VCE Methods Question Thread!

[TrueTears]

3. (bx)/(ba)+(ya)/(ba)=1 => bx+ya=ba => abx+a^2y=ba^2

(ax)/(ba) + (by)/(ba)=1 => ax+by=ba => abx+b^2y=b^2a

abx+a^2y-(abx+b^2y) = ba^2-b^2a

(a^2-b^2)y = ab(a-b)

(a+b)(a-b)y = ab(a-b)

y = (ab)/(a+b)

same shit for x

try the same method for the other q

btw i love the (ba) as the denominators, BoredAussie.

[Phy124]

ax + by = p (multiply by b)
bx - ay = q (multiply by a)

abx + b²y = bp
(-) abx - a²y = aq

b²y + a²y = bp - aq
y(b² + a²) = bp - aq

y = (bp - aq)/(b² + a²) = (bp - aq)/(a² + b²)

Do the same but eliminate the y not x for the other answer

edit: i.e. multiple top by a and bottom by b

[Hutchoo]

AHHH, lol, the solutions are very logical and straight forward Ty guys!

[Insa]

Hey guys, I tried to differentiate this equation and got "-10(1-2x)^3". The answer says "20(1-2x)^3". Can anyone help me? What did I do wrong?

Question is attached. Thanks 

[b^3]

Hey guys, I tried to differentiate this equation and got "-10(1-2x)^3". The answer says "20(1-2x)^3". Can anyone help me? What did I do wrong?

Question is attached. Thanks 

Remember the chain rule, you need to mulitply by the derivative of the inside of the bracket aswell, i.e. multiply by -2

i.e. -10(1-2x)³ *-2=20(1-2x)³

[Insa]

Ahh! Thanks for the help

[Zahta]

Need help please with questions about simultanous equations and using the paramater do you use the calculator because the example the teacher showed us before the holidays was using a calculator , the equations had infinetly many solutions .
What are the steps involved with simulatanous equations and parameters

[gumby97]

Hi,

Find all the values of x in the interval [o, 2pi]

a) sin2x= -0.5

and

c) root2 cos(x)= -1

Any help is much appreciated!

[chocolatedaddy]

Hi,

Find all the values of x in the interval [o, 2pi]

a) sin2x= -0.5

and

c) root2 cos(x)= -1

Any help is much appreciated!

By using exact values.
a) BA(base angle):pi/6,whilst sin is negative in the 3rd and 4th Quadrants
However since it is sin2x. The domain becomes [0,4pi]
So the values are pi+pi/6,2pi-pi/6,3pi+pi/6,4pi-pi/6
2x=7pi/6,11pi/6,19pi/6,23pi/6
x=7pi/12,11pi/12,19pi/12,23pi/12

Do the same thing for question b).
Hope i helped.
p.Smeone correct me if i'm wrong

[gumby97]

Hi,

Find all the values of x in the interval [o, 2pi]

a) sin2x= -0.5

and

c) root2 cos(x)= -1

Any help is much appreciated!

By using exact values.
a) BA(base angle):pi/6,whilst sin is negative in the 3rd and 4th Quadrants
However since it is sin2x. The domain becomes [0,4pi]
So the values are pi+pi/6,2pi-pi/6,3pi+pi/6,4pi-pi/6
2x=7pi/6,11pi/6,19pi/6,23pi/6
x=7pi/12,11pi/12,19pi/12,23pi/12

Do the same thing for question b).
Hope i helped.
p.Smeone correct me if i'm wrong

Thanks man!

[gumby97]

Find the values of x between o and 360^degrees for which 2sin(2x-30)^degrees=1


Thanks in advance

[dc302]

To the power of degrees?

[gumby97]

To the power of degrees?

no, just degrees, i just didn't know how to communicate it clearly over the internet, sorry

[dc302]

I'm confused about this part: 2sin(2x-30)^degrees=1

Do you mean 2sin(2x-30) = 1? (where 2x-30 is in degrees)

[gumby97]

I'm confused about this part: 2sin(2x-30)^degrees=1

Do you mean 2sin(2x-30) = 1? (where 2x-30 is in degrees)

yes, that's right