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secondary 3 | E Maths
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Need real help for this graph question!!
From sec 3 e math tb
Thankyou so much
Date Posted: 3 years ago
Views: 678
2x+3y = 6
(i)
y = (6-2x)/3
= - ⅔ x + 2
(ii)
k = - ⅔
(iii)
y = 0
x = 3
Ans: (x,y) = (3,0)
(i)
y = (6-2x)/3
= - ⅔ x + 2
(ii)
k = - ⅔
(iii)
y = 0
x = 3
Ans: (x,y) = (3,0)
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Make equation into the form of y=mx+c.
Sub y=0 to find the line that intersect x axis.
Sub y=0 to find the line that intersect x axis.
Date Posted:
3 years ago
Thank you
i)
x/3 + y/2 = 1
Multiply both sides by 2,
2x/3 + y = 2
y = -2x/3 + 2
y = -⅔x + 2
This is the equation of the line.
ii)
General equation of a straight line is y = mx + c, whereby :
m is the gradient
c is the y-intercept (the y-coordinate of the intersection point of the line and the y-axis. Any point on the y-axis has coordinate = 0)
So, looking at the result in i), the gradient is -⅔.
iii)
Anywhere on the x-axis, y = 0.
So substitute y = 0 into the line's equation.
0 = -⅔x + 2
⅔x = 2
x = 2/(⅔)
x = 2 × 3/2
x = 3
The coordinates of the intersection point are (3,0)
x/3 + y/2 = 1
Multiply both sides by 2,
2x/3 + y = 2
y = -2x/3 + 2
y = -⅔x + 2
This is the equation of the line.
ii)
General equation of a straight line is y = mx + c, whereby :
m is the gradient
c is the y-intercept (the y-coordinate of the intersection point of the line and the y-axis. Any point on the y-axis has coordinate = 0)
So, looking at the result in i), the gradient is -⅔.
iii)
Anywhere on the x-axis, y = 0.
So substitute y = 0 into the line's equation.
0 = -⅔x + 2
⅔x = 2
x = 2/(⅔)
x = 2 × 3/2
x = 3
The coordinates of the intersection point are (3,0)
Thanks so much
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