This page covers the parts of algebra that usually require more than one step of thinking: translating a situation, combining equations, or justifying that two expressions are always equal.
These topics appear frequently in CSEC reasoning questions. On Paper 02, define the variables, write the equation or relationship, solve step by step, and interpret the result in context.
What Are Non-Linear Equations?
Non-linear means at least one equation has a variable to a power higher than 1 (like x 2 x^2 x 2
When you solve non-linear simultaneous equations, you're finding where a line intersects a curve (like a parabola).
Strategy: Substitution (Exactly Like Linear)
The method is the SAME as linear simultaneous equations:
Rearrange one equation to get a variable by itself
Substitute into the other equation
Solve (may be quadratic now)
Find the other variable
Example / Example 1: Line + Parabola Solve:
y = x + 3 ...(1) y = x + 3 \quad \text{...(1)} y = x + 3 ...(1) y = x 2 − 2 x + 1 ...(2) y = x^2 - 2x + 1 \quad \text{...(2)} y = x 2 − 2 x + 1 ...(2)
Step 1: Equation (1) already has y y y y = x + 3 y = x + 3 y = x + 3
Step 2: Substitute into equation (2)
x + 3 = x 2 − 2 x + 1 x + 3 = x^2 - 2x + 1 x + 3 = x 2 − 2 x + 1
Step 3: Rearrange into standard form
0 = x 2 − 2 x + 1 − x − 3 0 = x^2 - 2x + 1 - x - 3 0 = x 2 − 2 x + 1 − x − 3 0 = x 2 − 3 x − 2 0 = x^2 - 3x - 2 0 = x 2 − 3 x − 2
Step 4: Check the discriminant
Δ = b 2 − 4 a c = ( − 3 ) 2 − 4 ( 1 ) ( − 2 ) = 9 + 8 = 17 \Delta = b^2 - 4ac = (-3)^2 - 4(1)(-2) = 9 + 8 = 17 Δ = b 2 − 4 a c = ( − 3 ) 2 − 4 ( 1 ) ( − 2 ) = 9 + 8 = 17
Since Δ > 0 \Delta > 0 Δ > 0
Step 5: Use the quadratic formula
x = − ( − 3 ) ± 17 2 ( 1 ) = 3 ± 17 2 x = \frac{-(-3) \pm \sqrt{17}}{2(1)} = \frac{3 \pm \sqrt{17}}{2} x = 2 ( 1 ) − ( − 3 ) ± 17 = 2 3 ± 17
x 1 = 3 + 17 2 ≈ 3.56 x_1 = \frac{3 + \sqrt{17}}{2} \approx 3.56 x 1 = 2 3 + 17 ≈ 3.56 x 2 = 3 − 17 2 ≈ − 0.56 x_2 = \frac{3 - \sqrt{17}}{2} \approx -0.56 x 2 = 2 3 − 17 ≈ − 0.56
Step 6: Find corresponding y y y
When x ≈ 3.56 x \approx 3.56 x ≈ 3.56 y = 3.56 + 3 = 6.56 y = 3.56 + 3 = 6.56 y = 3.56 + 3 = 6.56
When x ≈ − 0.56 x \approx -0.56 x ≈ − 0.56 y = − 0.56 + 3 = 2.44 y = -0.56 + 3 = 2.44 y = − 0.56 + 3 = 2.44
Solutions: ( 3.56 , 6.56 ) (3.56, 6.56) ( 3.56 , 6.56 ) ( − 0.56 , 2.44 ) (-0.56, 2.44) ( − 0.56 , 2.44 )
Or exactly: ( 3 + 17 2 , 9 + 17 2 ) \left(\frac{3 + \sqrt{17}}{2}, \frac{9 + \sqrt{17}}{2}\right) ( 2 3 + 17 , 2 9 + 17 ) ( 3 − 17 2 , 9 − 17 2 ) \left(\frac{3 - \sqrt{17}}{2}, \frac{9 - \sqrt{17}}{2}\right) ( 2 3 − 17 , 2 9 − 17 )
