Twitter: @Owen134866
www.mathsfreeresourcelibrary.com
, Prior Knowledge Check
1) Find the point of intersection 3) Make y the subject of each
for each pair of lines: equation:
a) 𝑦 = 4𝑥 + 7 and 5𝑦 = 2𝑥 − 1 a) 6𝑥 + 3𝑦 − 15 = 0
(−2, −1) 𝑦 = 5 − 2𝑥
b) 𝑦 = 5𝑥 − 1 and 3𝑥 + 7𝑦 = 11
9 26
,
19 19
b) 2𝑥 − 5𝑦 − 9 = 0
c) 2𝑥 − 5𝑦 = −1 and 5𝑥 − 7𝑦 = 14 2 9
𝑦= 𝑥−
(7,3) 5 5
c) 3𝑥 − 7𝑦 + 12 = 0
2) Simplify each of the following: 3 12
𝑦= 𝑥+
a) 80 4 5 7 7
b) 200 10 2
c) 125 5 5
,
, Straight-line graphs
You can find the gradient of a straight
line joining two points by considering
the vertical and horizontal distance
between them. It can also be found y (x2,y2)
from the equation of the graph.
(x1,y1) y2 - y1
You can work out the gradient of a line x2 - x1
if you know 2 points on it.
x
Let the first point be (x1,y1) and the
second be (x2,y2). The following
formula gives the gradient:
y2 − y1
m=
x2 − x1
‘The change in the y values, divided by
the change in the x values’
5A/B
www.mathsfreeresourcelibrary.com
, Prior Knowledge Check
1) Find the point of intersection 3) Make y the subject of each
for each pair of lines: equation:
a) 𝑦 = 4𝑥 + 7 and 5𝑦 = 2𝑥 − 1 a) 6𝑥 + 3𝑦 − 15 = 0
(−2, −1) 𝑦 = 5 − 2𝑥
b) 𝑦 = 5𝑥 − 1 and 3𝑥 + 7𝑦 = 11
9 26
,
19 19
b) 2𝑥 − 5𝑦 − 9 = 0
c) 2𝑥 − 5𝑦 = −1 and 5𝑥 − 7𝑦 = 14 2 9
𝑦= 𝑥−
(7,3) 5 5
c) 3𝑥 − 7𝑦 + 12 = 0
2) Simplify each of the following: 3 12
𝑦= 𝑥+
a) 80 4 5 7 7
b) 200 10 2
c) 125 5 5
,
, Straight-line graphs
You can find the gradient of a straight
line joining two points by considering
the vertical and horizontal distance
between them. It can also be found y (x2,y2)
from the equation of the graph.
(x1,y1) y2 - y1
You can work out the gradient of a line x2 - x1
if you know 2 points on it.
x
Let the first point be (x1,y1) and the
second be (x2,y2). The following
formula gives the gradient:
y2 − y1
m=
x2 − x1
‘The change in the y values, divided by
the change in the x values’
5A/B
