Lesson 35.2 · Writing Linear Equations

From Graph to Equation

Two points, one rate, one start value: write the rule for any straight line.

One recipe turns any straight graph into an equation

Last lesson you read the start and the rate from a graph. Now you will write them down as one rule. Every straight line has the form $y=mx+b$, where $m$ is the rate and $b$ is the start value (the $y$-value when $x=0$).

1 · Read two pointsPick two clear points on the line and put them in a table.

2 · Find $m$ and $b$$m=\frac{\Delta y}{\Delta x}$. Then $b$ is $y$ when $x=0$, or work back from a point.

3 · Write and checkPut $m$ and $b$ into $y=mx+b$, then test a third point.

Worked example · Pony trekking in Andalusia

A trekking company charges a booking fee plus an hourly price. The graph shows the total cost.

Hours, $x$	0	1	2	3
Cost, $y$ (€)	20	35	50	65

Rate. $m=\frac{\Delta y}{\Delta x}=\frac{35-20}{1-0}=15$ € per hour.

Start. $b=20$, the cost at $x=0$ (the booking fee).

Equation. $y=15x+20$.

Check with a new point: $x=4$ gives $y=15(4)+20=80$.

Practice · From graph to equation

Guided

1 · An Iceland riding tour Another company's total cost is shown below, where $x$ is hours of riding and $y$ is euros.Derive the tour equation $y=mx+b$ directly from the graph. Support it with two exact points, the rate with units, and an interpretation of the starting value.

select exact points → determine m and b → equation → interpret the start

2 · Use the tour model Use the equation you derived in Problem 1.
(a) Predict the cost of a $2$-hour tour and use the graph to verify it.
(b) Determine how many riding hours cost €140, showing the reverse calculation.

forward prediction → graph check → reverse the equation for a target cost

3 · The draining trough Water drains from a horse trough. $x$ is minutes since the tap opened, $y$ is litres left.Derive and verify the trough equation from the graph. Show the signed rate and start value as evidence, then check the equation with $(1,48)$.

the line goes down, so m is negative → b is the start amount → test a different point

From here, you choose the points and write the whole equation.

Independent

4 · The window starts at 2 Mia collects stamps for practising judo. This graph does not show $x=0$. $x$ is days of practice and $y$ is stamps earned.Derive and verify the equation even though the graph window hides $x=0$. Show how two visible points determine both $m$ and $b$, then check with $(6,41)$.

choose visible points → determine m → recover the hidden b → write and check

5 · Which equation matches? Which equation matches this line?A: $y=2x+10$ B: $y=-2x+10$ C: $y=10x-2$ D: $y=-10x+2$
(a) Choose the correct equation.
(b) Prove it by substituting the point $(4,2)$.
(c) Explain why A is wrong, using the direction of the line.

start value first, then the sign of the rate, then test a point

6 · Design a lesson plan A riding school charges a €30 joining fee plus €12 per lesson. $x$ is lessons booked and $y$ is total cost in euros.(a) Write the equation.
(b) Plot the points for $x=0$, $3$, and $5$ on the grid, then draw the line.
(c) Explain what the $30$ and the $12$ mean in the school.

words → equation → table of three points → plot and join → meaning of b and m

Spaced review

7 · Mountain trail points A trail passes through $(-3,4)$ and $(5,-12)$. Calculate the signed slope, showing the coordinate changes in a consistent order.

subtract in the same order for both coordinates → divide → check the sign

8 · Missing x-coordinate A line has slope $\frac23$ and passes through $(3,4)$ and $(x,10)$. Determine the missing coordinate and show how the slope connects the two points.

vertical change → use slope to determine horizontal change → complete the coordinate

9 · Ribbon for a rosette Mia cuts $x$ cm of ribbon. Five sixths of it, minus $4$ cm for the knot, leaves $16$ cm: $\frac56x-4=16$. (a) Add $4$ to both sides. (b) Solve for $x$.

undo the −4 first → multiply by the reciprocal 6/5

10 · Using the trekking rule Use the trekking equation $y=15x+20$. (a) Calculate $y$ when $x=7$. (b) Solve for $x$ when $y=155$.

forward: substitute → backward: balance

Show answers

Problem 1a · cost at x = 0

€40.

Problem 1a · cost at x = 1

€65.

Problem 1b

$m=\frac{115-65}{3-1}=\frac{50}{2}=25$ € per hour.

Problem 1c

$b=40$: the fixed starting cost (booking fee) is €40.

Problem 1 · equation

$y=25x+40$.

Problem 2a

$y=25(2)+40=90$, so a $2$-hour tour costs €90; the graph contains $(2,90)$.

Problem 2b

$140=25x+40$, so $100=25x$ and $x=4$ hours.

Problem 3a

$m=\frac{32-56}{3-0}=\frac{-24}{3}=-8$ litres per minute.

Problem 3b

$b=56$ litres.

Problem 3c

$y=-8x+56$. Check: $-8(1)+56=48$, matching $(1,48)$.

Problem 4a

$m=\frac{29-17}{4-2}=\frac{12}{2}=6$ stamps per day.

Problem 4b

$$17=6(2)+b\quad\mid\text{simplify}$$$$17=12+b\quad\mid-12$$$$5=b$$

Problem 4c

$y=6x+5$. Check: $6(6)+5=41$, matching $(6,41)$.

Problem 5a

B: $y=-2x+10$.

Problem 5b

$-2(4)+10=-8+10=2$, so $(4,2)$ is on the line.

Problem 5c

A has a positive rate and would rise, but the graph falls as $x$ increases. Also $2(4)+10=18$, not $2$.

Problem 6a

$y=12x+30$.

Problem 6b

Points $(0,30)$, $(3,66)$, $(5,90)$ on a straight line.

Problem 6c

$30$ is the €30 joining fee (cost with $0$ lessons). $12$ is the extra €12 for each lesson.

Problem 7c

$m=\frac{-16}{8}=-2$.

Problem 8c

$x=3+9=12$.

Problem 9a

$\frac56x=20$.

Problem 9b

$$\frac56x=20\quad\mid\times\frac65$$$$x=24$$So the ribbon is $24$ cm long.

Problem 10a

$y=15(7)+20=125$.

Problem 10b

$$155=15x+20\quad\mid-20$$$$135=15x\quad\mid\div15$$$$9=x$$

Next upNext, you will use equations like these to compare plans, decide which option is cheaper, and explain your choice to someone else.

One recipe turns any straight graph into an equation

Worked example · Pony trekking in Andalusia

Practice · From graph to equation

Spaced review

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