y = 2x + 1 (line) and y = x² - 2 (parabola) Example / Example 2: Two Parabolas (Quadratic + Quadratic) Solve:
y = x 2 + 2 ...(1) y = x^2 + 2 \quad \text{...(1)} y = x 2 + 2 ...(1) y = − x 2 + 4 x − 1 ...(2) y = -x^2 + 4x - 1 \quad \text{...(2)} y = − x 2 + 4 x − 1 ...(2)
Step 1: Both have y y y
Step 2: Set them equal
x 2 + 2 = − x 2 + 4 x − 1 x^2 + 2 = -x^2 + 4x - 1 x 2 + 2 = − x 2 + 4 x − 1
Step 3: Rearrange
x 2 + x 2 − 4 x + 2 + 1 = 0 x^2 + x^2 - 4x + 2 + 1 = 0 x 2 + x 2 − 4 x + 2 + 1 = 0 2 x 2 − 4 x + 3 = 0 2x^2 - 4x + 3 = 0 2 x 2 − 4 x + 3 = 0
Step 4: Check discriminant
Δ = ( − 4 ) 2 − 4 ( 2 ) ( 3 ) = 16 − 24 = − 8 \Delta = (-4)^2 - 4(2)(3) = 16 - 24 = -8 Δ = ( − 4 ) 2 − 4 ( 2 ) ( 3 ) = 16 − 24 = − 8
Since Δ < 0 \Delta < 0 Δ < 0 NO real solutions .
This means the two parabolas don't intersect.
How Many Solutions?
The discriminant tells you:
Δ > 0 \Delta > 0 Δ > 0 Δ = 0 \Delta = 0 Δ = 0 Δ < 0 \Delta < 0 Δ < 0
Common mistake: After substituting and getting a quadratic, students forget to solve it completely.
Remember: you need BOTH x x x y y y
Solve the quadratic to find all x x x
For EACH x x x y y y
Write both coordinates: ( x , y ) (x, y) ( x , y )
Word Problems
Setting Up Equations from Words
Read carefully and define variables!
"The sum of two numbers is 15. Their difference is 3. Find the numbers."
Let x x x y y y
x + y = 15 ...(1) x + y = 15 \quad \text{...(1)} x + y = 15 ...(1) x − y = 3 ...(2) x - y = 3 \quad \text{...(2)} x − y = 3 ...(2)
Add equations: 2 x = 18 ⇒ x = 9 2x = 18 \Rightarrow x = 9 2 x = 18 ⇒ x = 9
From (1): y = 6 y = 6 y = 6
Numbers are 9 and 6.
"A rectangle's length is 3 cm more than its width. Its area is 40 cm². Find dimensions."
Let w w w l = w + 3 l = w + 3 l = w + 3
w ( w + 3 ) = 40 w(w + 3) = 40 w ( w + 3 ) = 40 w 2 + 3 w − 40 = 0 w^2 + 3w - 40 = 0 w 2 + 3 w − 40 = 0 ( w + 8 ) ( w − 5 ) = 0 (w + 8)(w - 5) = 0 ( w + 8 ) ( w − 5 ) = 0
w = 5 w = 5 w = 5 l = 8 l = 8 l = 8
Dimensions: 5 cm × 8 cm
Direct and Inverse Variation
Direct Variation
Direct variation happens when one quantity increases and another quantity also increases at the same rate.
Real-world examples:
More workers = More work gets done
Longer you drive = Further distance you travel
More pizzas you buy = Higher your bill
More study hours = Higher your test score (usually!)
Key idea: If you double one thing, the other thing doubles too. If you triple one thing, the other thing triples too.
