In this unit we have drawn equations of straight lines where y is on its own (examples: y = 2x + 1, y = 3x – 2)
If x and y
are on the same side of the ‘=’ sign (examples: x + y = 2, 2x – 3y
= 6), we can find the point to
join up on a graph by using the cover-up method.
EXAMPLE:
Draw 2x + 3y = 6.
SOLUTION:
SOLUTION:
Use x = 0
so 2x = 0
Cover up 2x in the equation.
2x + 3y
= 6 becomes 0 + 3y = 6 so y = 2
Use y = 0
so 3y = 0
Cover up 3y in the equation.
2x + 3y
= 6 becomes 2x + 0 = 6 so x = 3
Always use x = 0 then y = 0.
Plot the points x = 0, y = 2 and x = 3, y = 0 on the graph and join them up to get your straight line.
Plot the points x = 0, y = 2 and x = 3, y = 0 on the graph and join them up to get your straight line.
EXAMPLE:
Suppose salami and sausage cost £6 and £3 per kilogram,
and we wish to buy £12 worth. How much
of each can we purchase?
SOLUTION:
Letting x
and y be the weights of salami and sausage, the total cost is
6x + 3y = 12.
Solving for y gives the point-slope form
y = –2x + 4,
as above. That is, if we fist choose the amount of
salami x,
the amount of sausage can be computed as a function
y = f(x)
= –2x + 4.
Since salami costs twice as much as sausage, adding
one kilo of salami decreases the sausage by 2 kilos:
f(x + 1) = f(x) – 2, and the slope
is –2.
The y-intercept
point (x, y) = (0, 4) corresponds to buying only 4
kg of sausage; while the x-intercept
point (x, y) = (2, 0) corresponds to buying only 2
kg of salami.
Note that the graph includes points with negative
values of x
or y,
which have no meaning in terms of the original variables (unless we
imagine selling meat to the butcher).
Thus we should restrict our function f(x) to the domain 0 ≤ x ≤ 2.
Also, we could choose y as the independent variable,
and compute x
by the inverse linear function:
x = q(y)
= – 1/2y + 2
over the domain 0 ≤ y ≤ 4.
Lesson 8
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