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November 08, 2025, 05:45:51 am
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Author Topic: VCE Methods Question Thread!  (Read 5782208 times)  Share 

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Yertle the Turtle

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Re: VCE Methods Question Thread!
« Reply #15885 on: January 16, 2018, 08:57:46 pm »
0
Can anyone explain this literal equation to me, I have always hated literal equations >:(

1/(x+a) + 1/(x+2a) = 2/(x+3a)

Thanks
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zhen

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Re: VCE Methods Question Thread!
« Reply #15886 on: January 16, 2018, 09:35:14 pm »
+1
Quote from: Marvin K. Mooney on January 16, 2018, 08:57:46 pm
Can anyone explain this literal equation to me, I have always hated literal equations >:(

1/(x+a) + 1/(x+2a) = 2/(x+3a)

Thanks
Generally these equations make you solve for x in terms of a. Basically a is constant so treat it like a normal number and we want to find the value of x in terms of a. So x could be a+1 or something like that.
I would personally start off by multiplying everything out

Then I would expand it out and from that you can solve the equation. I’ll leave the rest for you to do.
« Last Edit: January 16, 2018, 09:40:16 pm by zhen »
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Lear

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Re: VCE Methods Question Thread!
« Reply #15887 on: January 16, 2018, 09:59:21 pm »
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Are literal equations common in vcaa exams? I haven’t seen many
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Bri MT

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Re: VCE Methods Question Thread!
« Reply #15888 on: January 16, 2018, 10:41:53 pm »
+1
Quote from: Lear on January 16, 2018, 09:59:21 pm
Are literal equations common in vcaa exams? I haven’t seen many
What do you mean?
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TheAspiringDoc

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Re: VCE Methods Question Thread!
« Reply #15889 on: January 16, 2018, 10:42:55 pm »
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Quote from: miniturtle on January 16, 2018, 10:41:53 pm
What do you mean?
Quote from: Marvin K. Mooney on January 16, 2018, 08:57:46 pm
Can anyone explain this literal equation to me, I have always hated literal equations >:(

1/(x+a) + 1/(x+2a) = 2/(x+3a)

Thanks
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Bri MT

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Re: VCE Methods Question Thread!
« Reply #15890 on: January 16, 2018, 10:44:50 pm »
+1
Quote from: TheAspiringDoc on January 16, 2018, 10:42:55 pm


No,  these aren't particularly common. 
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Yertle the Turtle

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Re: VCE Methods Question Thread!
« Reply #15891 on: January 16, 2018, 11:27:42 pm »
0
Quote from: miniturtle on January 16, 2018, 10:41:53 pm
What do you mean?
Literal equations are equations where all the coefficients of the main terms (e.g. x,y) are pronumerals
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Sine

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Re: VCE Methods Question Thread!
« Reply #15892 on: January 17, 2018, 01:47:48 am »
+2
Quote from: Lear on January 16, 2018, 09:59:21 pm
Are literal equations common in vcaa exams? I haven’t seen many
As miniturtle said no, they won't come up on exams and from memory they never have come up as a standalone question as far as i can remember. But you will need to know how to manipulate and work with them.
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Phoenix11

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Re: VCE Methods Question Thread!
« Reply #15893 on: January 17, 2018, 03:45:12 pm »
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Hello guys!
How are you all?
I was just here to ask questions that I've had for a long time.
I was wondering if anyone knew how to solve these questions?
1. Solve [a/(x+a)]+[b/(x-b)] for x.

2.Make v the subject of the formula: \[\frac{\text{}x}{\text{}y}=\sqrt{1-\frac{\text{}v^{2}}{\text{}u^{2}}}\]

Thank you in advance!
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zhen

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Re: VCE Methods Question Thread!
« Reply #15894 on: January 17, 2018, 04:10:23 pm »
+2
Quote from: Phoenix11 on January 17, 2018, 03:45:12 pm
Hello guys!
How are you all?
I was just here to ask questions that I've had for a long time.
I was wondering if anyone knew how to solve these questions?
1. Solve [a/(x+a)]+[b/(x-b)] for x.

2.Make v the subject of the formula: \[\frac{\text{}x}{\text{}y}=\sqrt{1-\frac{\text{}v^{2}}{\text{}u^{2}}}\]

Thank you in advance!
The first one isn’t an equation, so you can’t solve it. Did you mean those things =0?

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skrt skrt

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Re: VCE Methods Question Thread!
« Reply #15895 on: January 17, 2018, 05:33:47 pm »
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Hey guys, just need help with this question.
Find the values of a for which the equation (a-3)x^2 + 2ax + (a+2)=0 has no solutions for x.
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chooby

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Re: VCE Methods Question Thread!
« Reply #15896 on: January 17, 2018, 05:46:34 pm »
+3
Quote from: skrt skrt on January 17, 2018, 05:33:47 pm
Hey guys, just need help with this question.
Find the values of a for which the equation (a-3)x^2 + 2ax + (a+2)=0 has no solutions for x.
You would use discriminant here. When x has no solutions, the discriminant is less than 0: Now you would sub in the values for a,b and c from your quadratic. . Then you just solve for a. Hope this helps :D
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Phoenix11

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Re: VCE Methods Question Thread!
« Reply #15897 on: January 18, 2018, 10:21:19 am »
0
Quote from: zhen on January 17, 2018, 04:10:23 pm
The first one isn’t an equation, so you can’t solve it. Did you mean those things =0?



Hi Zhen.
Thank you.
Your answer seems right to me but the answer says \[v=\pm\sqrt{u^2(1-\frac{x^{2}}{y^{2}}})\] for some reason?

Also the first question does equal 0.
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RuiAce

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Re: VCE Methods Question Thread!
« Reply #15898 on: January 18, 2018, 10:23:25 am »
+3
Quote from: Phoenix11 on January 18, 2018, 10:21:19 am
Hi Zhen.
Thank you.
Your answer seems right to me but the answer says \[v=\pm\sqrt{u^2(1-\frac{x^{2}}{y^{2}}})\] for some reason?

Also the first question does equal 0.

Recall: \( \sqrt{ab} = \sqrt{a}\sqrt{b} \)
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Phoenix11

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Re: VCE Methods Question Thread!
« Reply #15899 on: January 18, 2018, 10:27:08 am »
0
Thank you RuiAce!
So both answers are correct?
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