The Pattern
When y y y x x x
y = k x y = kx y = k x
Where:
y y y x x x k k k constant of proportionality (the "multiplier")k k k same ratio : k = y x k = \frac{y}{x} k = x y
In other words: The ratio y x \frac{y}{x} x y
How to Solve Direct Variation Problems
Step 1: Write the formula: y = k x y = kx y = k x
Step 2: Find k k k
Step 3: Write the specific equation with that k k k
Step 4: Use it to answer questions
"It costs 4 dollars per item. If you buy 3 items, it costs 12 dollars. How much for 5 items?"
Step 1: Cost varies directly with number of items
Cost = k × Items \text{Cost} = k \times \text{Items} Cost = k × Items
Step 2: Find k k k 12 = k × 3 12 = k \times 3 12 = k × 3 k = 4 dollars per item k = 4 \text{ dollars per item} k = 4 dollars per item
This makes sense! Each item costs 4 dollars.
Step 3: The equation is:
Cost = 4 × Items \text{Cost} = 4 \times \text{Items} Cost = 4 × Items
Step 4: For 5 items:
Cost = 4 × 5 = 20 dollars \text{Cost} = 4 \times 5 = 20 \text{ dollars} Cost = 4 × 5 = 20 dollars
Notice: When items went from 3 to 5 (multiply by 5/3), cost went from 12 to 20 (also multiply by 5/3). They move together!
Graph of Direct Variation
A direct variation graph is always a straight line through the origin .
Why through the origin? Because when x = 0 x = 0 x = 0 y = k ( 0 ) = 0 y = k(0) = 0 y = k ( 0 ) = 0 Steeper line = larger k k k Flatter line = smaller k k k
Direct variation: y = kx through origin Inverse Variation
Inverse variation happens when one quantity increases and another quantity decreases . They work in opposite directions.
Real-world examples:
More workers on a job = Less time the job takes
Faster you drive = Less time to reach destination
More savings means fewer days of work required
A larger pizza means fewer pieces needed to feed the same number of people
Higher the price = Fewer people will buy it
Key idea: If you double one thing, the other thing gets cut in half. If you triple one thing, the other thing gets divided by 3.
The Pattern
When y y y x x x
y = k x y = \frac{k}{x} y = x k
Where:
y y y x x x k k k constant of proportionality The product x × y x \times y x × y x y = k xy = k x y = k
In other words: When you multiply x x x y y y k k k
How to Solve Inverse Variation Problems
Step 1: Write the formula: y = k x y = \frac{k}{x} y = x k x y = k xy = k x y = k
Step 2: Find k k k
Step 3: Write the specific equation with that k k k
Step 4: Use it to answer questions
"A job takes 12 hours with 2 workers. How long with 4 workers?"
Step 1: Time varies inversely with number of workers
Time = k Workers \text{Time} = \frac{k}{\text{Workers}} Time = Workers k
Step 2: Find k k k 12 = k 2 12 = \frac{k}{2} 12 = 2 k k = 12 × 2 = 24 worker-hours k = 12 \times 2 = 24 \text{ worker-hours} k = 12 × 2 = 24 worker-hours
This makes sense! The total amount of work is 24 worker-hours. No matter how many workers, the work stays the same.
Step 3: The equation is:
Time = 24 Workers \text{Time} = \frac{24}{\text{Workers}} Time = Workers 24
Step 4: For 4 workers:
Time = 24 4 = 6 hours \text{Time} = \frac{24}{4} = 6 \text{ hours} Time = 4 24 = 6 hours
Notice: When workers doubled (2 → 4), time halved (12 → 6). They move opposite ways!
Check the product:
With 2 workers: 2 × 12 = 24 2 \times 12 = 24 2 × 12 = 24
With 4 workers: 4 × 6 = 24 4 \times 6 = 24 4 × 6 = 24
The product is always 24!
Graph of Inverse Variation
An inverse variation graph is a hyperbola (curved shape, like a stretched smile).
