Maths help-equations
What is the answer to this? Can you show working please?!
1/x-2 + 3/x+6 =1/2
1/x-2 + 3/x+6 =1/2
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Original post by isabel12144
What is the answer to this? Can you show working please?!
1/x-2 + 3/x+6 =1/2
1/x-2 + 3/x+6 =1/2
You should remove ambiguity by using brackets, so it would be, I'm assuming 1/(x-2) + 3(x+6) = 1/2, right, that is:
.
Multiply throughout by and then expand brackets and solve the remaining quadratic:
- multiply both sides by two, re-arrange, etc... and then solve the quadratic.
Original post by isabel12144
What is the answer to this? Can you show working please?!
1/x-2 + 3/x+6 =1/2
1/x-2 + 3/x+6 =1/2
What do you think you should do? (hint: you've got some fractions there ...)
Original post by SeanFM
What do you think you should do? (hint: you've got some fractions there ...)
I did this:
x+6/(x-2)(x+6) + 3(x-2)/(x-2)(x+6)
x+6+3x-6/(x-2)(x+6) = 4x/(x-2)(x+6)
4x/(x-2)(x+6)=1/2
Then I didn't know what to do.
Original post by isabel12144
I did this:
x+6/(x-2)(x+6) + 3(x-2)/(x-2)(x+6)
x+6+3x-6/(x-2)(x+6) = 4x/(x-2)(x+6)
4x/(x-2)(x+6)=1/2
Then I didn't know what to do.
x+6/(x-2)(x+6) + 3(x-2)/(x-2)(x+6)
x+6+3x-6/(x-2)(x+6) = 4x/(x-2)(x+6)
4x/(x-2)(x+6)=1/2
Then I didn't know what to do.
Good work
Now it may help to 'get rid' of the fractions on each side so you can form a quadratic equation.. any ideas?
Original post by SeanFM
Good work
Now it may help to 'get rid' of the fractions on each side so you can form a quadratic equation.. any ideas?
Now it may help to 'get rid' of the fractions on each side so you can form a quadratic equation.. any ideas?
Do you do?
4x=1/2*(x-2)(x-6)
Original post by isabel12144
Do you do?
4x=1/2*(x-2)(x-6)
4x=1/2*(x-2)(x-6)
Careful with how you write things
but yes, as long as that's that's fine, and you can also 'get rid' of the 1/2 on the right hand side...What next?
Original post by SeanFM
Careful with how you write things but yes, as long as that's that's fine, and you can also 'get rid' of the 1/2 on the right hand side...
What next?
but yes, as long as that's that's fine, and you can also 'get rid' of the 1/2 on the right hand side...What next?
8x=(x-2)(x+6)
?
Original post by isabel12144
8x=(x-2)(x+6)
?
?
Correct
now what can you do?
Original post by isabel12144
8x=(x-2)(x+6)
?
?
Keep going...
Hint (only if you need it)
Spoiler
Original post by SeanFM
Correct now what can you do?
now what can you do?8x=x^2+4x-12
Original post by isabel12144
8x=x^2+4x-12
Yep, so it almost looks like a quadratic that you can solve. What's next?
Original post by SeanFM
Yep, so it almost looks like a quadratic that you can solve. What's next?
8x+12=x^2+4x
12=x^2-4x
Original post by isabel12144
8x+12=x^2+4x
12=x^2-4x
12=x^2-4x
If you rearrange the equation so it equals 0, it might be easier to solve
Original post by Andy98
If you rearrange the equation so it equals 0, it might be easier to solve
is it this then?
x^2-4x-12=0
(x-6)(x+2)=0
x=-6 or 2
Original post by isabel12144
is it this then?
x^2-4x-12=0
(x-6)(x+2)=0
x=-6 or 2
x^2-4x-12=0
(x-6)(x+2)=0
x=-6 or 2
You got the factorisation right, but the solutions incorrect
Original post by isabel12144
is it this then?
x^2-4x-12=0
(x-6)(x+2)=0
x=-6 or 2
x^2-4x-12=0
(x-6)(x+2)=0
x=-6 or 2
Almost - remember, you have to work out:
x-6=0 and x+2=0
Original post by isabel12144
is it this then?
x^2-4x-12=0
(x-6)(x+2)=0
x=-6 or 2
x^2-4x-12=0
(x-6)(x+2)=0
x=-6 or 2
for (x-6)(x+2)=0
either x-6=0 or x+2=0
if you rearrange these you should get the correct values for x
Original post by Andy98
You got the factorisation right, but the solutions incorrect
x=-2 or 6
Original post by isabel12144
x=-2 or 6
correct.
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