# 1968 JMB GCE O-level question

#### Sean in France

##### New member
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

##### Senior Member
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

##### Elite Member
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

##### New member
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

##### Elite Member
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

##### Full Member
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 .