1968 JMB GCE O-level question

Sean in France

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I took my old Maths GCE O-level one year early in 1973. I'm quite interested in the debate about educational standards and whether they have slipped over the decades. I found this question on an old 1968 JMB paper, and I haven't found a similar question in any modern GCSE Higher Tier paper. I must admit, the question has me flummoxed. I've tried answering it, but with no success. Here is the question:

Find the value of x and the value of y which satisfy the equations
\(\displaystyle 2x - y = \frac{3}{7}(2x+y) = 6x-7y + 4\)

I'd be grateful to anyone who could demonstrate a model solution to the question. Many thanks in advance.
 

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lev888

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I took my old Maths GCE O-level one year early in 1973. I'm quite interested in the debate about educational standards and whether they have slipped over the decades. I found this question on an old 1968 JMB paper, and I haven't found a similar question in any modern GCSE Higher Tier paper. I must admit, the question has me flummoxed. I've tried answering it, but with no success. Here is the question:

Find the value of x and the value of y which satisfy the equations
\(\displaystyle 2x - y = \frac{3}{7}(2x+y) = 6x-7y + 4\)

I'd be grateful to anyone who could demonstrate a model solution to the question. Many thanks in advance.
Start by simplifying.
 

HallsofIvy

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Jan 27, 2012
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That looks like a pretty basic "High School Algebra" question. First, of course, it is two equations:
\(\displaystyle 2x- y= \frac{3}{7}(2x+ y)\) and \(\displaystyle \frac{3}{7}(2x+ y)= 6x-7y+ 4\);
(The third equation, \(\displaystyle 2x- y= 6x- 7y+ 4\), is implied but gives no new information.)

The first thing I would do is simplify both equations. Multiply both sides of the first equation by 7:
7(2x- y)= 14x- 7y= 3(2x+ y)= 6x+ 3y
14x- 7y= 6x+ 3y
Subtract 6x from both sides and add 7y to both sides:
8x= 10y. Divide both sides by 2: 4x= 5y.
That's about as simple as you can get it.

Multiply both sides of the second equation by 7: 3(2x+ y)= 6x+ 3y= 7(6x- 7y+ 4)= 42x- 49y+ 28
6x+ 3y= 42x- 49y+ 28
Subtract 6x from both sides and add 49y to both sides:
52y= 36x+ 28
Divide both sides by 4: 13y= 9x+ 7
That's about as simple as you can get that.

We can divide both sides of 4x= 5y by 4 to get \(\displaystyle x= \frac{5}{4}y\).
Replace x by that in 13y= 9x+7: \(\displaystyle 13y= \frac{45}{4}y+ 7\)
so that we have a single equation in the single unknown, y.

Subtract \(\displaystyle \frac{45}{4}y\) from both sides. \(\displaystyle 13= 13\frac{4}{4}= \frac{52}{4}\) so
\(\displaystyle 13y- \frac{45}{4}y= \left(\frac{52}{4}- \frac{45}{4}\right)y= \frac{7}{4}y=7\).

Multiply both sides by \(\displaystyle \frac{4}{7}\): \(\displaystyle y= 4\).

Since \(\displaystyle x= \frac{5}{4}y\) from before, \(\displaystyle x= \frac{5}{4}(4)= 5\).
x= 5, y= 4.

And, of course, check those answers in the original equations,
\(\displaystyle 2x- y= \frac{3}{7}(2x+ y)= 6x-7y+ 4\).

If x= 5, y= 4 then
\(\displaystyle 2x- y= 2(5)- 4= 10- 4= 6\)
\(\displaystyle \frac{3}{7}(2x+ y)= \frac{3}{7}(2(5)+ 4)= \frac{3}{7}(14)= 6\)
\(\displaystyle 6x- 7y+ 4= 6(5)- 7(4)+ 4= 30- 28+ 4= 2+ 4= 6\).

Yes, those are all equal so x= 5, y= 4 satisfy the equation.
 

Sean in France

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Thank you very much. I'm impressed.

You say that it looks like basic "high school" algebra. Your use of the expression "high school" tells me you are American and when I looked again at the name of the site it says "Math" help, not "Maths" help, which seems to confirm this is an American forum. I don't know if you know, but GCE O-level was the old version of GCSE, which are UK examinations for 16-year-olds. Would you say that an American 16-year-old would be asked a question like this in a maths exam, or in a maths test (I'm not sure that there are formal exams for 16-year-olds in the US)?
 

HallsofIvy

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Actually, I would expect a question like this to be on a "freshman algebra" secondary school test which would mean that the students would be about 14 years old.
 

yoscar04

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Thank you very much. I'm impressed.

You say that it looks like basic "high school" algebra. Your use of the expression "high school" tells me you are American and when I looked again at the name of the site it says "Math" help, not "Maths" help, which seems to confirm this is an American forum. I don't know if you know, but GCE O-level was the old version of GCSE, which are UK examinations for 16-year-olds. Would you say that an American 16-year-old would be asked a question like this in a maths exam, or in a maths test (I'm not sure that there are formal exams for 16-year-olds in the US)?
I studied this kind of problem during 8th and 9th year (South America). In Israel it is standard material for 10th grade (upper level) math .
 
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