Never touches the axes: The graph never crosses x = 0 x = 0 x = 0 y = 0 y = 0 y = 0 Closer to one axis, farther from other: When x x x y y y Two branches: One in upper-right quadrant, one in lower-left quadrant (for positive k k k Curves smooth, never straight: The relationship changes as you move along the curve
Inverse variation: y = 4/x Comparing Direct vs. Inverse
Recognizing variation in exams:
Look for these phrases:
"directly proportional" = direct variation: y = k x y = kx y = k x
"varies directly" = direct variation: y = k x y = kx y = k x
"inversely proportional" = inverse variation: y = k / x y = k/x y = k / x
"varies inversely" = inverse variation: y = k / x y = k/x y = k / x
Strategy:
Identify which type (direct or inverse)
Write appropriate formula
Find k k k
Answer the question using that specific equation
Proving Algebraic Identities
What's an Identity?
An identity is an equation that is TRUE for ALL possible values of the variable(s).
Compare:
Equation: 2 x + 3 = 7 2x + 3 = 7 2 x + 3 = 7 x = 2 x = 2 x = 2 Identity: 2 ( x + 1 ) = 2 x + 2 2(x + 1) = 2x + 2 2 ( x + 1 ) = 2 x + 2 x x x
How to Prove an Identity
Goal: Show that the left side and right side are exactly the same thing.
Method: Transform one side (usually the more complicated side) into the other side using algebraic rules.
Key principle: You can only manipulate ONE side at a time. You cannot add/subtract the same thing from both sides like in equations.
Strategy
Look at both sides, which one looks more complicated?
Start with the complicated side
Use algebraic operations to simplify it
Try to make it look like the other side
When you've transformed one side into the other, you're done!
Example / Example 1: Expanding to Prove Prove: ( a + b ) 2 = a 2 + 2 a b + b 2 (a + b)^2 = a^2 + 2ab + b^2 ( a + b ) 2 = a 2 + 2 ab + b 2
Which side is more complex? The left side has a squared binomial (more complex).
Start with left side and expand:
Step 1: Write the square as multiplication
( a + b ) 2 = ( a + b ) ( a + b ) (a+b)^2 = (a+b)(a+b) ( a + b ) 2 = ( a + b ) ( a + b )
Step 2: Use FOIL
= a ( a ) + a ( b ) + b ( a ) + b ( b ) = a(a) + a(b) + b(a) + b(b) = a ( a ) + a ( b ) + b ( a ) + b ( b ) = a 2 + a b + a b + b 2 = a^2 + ab + ab + b^2 = a 2 + ab + ab + b 2
Step 3: Combine like terms
= a 2 + 2 a b + b 2 = a^2 + 2ab + b^2 = a 2 + 2 ab + b 2
Step 4: This is exactly the right side!
Left side = Right side ✓ Identity proven
What this proves: No matter what a a a b b b ( a + b ) 2 (a+b)^2 ( a + b ) 2 a 2 + 2 a b + b 2 a^2 + 2ab + b^2 a 2 + 2 ab + b 2
Example / Example 2: Difference of Squares Pattern Prove: ( x − y ) ( x + y ) = x 2 − y 2 (x-y)(x+y) = x^2 - y^2 ( x − y ) ( x + y ) = x 2 − y 2
Which side is more complex? The left side has two binomials (more complex).
Start with left side and expand:
Step 1: Use FOIL on ( x − y ) ( x + y ) (x-y)(x+y) ( x − y ) ( x + y ) = x ( x ) + x ( y ) + ( − y ) ( x ) + ( − y ) ( y ) = x(x) + x(y) + (-y)(x) + (-y)(y) = x ( x ) + x ( y ) + ( − y ) ( x ) + ( − y ) ( y ) = x 2 + x y − x y − y 2 = x^2 + xy - xy - y^2 = x 2 + x y − x y − y 2
Step 2: Combine like terms (notice x y − x y = 0 xy - xy = 0 x y − x y = 0 = x 2 − y 2 = x^2 - y^2 = x 2 − y 2
Step 3: This is exactly the right side!
Left side = Right side ✓ Identity proven
What this proves: Whenever you multiply a sum by a difference, the middle terms always cancel!
