+6=0x^2 - 5x + 6 = 0.

Step 2 — factor: (x−2)(x−3)=0(x-2)(x-3) = 0, so x=2x = 2 or x=3x = 3.

Step 3 — check x=2x=2: 4=10−64 = 10 - 6 ✓.