A line has equation 3x+5y=12.
Find it's gradient and the coordinates of the points where it crosses the axes.
A line has equation 3x+5y=12.?
2008-12-17 9:13 am
回答 (11)
2008-12-17 9:26 am
✔ 最佳答案
3x + 5y = 125y = 12 - 3x
y = 12/5 - 3x/5
The gradient is -3/5
The y intercept is 12/5
The x intercept is 4
2008-12-17 9:36 am
gradient:
3x + 5y = 12
5y = 12 - 3x
y = (12 - 3x)/5
y = 12/5 - 3x/5
gradient: -3/5 (-0.8)
= = = = = = = =
y-intercept:
3x + 5y = 12
5y = 12 - 3x
y = (12 - 3x)/5
Let x = 0:
y = (12 - 3x)/5
y = (12 - 3*0)/5
y = 12/5 (2.4)
= = = = = = = =
x-intercept:
3x + 5y = 12
3x = 12 - 5y
x = (12 - 5y)/3
Let y = 0:
x = (12 - 5y)/3
x = 12/3 - 5(0)/3
x = 4
3x + 5y = 12
5y = 12 - 3x
y = (12 - 3x)/5
y = 12/5 - 3x/5
gradient: -3/5 (-0.8)
= = = = = = = =
y-intercept:
3x + 5y = 12
5y = 12 - 3x
y = (12 - 3x)/5
Let x = 0:
y = (12 - 3x)/5
y = (12 - 3*0)/5
y = 12/5 (2.4)
= = = = = = = =
x-intercept:
3x + 5y = 12
3x = 12 - 5y
x = (12 - 5y)/3
Let y = 0:
x = (12 - 5y)/3
x = 12/3 - 5(0)/3
x = 4
2016-04-03 9:20 pm
3x + 5y - 12 = 0 ( This is in the form Ax + By + C = 0 ) slope = - A / B
2008-12-17 9:35 am
The gradient is:
3x + 5y = 12 ............................[1]
5y = -3x + 12
y = -3x/5 + 12/5
Therefore gradient is: -3 / 5 Answer.
For crossing the axes : For y = 0 in eqn[1] x = 4,
therefore it crosses the y-axis at point (4, 0)
For x = 0, in eqn[1] y = 12/5 = 2.4
therefore the line crosses the x-axis at point (0, 2.4)
Thus (4, 0) and (0, 2.4) Answer.
3x + 5y = 12 ............................[1]
5y = -3x + 12
y = -3x/5 + 12/5
Therefore gradient is: -3 / 5 Answer.
For crossing the axes : For y = 0 in eqn[1] x = 4,
therefore it crosses the y-axis at point (4, 0)
For x = 0, in eqn[1] y = 12/5 = 2.4
therefore the line crosses the x-axis at point (0, 2.4)
Thus (4, 0) and (0, 2.4) Answer.
參考: analytical book
2008-12-17 9:29 am
3X+5Y=12
=> Y= (-3/5)X + 12/5
So the gradient is (-3/5)
3X + 5Y = 12
=> 3X/12+ 5Y/12=1
=> X/4+ Y/(12⁄5)=1
therefore it crosses the X and Y axis at ( 4 , 0 ) and ( 0 , 12/5 ) respectibly ..
=> Y= (-3/5)X + 12/5
So the gradient is (-3/5)
3X + 5Y = 12
=> 3X/12+ 5Y/12=1
=> X/4+ Y/(12⁄5)=1
therefore it crosses the X and Y axis at ( 4 , 0 ) and ( 0 , 12/5 ) respectibly ..
2008-12-17 9:28 am
Hi there
The way I would do it is to get it in a form y = ?x + c so
subtract 3x from both sides and we get
5y = -3x + 12..Now divide both sides by 5
y = (-3/5)x + 12/5
In this form the x value (-3/5) is the slope or gradient and the 12/5 is the point that the line crosses the y-axis
To find out where it crosses the x axis make the equation
0 = (-3/5)x + 12/5
subtract 12/5 from both sides
-12/5 = (-3/5)x
now we want to divide both sides by (-3/5), you will remember that to divide one fraction by another you turn the second upside down the multiply. so we get
-12/5 * -5/3 = x giving x = 60/15 so x = 4
so the line crosses the x axis at 4
Check my arithmetic, but I think it's correct
John B
The way I would do it is to get it in a form y = ?x + c so
subtract 3x from both sides and we get
5y = -3x + 12..Now divide both sides by 5
y = (-3/5)x + 12/5
In this form the x value (-3/5) is the slope or gradient and the 12/5 is the point that the line crosses the y-axis
To find out where it crosses the x axis make the equation
0 = (-3/5)x + 12/5
subtract 12/5 from both sides
-12/5 = (-3/5)x
now we want to divide both sides by (-3/5), you will remember that to divide one fraction by another you turn the second upside down the multiply. so we get
-12/5 * -5/3 = x giving x = 60/15 so x = 4
so the line crosses the x axis at 4
Check my arithmetic, but I think it's correct
John B
2008-12-17 9:28 am
to find the gradient of any line, you need to arrange in the form of y = mx + b, i.e. make y the subject. Note: m will be your gradient.
so make y the subject of 3x + 5y = 12
3x + 5y = 12
5y = 12 - 3x (subtract 3x on both sides)
y = 12/5 - (3/5)x (divide by 5 on both sides)
y = (-3/5)x + (12/5) (rearrange right hand side so it is in the form of mx + b)
therefore m = -3/5 (so the gradient is (-3/5))
to find where it crosses the axes, i.e. the intercepts:
to find the x-intercept (Let y = 0)
3x + 5y = 12
3x + 5*0 = 12
3x + 0 = 12
3x = 12
x = 4
so, when y = 0, x = 4
therefore the x-intercept is (4,0)
to find the y-intercept (Let x = 0)
3x + 5y = 12
3*0 + 5y = 12
5y = 12
y = 12/5
so, when x = 0, y = 12/5
therefore the y-intercept is (0, 12/5)
Hope that helps. :)
so make y the subject of 3x + 5y = 12
3x + 5y = 12
5y = 12 - 3x (subtract 3x on both sides)
y = 12/5 - (3/5)x (divide by 5 on both sides)
y = (-3/5)x + (12/5) (rearrange right hand side so it is in the form of mx + b)
therefore m = -3/5 (so the gradient is (-3/5))
to find where it crosses the axes, i.e. the intercepts:
to find the x-intercept (Let y = 0)
3x + 5y = 12
3x + 5*0 = 12
3x + 0 = 12
3x = 12
x = 4
so, when y = 0, x = 4
therefore the x-intercept is (4,0)
to find the y-intercept (Let x = 0)
3x + 5y = 12
3*0 + 5y = 12
5y = 12
y = 12/5
so, when x = 0, y = 12/5
therefore the y-intercept is (0, 12/5)
Hope that helps. :)
2008-12-17 9:23 am
hello.
2/3 = y=2
7/3 = y=1
12/3 = y=0
so the gradient is -1/2.6666666
the x intercept is 4 and the y intercept is 12/5.
there you go
2/3 = y=2
7/3 = y=1
12/3 = y=0
so the gradient is -1/2.6666666
the x intercept is 4 and the y intercept is 12/5.
there you go
2008-12-17 9:18 am
5y=-3x+12,m=-3/5
x=0 y=12/5
x=4 y=0
x=0 y=12/5
x=4 y=0
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