New Maths GCSE 9-1 Sample Question Confused?!

Anyone know how to solve this step by step

Anyone know how to solve this step by step

Anyone know how to solve this step by step

Thanks for the help!

3(x-1) + 4(x+2) = 12x(5)
3x - 3 + 4x +8 = 60x
5 = 53x
x =5/53

But doesnt it turn into a quadratic?

Where does the "x" go when you do
3(x-1) + 4(x+2)???

You are correct, to get a common denominator you need to times the tops and bottoms by the denominator of the opposing side it should be:

Numerators: 3x(x-1) + 4x(x+2)
Denominator: 4x(3x) + 3x(4x)

So you should get up with (after multiplying out the brackets):

3x²-3x/12x² + 4x²+8/12² = 5

Then, I recommend you add up the fractions, and get rid of the denominator. Then you end up with a quadratic

What should I do after i get:
7x²+5x=60x²

3(x-1) + 4(x+2) = 12x(5)
3x - 3 + 4x +8 = 60x
5 = 53x
x =5/53

Have you learn't how to solve a quadratic? If you haven't, you really need to get on that, perhaps one of the first things really. There are a few ways to solve a quadratic.

This is the format of a quadratic: ax²+bx+c=0 (Where a,b and c are integers, you must rearrange an equation to this format before trying to solve it)

According to the quadratic, it need to be solved in different ways - dependant on the situation:

1) When a=1 (eg. x²-3x-10=0)
2) When a is not = 1 (eg. 2x²-9x+4=0)
3) Completing the square
4) Graphically

Note: 1 and 2 is just a case of factorising, 3 is largely the same thing, but looks a little fancier and usually used in function questions, an A* topic

If you've learn't how to solve quadratics, reply to this comment and I'll go through it with you step by step. If you have, you can go ahead and attempt your original question.

You are correct, to get a common denominator you need to times the tops and bottoms by the denominator of the opposing side it should be:

Numerators: 3x(x-1) + 4x(x+2)
Denominator: 4x(3x) + 3x(4x)

So you should get up with (after multiplying out the brackets):

3x²-3x/12x² + 4x²+8/12² = 5

Then, I recommend you add up the fractions, and get rid of the denominator. Then you end up with a quadratic

No need to form any quadratic's here. That would literally take longer, and one of the solutions wouldn't be valid when you solve it.

Of course! I've totally forgotten the last way! So sorry!

5) The Quadratic Equation, You must learn this off by heart, learn it better than your own name (In my maths teacher words)

Where a,b and c are integers from the original ax²+bx+c=0 format

If you need help explaining, I'm glad to help.

I'm sorry but I think you missed my point?

Why are you explaining quadratics to him when he isn't even required to solve one?

Sorry, but I'm sure in GCSE maths, you'll need to solve a quadratic.

If you're referring to this specific question, perhaps you can give me some insight, I'm always happy to learn! (Even though it's summer break!)

Yes, at GCSE you are required to solve quadratics, but not for this particular question.

Since OP already has the answer spoilt for him, here it is:

$\frac{x-1}{4x}+\frac{x+2}{3x}=5$

$\rightarrow \frac{x-1}{4}+\frac{x+2}{3}=5x$

$\rightarrow 3(x-1)+4(x+2)=5x \cdot 12$

and solve it from there without much fuss of a quadratic.


IF we are going with a quadratic like your's and get to the stage of $53x^2-5x=0$ we can simply divide by x and get $53x-5=0$.

Before anyone jumps and says that we cannot divide by x, well we can. Only because in this particular scenario $x\not= 0$ otherwise the original equation wouldn't make sense due to division by 0.

Anyone know how to solve this step by step

But doesnt it turn into a quadratic?

You need to find the common denominator, which is 12x (response to

Where does the "x" go when you do
3(x-1) + 4(x+2)???