Please help me solve a GCSE equation?

Question

Hey, my GCSE maths exam is this monday and i can't get my head around these equations on my NC Past paper.

a) Solve 3/x + 3/2x = 2

b) Using your answer to part (a), or otherwise, solve

3/ (y-2)^2 +3/2(y-1)^2 = 2

Can someone show me step by step on how to derive x for (a) and y for (b)?? Thanks!

Ah sorry,

for b) it is 3/(y-1)^2 + 3/2(y-1)^2

However for b) it says there are 2 possibilities for what y could be e.g y= or y=

Thank you for replying !!!

Answer 1

a) 3/x + 3/2x = 2
or 6/2x + 3/2x = 2
or (6+3)/2x = 2
or 9 = 4x
or x = 9/4.....(i)

Leave the answer as is.

for b), compare 3/x + 3/2x = 2 to the question given, i.e. 3/(y-2)^2 + 3/2(y-2)^2 = 2

therefore, (y-2)^2 = x

Substituting, the value of x in equation (i),

(y-2)^2 = 9/4
hence, y-2 = ±3/2
or y= ±3/2 + 2

y= 3.5, 0.5

Answer 2

a) Solve 3/x + 3/2x = 2
If you mean 3/x + 3/(2x) = 2 , multiply both sides of the equation by x
3 + 3/2 = 2x
So 9/2 = 2x
Multiply both sides by 1/2
9/4=x

If you mean 3/x + (3/2)x = 2 multiply both sides by x
3 + 3x²/2 = 2x
Add -2x to both sides
3x²/2 + -2x + 3 = 0
Right away I can tell you’re going to have complex roots.
Multiply both sides by 2
3x² + -4x + 6 = 0
Using the quadratic formula…
x= {4 ± √[(-4)² - (4)(3)(6)]}/([2(3)]
={4 ± √[(16) – (72)]}/6
={4 ± √(-56)}/6
={4 ± 2i√14}/6
=2/3 ± (i√14)/3
If this is the case, you’re going to have fun with (b)

b) Using your answer to part (a), or otherwise, solve
3/ (y-2)^2 +3/2(y-1)^2 = 2

Let (y-1)² = whatever result you get for x in the first part of this problem.
(y-1)²=9/4 [This is easy because 9/4 is a perfect square] or
(y-1)²= 2/3 ± (i√14)/3… good luck.

Remember, since you're taking the square root of both sides, you have to consider both the positive square root and the negative square root.

Answer 3

For a) multiply both sides of the equation by 2x.
You get:
6+3 = 4x (for x not equal to 0)
x = 4/9

Now, part b is much harder, since the denominators on the fraction don't have any like terms. I assume that you meant:
3/ (y-2)^2 +3/2(y-2)^2 = 2
Which is the analog to the first equation if x=y-2.
Then you'd get (y-2) = 4/9
y = 22/9

Answer 4

Well first multiply both sides of the first fraction by 2 so (2(3x-2)) - (x + 1) / 4 = 2 we multiply both sides by 4 2(3x-2) - 1(x+1) = 8 Expand the brackets 6x-4 -x -1 = 8 (remember the effect of the minus in front of the second fraction) Simplify 5x-5=8 Solve 5x= 13 x = 13/5

Answer 5

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

make your denominators the same
i.e.

6/2x + 3/2x = 2
(6+3)/2x = 2
9/2x = 2
9/2 = 2x
x = 9/4

b) are you sure you got the question correct?

Answer 6

3/x + 3/2x = 2

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

6 + 3 = 4x

9 = 4x

9/4 = 4x/4

9/4 = x

2 1/4 = x

2.25 = x

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