Please help!O.L maths problem:

FunkyOxygen

Okay this problem is coming from a 2011 mock (O.L) don't know how to tell which company but it just has a small barcode in the top right corner with Y 5 underneath? Paper 1.

Basically:
Q.5 c (i) is "express 3x+11(OVER)5 - x+1(OVER)2 as a single fraction.

Done that bit^ got x+17(OVER)10

However part ii)
Verify your answer to part i by substituting x=3 into the sum ^^, and into your answer to part (i) My answer comes out weird and doesen't seem right

Then how do i do part (iii) "Solve the equation x(SQUARED)-3x-15=0

?


I would REALLY appreciate help as my maths teacher has gone off with transition year and I'm really stuck! Thank you to anyone who can help even with only one bit of the 2 questions above!

ciara95

I'm no maths genius but I tried the Q and this is what i got:
(i) same as you: 1x + 17((over)) 10

(ii) 1(3)+17 ((over)) 10
=3+17 ((over)) 10
= 20 ((over)) 10
= 2

(iii) quadratic equation so use the formula
-b ((PLUS OR MINUS)) the square root of b(Squared) -4ac

And for that I got x=3(PLUS OR MINUS) square root of 69 (OVER) 2

Hope this helps, sorry if it's wrong!
I don't know how to do all the Mathsy signs lol!

ciara95

And for part (iii) go on to solve the answer I wrote to get the two roots (put them equal to zero and solve)

electoralshock

FunkyOxygen wrote: »

Okay this problem is coming from a 2011 mock (O.L) don't know how to tell which company but it just has a small barcode in the top right corner with Y 5 underneath? Paper 1.

Basically:
Q.5 c (i) is "express 3x+11(OVER)5 - x+1(OVER)2 as a single fraction.

Done that bit^ got x+17(OVER)10

However part ii)
Verify your answer to part i by substituting x=3 into the sum ^^, and into your answer to part (i) My answer comes out weird and doesen't seem right

Then how do i do part (iii) "Solve the equation x(SQUARED)-3x-15=0

?


I would REALLY appreciate help as my maths teacher has gone off with transition year and I'm really stuck! Thank you to anyone who can help even with only one bit of the 2 questions above!

Hi,
from part (i) you got x + 17 (over) 10

from there let x=3 so substitute the x with a 3.

it becomes 3 + 17 (over) 10

which is 20 (over) 10
which is 2.

Part (iii) X (SQUARED) -3X -15 = 0
this can be broken down:

(X + 3) multplied by (X -5) = 0

X +3 = 0 OR X -5 =0

THEREFORE:
X= -3 or x= 5


(your text should have examples on how to this! hope it helps! gud luck!!)

MathsManiac

Both Ciara95 and ElectoralShock have given you one half of part (ii), but it's not finished.

The point is that the expression you got in part (i) is supposed to be identical to the expression given in the question, (i.e., the two fractions). That means that, no matter what the value of x is, if you substitute it into the original expression and into your simplified one, you should get the same answer.

So, as well as putting x=3 into your answer to part (ii), you also have to put it into the original expression (3x+11)/5 - (x+1)/2, and show that you end with the same answer either way, which you do.

mathsmate

electoralshock wrote: »

Hi,
from part (i) you got x + 17 (over) 10

from there let x=3 so substitute the x with a 3.

it becomes 3 + 17 (over) 10

which is 20 (over) 10
which is 2.

Part (iii) X (SQUARED) -3X -15 = 0
this can be broken down:

(X + 3) multplied by (X -5) = 0

X +3 = 0 OR X -5 =0

THEREFORE:
X= -3 or x= 5


(your text should have examples on how to this! hope it helps! gud luck!!)

Part (3) is incorrect. (X + 3) multplied by (X -5) = 0
x squared-3x-15
You would need to use the quadratic formula to solve this equation.
as -5x +3x does not equal -3x
When solving these equations the factors of -15 must add to give the middle term ie -3x if not the quadratic formula must be used to solve for x

jhayden

Hi there,
Its been a long since I did Junior cert maths and I think there is a problem in the first part of the equation which means the second part is wrong. Here is how I would go about solving the equation.

Part (i)
((3x+11)/5)-((x-1))/2=y
10((3x+11)/5)-10((x-1))/2=10y
2((3x+11))-5((x-1))/2=y
(6x+22)-(5x-5)=10y
6x+22-5x+5=10y
X+27=10y
(X+27)/10=y

Part (ii)
((3x+11)/5)-((x-1))/2=y
=>((3(3)+11)/5)-((x-1))/2=(X+27)/10
=>(20/5)-(2/2) = (30/10)
=>(4)-(1) = (3)
=>3 = 3

Part (iii)
x2-3x-15
(-b+-(√(b)2 -4ac)) / (2a)

A= 1
B=-3
C=-15

(--3+-(√(-3)2 -4(1)(-15))) / 2(1)
((3+-(√9+60) / 2(1)))
((3+-(√69) / 2(1)))


=> (3+ (√69))/2
=> (3 - (√69)) /2

Hope this helps. Good luck in your exams.

OBrienJoey

put 3 in for 3(3)+11/5 - (3)+1/2 to get 20/5 - 4/2 which = 2

then put it in for your answer from (i) (3)+17/10 which equals 2 aswell so your answer is right

jhayden

OBrienJoey wrote: »

put 3 in for 3(3)+11/5 - (3)+1/2 to get 20/5 - 4/2 which = 2

then put it in for your answer from (i) (3)+17/10 which equals 2 aswell so your answer is right

Sorry I see where I went wrong misread the question. Lucky I don't have to do the exam again. I put x-1 in brackets instead of x+1.