Sec 2 Maths Problem

kkhing SG

Please help to solve these 2 Sec 2 Maths problem.

1) Given that 3a + e = 7 and 4ae/3 = 2, find the value of e-3a.

2) Given that m^2 + 8m = v^2 + 8v and that m is not equal to v, find the value of 5(m + v).

Thanks.

Darkness_hacker99 SG

wee_ws SG

Hi,

What is the difficulty you are facing please? Thanks.

Cheers,
Wen Shih

theawesomejo SG

Not sure about this , maybe you can work it out ?

1) Use simultaneous eqn method, then get the answer and then get the ans.

Sakuragi86 SG

Not sure about this , maybe you can work it out ?

1) Use simultaneous eqn method, then get the answer and then get the ans.

simultaneous eqn method cannot get any answer.

wee_ws SG

Hi,

It seems to me that Q1 is not a reasonable question for a Sec 2 student. It is not worth solving. Thanks.

Cheers,
Wen Shih

wee_ws SG

Hi,

Let e - 3a = k, then 3a = e - k -- (1).

Given 3a + e = 7, we obtain

e - k + e = 7, because of (1)

=> 2e - k = 7

=> e = (k + 7)/2 -- (2).

Given 4ae/3 = 2, we obtain ae = 3/2

=> 3ae = 9/2

=> (e - k)e = 9/2, because of (1)

=> [ (k + 7)/2 - k ] (k + 7)/2 = 9/2, because of (2)

=> (7 - k)(7 + k) / 4 = 9/2

=> 49 - k^2 = 18

=> k = +/- sqrt(31).

The working is a little demanding at Sec 2 level, unless there is really a more elegant method that I have missed out?

Thanks.

Cheers,
Wen Shih

Chemfreak022 SG

3a + e = 7

(3a + e)^2 = 49

(3a + e)^2 = 9a^2 + 6ae + e^2

4ae/3 = 2 ..... 2ae = 3 .... 6 ae = 9

(e-3 a)^2 = e^2 - 6ae + 9a^2= 9a^2 + 6ae + e^2 - 12ae = 49 -18 = 31

Therefore (e-3a) = sqrt(31)

Chemfreak022 SG

I think mine is shorter, although i forgot the -ve root

wee_ws SG

Hi Chemfreak022,

Thanks for sharing an elegant solution :)

Students can rest assure that Cambridge will provide the suggestion to lead them to think in that direction. It may be phrased like this:

It is given that e + 3a = 7 and ae = 3/2.

Express (e - 3a)^2 in terms of (e + 3a) and ae. Hence find, exactly, the possible values of (e - 3a).

Thanks.

Cheers,
Wen Shih

wee_ws SG

Hi,

As a result of Q1 and the discussion that followed, it is worthwhile for the keen student to take note of the relationships between (a + b)^2 and (a - b)^2:

1. (a - b)^2 = (a + b)^2 - 4ab,

2. (a + b)^2 = (a - b)^2 + 4ab.

Thanks.

Cheers,
Wen Shih

syncopation_music SG

agree with you .Master Wenshih , Cambridge do not wish to slap themselves by writing ambiguous question in their papers.

I read through some Examiner Report of IGCSE .They shared some insight in the type of question which will come out .

What i worried is Cambridge could set english-based Mathematics question

for example :subtract 3x-4 from x -2

common folks will think that the working is (3x-4) - (x-2) = 2x+2

however , the interpretation is wrong

its (x-2) - (3x-4)

which gives -2x +2

wee_ws SG

Hi syncopation_music,

Thanks for the respectful address.

We are learning from one another, so call me Wen Shih :)

Cheers,
Wen Shih