Example / Example 3: Simplification to Prove Prove: x 2 − 1 x − 1 = x + 1 \frac{x^2 - 1}{x - 1} = x + 1 x − 1 x 2 − 1 = x + 1 x ≠ 1 x \neq 1 x = 1
Which side is more complex? The left side has a fraction (more complex).
Start with left side and simplify:
Step 1: Factor the numerator (difference of squares)
x 2 − 1 x − 1 = ( x − 1 ) ( x + 1 ) x − 1 \frac{x^2 - 1}{x - 1} = \frac{(x-1)(x+1)}{x-1} x − 1 x 2 − 1 = x − 1 ( x − 1 ) ( x + 1 )
Step 2: Cancel the common factor ( x − 1 ) (x-1) ( x − 1 ) = ( x − 1 ) ( x + 1 ) x − 1 = x + 1 = \frac{\cancel{(x-1)}(x+1)}{\cancel{x-1}} = x+1 = x − 1 ( x − 1 ) ( x + 1 ) = x + 1
Step 3: This is exactly the right side!
Left side = Right side ✓ Identity proven
Important: We require x ≠ 1 x \neq 1 x = 1
Example / Example 4: Factoring to Prove Prove: a b + 3 a + 2 b + 6 = ( a + 2 ) ( b + 3 ) ab + 3a + 2b + 6 = (a + 2)(b + 3) ab + 3 a + 2 b + 6 = ( a + 2 ) ( b + 3 )
Which side is more complex? The left side has four separate terms (more complex).
Start with left side and factor:
Step 1: Group the terms
a b + 3 a + 2 b + 6 = ( a b + 3 a ) + ( 2 b + 6 ) ab + 3a + 2b + 6 = (ab + 3a) + (2b + 6) ab + 3 a + 2 b + 6 = ( ab + 3 a ) + ( 2 b + 6 )
Step 2: Factor out common factors from each group
= a ( b + 3 ) + 2 ( b + 3 ) = a(b + 3) + 2(b + 3) = a ( b + 3 ) + 2 ( b + 3 )
Step 3: Factor out the common binomial
= ( a + 2 ) ( b + 3 ) = (a + 2)(b + 3) = ( a + 2 ) ( b + 3 )
Step 4: This is exactly the right side!
Left side = Right side ✓ Identity proven
What NOT to Do
Wrong approach: Don't start with the equality and try to manipulate both sides.
❌ Wrong:
x 2 + 2 x + 1 = ( x + 1 ) 2 x^2 + 2x + 1 = (x+1)^2 x 2 + 2 x + 1 = ( x + 1 ) 2 (subtract x 2 ) 2 x + 1 = ( x + 1 ) 2 − x 2 \text{(subtract } x^2 \text{)} \quad 2x + 1 = (x+1)^2 - x^2 (subtract x 2 ) 2 x + 1 = ( x + 1 ) 2 − x 2
✓ Right:
Start with ( x + 1 ) 2 and show it simplifies to x 2 + 2 x + 1 \text{Start with } (x+1)^2 \text{ and show it simplifies to } x^2 + 2x + 1 Start with ( x + 1 ) 2 and show it simplifies to x 2 + 2 x + 1 ( x + 1 ) 2 = ( x + 1 ) ( x + 1 ) = x 2 + x + x + 1 = x 2 + 2 x + 1 (x+1)^2 = (x+1)(x+1) = x^2 + x + x + 1 = x^2 + 2x + 1 ( x + 1 ) 2 = ( x + 1 ) ( x + 1 ) = x 2 + x + x + 1 = x 2 + 2 x + 1
Proof Strategy Summary
Identify the complex side (usually the one with brackets, fractions, or more terms)Transform that side using:
Expanding (FOIL, distributive law)
Factoring
Canceling common factors
Combining like terms
Simplify step-by-step until it matches the other sideConclude: "Left side = Right side ✓"