Every linear relationship — savings over time, feed costs per day, riding lesson revenue — can be captured in one compact equation: $y = mx + b$. The slope $m$ tells you how fast things change. The y-intercept $b$ tells you where they start. Together, they let you draw a precise picture of any linear situation.
Slope-intercept form:
$$y = mx + b$$
$m$ = slope (rate of change — rise per unit of run)
$b$ = y-intercept (value of $y$ when $x = 0$; the starting value)
To graph: (1) Plot the y-intercept $(0, b)$. (2) From that point, apply slope: go right $1$ unit, go up $m$ units (or down if $m < 0$). (3) Draw the line.
To write the equation: Identify $m$ (slope) and $b$ (starting value) from the context, then substitute.
Worked Example
Worked Example — Word problem to equation to graph
Mia starts with €50 in savings and saves €12 per week. Write the equation for her total savings $y$ after $x$ weeks, and describe the graph.
Starting value (y-intercept): $b = 50$ (she already has €50).
Rate of change (slope): $m = 12$ (adds €12 each week).
$$y = 12x + 50$$
Graph description: Starts at $(0, 50)$. Rises $12$ units for each $1$ unit to the right. At $x = 4$: $y = 12(4)+50 = 98$.
Warm-Up — Describe and use equations (equations given)
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Problem 1
For the equation $y = 3x + 7$:
(a) State the slope and y-intercept.
(b) Where does the line start on the $y$-axis?
(c) Find $y$ when $x = 5$.
(d) Does the line rise or fall as $x$ increases?
$m = 3$, $b = 7$
read off m and b → substitute x=5 → direction →
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Problem 2
For the equation $y = -2x + 10$:
(a) State the slope and y-intercept.
(b) Find $y$ when $x = 0$, $x = 3$, $x = 5$.
(c) For what value of $x$ does $y = 0$? (Set $y = 0$ and solve.)
$m = -2$, $b = 10$
substitute each x → solve for x-intercept →
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Problem 3
Mia's stable has a fixed monthly cost of €150 (rent and utilities) plus €8 per horse per day for feed. Write the equation for total monthly cost $y$ in terms of number of days $x$ the stable operates, if there are 3 horses.
$y = (\text{feed cost per day}) \times x + (\text{fixed cost})$
find m (feed/day × horses) → b = 150 → write equation →
Core Problems — write equation from context, then use it
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Problem 4
The arithmetic sequence $5, 8, 11, 14, \ldots$ has $a_1 = 5$ and common difference $d = 3$.
(a) Write the equation for the $n$-th term $y$ in terms of position $x$: think of $d$ as slope and $a_1 - d$ as y-intercept.
(b) Verify: when $x = 1$, $y = 5$? When $x = 3$, $y = 11$?
(c) What is the slope of the graph of this sequence? How does it relate to $d$?
general term y = dx + (a₁ - d) → verify → slope = d →
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Problem 5
A line has slope $\frac{3}{2}$ and passes through the point $(4, 11)$. Find the y-intercept $b$ and write the full equation.
(Hint: substitute the known point into $y = mx + b$ and solve for $b$.)
11 = (3/2)(4) + b → solve for b → write equation →
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Problem 6
A line passes through $(2, 5)$ and $(6, 13)$.
(a) Find the slope $m$.
(b) Use one point to find $b$.
(c) Write the equation.
(d) Find $y$ when $x = 10$.
slope formula → substitute for b → equation → extend →
From here — write the equation yourself, then solve
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Problem 7
Mia's savings account has €50 already and grows by €12 per week. A second account has €200 but grows by only €5 per week.
(a) Write an equation for each account.
(b) After how many weeks do they have the same amount? (Set the two equations equal and solve.)
write y = 12x + 50 and y = 5x + 200 → set equal → solve for x →
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Problem 8 Challenge
Write two different linear equations $y = m_1 x + b_1$ and $y = m_2 x + b_2$ that both pass through the point $(4, 20)$.
(a) Choose any two different slopes. Find the corresponding y-intercepts.
(b) If the two lines have the same slope but different y-intercepts, would they cross? Why or why not?
substitute (4,20) → solve for b with each slope → parallel lines discussion →
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Problem 9 Open
Design a business scenario for Mia where: (1) the starting cost (y-intercept) is positive, meaning she invests something upfront; (2) the slope is positive, meaning she earns more over time; (3) there is a "break-even point" where her earnings equal her costs.
• Write the equation.
• Find the break-even point algebraically.
• Describe the scenario in one or two sentences.
choose m and b → equation → break-even at y=0 or set revenue = cost → solve →
Show answers
Problem 1
(a) $m = 3$, $b = 7$ · (b) Starts at $y = 7$ · (c) $y = 3(5)+7 = 22$ · (d) Rises (positive slope)
Problem 2
(a) $m = -2$, $b = 10$ · (b) $y(0)=10$, $y(3)=4$, $y(5)=0$ · (c) $0 = -2x+10$ → $x = 5$
Problem 3
$m = 3 \times 8 = 24$ €/day · $b = 150$ · Equation: $y = 24x + 150$
Problem 4
(a) $y = 3x + 2$ (since $a_1 - d = 5 - 3 = 2$) · (b) $x=1$: $3+2=5$ ✓; $x=3$: $9+2=11$ ✓ · (c) Slope $= 3 = d$ — the common difference is the slope.
Problem 5
$11 = \frac{3}{2}(4) + b = 6 + b$ → $b = 5$ · Equation: $y = \frac{3}{2}x + 5$
Problem 6
(a) $m = (13-5)/(6-2) = 2$ · (b) $5 = 2(2)+b$ → $b = 1$ · (c) $y = 2x+1$ · (d) $y = 2(10)+1 = 21$
Problem 7
$y_1 = 12x+50$, $y_2 = 5x+200$ · Set equal: $12x+50 = 5x+200$ → $7x = 150$ → $x \approx 21.4$ weeks. After about $22$ weeks the accounts are equal.
Problem 8
Open — example: slope $= 3$: $20 = 3(4)+b$ → $b = 8$, equation $y = 3x+8$; slope $= -1$: $20 = -(4)+b$ → $b = 24$, equation $y = -x+24$. Both pass through $(4,20)$ ✓ · Same slope, different y-intercept → parallel lines — never cross (no solution exists where they're equal).
Problem 9
Open — example: Mia spends €90 on supplies. She earns €15 per lesson. Net profit after $x$ lessons: $y = 15x - 90$. Break-even: $0 = 15x-90$ → $x = 6$ lessons. After 6 lessons she has recovered her costs.
Coming up next → Lesson 23: Python — Plot a Line
You've done the maths. Now see it come alive in code. Lesson 23 is a hands-on programming session: $5$ lines of Python that plot $y = mx + b$ using matplotlib. No experience needed — just curiosity and a laptop.