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coordinates, graphs

for part a you sub x=0 so y=3 (0,3)?

for b, you can but them equal to each other so 2x-3=ax+2
2x-ax=5
then im not sure what to do nextScreenshot 2021-01-22 at 13.35.06.png
(edited 3 years ago)
Have you considered writing an equation for where they meet on the modulus side of the graph. 2x-3 will equal the negative version of the other equation I believe.
Original post by dnejfn
for part a you sub x=0 so y=3 (0,3)?

for b, you can but them equal to each other so 2x-3=ax+2
2x-ax=5
then im not sure what to do nextScreenshot 2021-01-22 at 13.35.06.png

a. Shouldn't this be y=0 as opposed to x = 0?
Anyway, the neat trick I picked up during A Levels is to set the modulous part negative and let them equal to each other i.e.
-(2x-3) = 2x-3

b.i. Again, similar to the above:
2x -3 = ax+2 - solve for x
-(2x-3) = ax+2 - solve for x
Find a.

b.ii. using b.i, solve for x
Original post by MindMax2000
a. Shouldn't this be y=0 as opposed to x = 0?
Anyway, the neat trick I picked up during A Levels is to set the modulous part negative and let them equal to each other i.e.
-(2x-3) = 2x-3

b.i. Again, similar to the above:
2x -3 = ax+2 - solve for x
-(2x-3) = ax+2 - solve for x
Find a.

b.ii. using b.i, solve for x

oh yeahh it should be y=0 so then x= 3/2 so (3/2,0)

but then for b how do i find out what a is when i dont know what x is? or would you use x=3/2
so then sub x in to get
3-3 = 3/2a +2
-2=3/2a
a=-4/3???
Just actually read the question I think it wants a range of possible values for a. As it crosses at two distinct points there are values for a which it wouldn't cross at all or only cross once. Such as if a=2 it would run parallel to the second half of the graph and cross once.
Original post by dnejfn
oh yeahh it should be y=0 so then x= 3/2 so (3/2,0)

but then for b how do i find out what a is when i dont know what x is? or would you use x=3/2
so then sub x in to get
3-3 = 3/2a +2
-2=3/2a
a=-4/3???

Nowhere in the question did it said that y = ax+2 crosses where the modulous equations cross.

The question specifically states that y = ax+2 crosses at 2 points on the modulous equation.
2x -3 = ax+2 - solve for x
-(2x-3) = ax+2 - solve for x
Find a.

using b.i, solve for x for part ii.

Also nowhere in the question did it states they want the values for x or a, just a in terms of x and vice versa
Original post by MindMax2000
Nowhere in the question did it said that y = ax+2 crosses where the modulous equations cross.

The question specifically states that y = ax+2 crosses at 2 points on the modulous equation.
2x -3 = ax+2 - solve for x
-(2x-3) = ax+2 - solve for x
Find a.

using b.i, solve for x for part ii.

Also nowhere in the question did it states they want the values for x or a, just a in terms of x and vice versa

(b) i,

if i find a in terms of x then,
2x-3=ax+2
2x-5=ax
a= 2x-5/x?
im so confused lol
Original post by dnejfn
(b) i,

if i find a in terms of x then,
2x-3=ax+2
2x-5=ax
a= 2x-5/x?
im so confused lol

They're simultaneous equations.
It might be easier to see if you draw a hypothetical line across both parts of the modulous function and label it y = ax+2
In your case, you want to eliminate one of the unknowns, so I recommend eliminating x in both equations in order to find a.

I have a = -4/3
Original post by MindMax2000
They're simultaneous equations.
It might be easier to see if you draw a hypothetical line across both parts of the modulous function and label it y = ax+2
In your case, you want to eliminate one of the unknowns, so I recommend eliminating x in both equations in order to find a.

I have a = -4/3

yeah, i got a=-4/3 for B (i). But then now what do i do for B(ii)?
Original post by dnejfn
yeah, i got a=-4/3 for B (i). But then now what do i do for B(ii)?

2x -3 = ax+2 - first equation
-(2x-3) = ax+2 - second equation

sub a = -4/3 into both equatios. Solve for the 2 x coordinates.
Original post by MindMax2000
2x -3 = ax+2 - first equation
-(2x-3) = ax+2 - second equation

sub a = -4/3 into both equatios. Solve for the 2 x coordinates.

i got x=3/2 for both equations when i subbed a in
Original post by dnejfn
i got x=3/2 for both equations when i subbed a in

Sorry, it seems I led you astray on part B. Loubielou was right in the sense that you have a range of values and not one for a.

Arranging the equations for one formula yields a = (2x-5)/x and x = 5/(2-a); x cannot be 0 and a cannot be 2 (you should have this)
For the other, a = (1-2x)/x and x = 1/(a+2); x cannot be 0 and a canot be -2 (you might have this)
So from the above, a can be any number but 2 and -2

bii. the x values are provided above.

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