solve the equation Watch

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thewebmaster
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#21
Report Thread starter 13 years ago
#21
I WILL NOT GIVE UP ON THIS EQUATION! STICK WITH ME HERE..

x²+x-2x²+3x-1+2x = 4
TAKE -1 across just gets in way

x²+x-2x²+3x+2x = 5
FACTORISE
x(x + 1 - 2x + 3 + 2) = 5
SIMPLY ADDITION OF NUMBERS IN BRACKETS
x(-x + 1 + 2 + 3) DOES NOT EQUAL 5!!
x(-x + 6) = 5
TAKE ACROSS THE 6
x(-x) = 5 - 6
MULTIPLY BY -1
x(x) = -5 +6
x(x) = +1

WE AREN'T FINISHED YET PEOPLE....
SUBSTITUTE IN ORIGINAL EQUATION

2 * 1 + y = 1
2 * 1 = 2 + y = 1
therefore y = 1 -2
y = -1

AM I RIGHT? DID I TWIST THE RULES,
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DaveManUK
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#22
Report 13 years ago
#22
x(-x + 6) = 5
TAKE ACROSS THE 6 <<<------------------- HERE. WHAT THE HELL?!

expand the brackets first!
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thewebmaster
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#23
Report Thread starter 13 years ago
#23
you finish up with x^2 - 6x + 5 = 0
HOW?

x²+x(1-2x)+3x-(1-2x) -->

x²+x-2x+3x-1-2=4

can't see how you get to the quadratic above with that multiplied out?
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thewebmaster
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#24
Report Thread starter 13 years ago
#24
x(-x + 6) = 5

expand brackets

-x² + 6x = 5

take across 5 then
-x² + 6x -5 = 0

multiply by -1
x² -6x +5 = 0
factorise to give
(x-1)(x-5)
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ciara
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#25
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#25
I don't understand why you did that..
(Original post by thewebmaster)
I WILL NOT GIVE UP ON THIS EQUATION! STICK WITH ME HERE..

x²+x-2x²+3x-1+2x = 4
TAKE -1 across just gets in way

x²+x-2x²+3x+2x = 5
FACTORISE
x(x + 1 - 2x + 3 + 2) = 5
SIMPLY ADDITION OF NUMBERS IN BRACKETS
x(-x + 1 + 2 + 3) DOES NOT EQUAL 5!!
x(-x + 6) = 5
TAKE ACROSS THE 6 - you can't do this. the 6 is in brackets which need to be mulitiplied out before you can move it.
x(-x) = 5 - 6
MULTIPLY BY -1
x(x) = -5 +6
x(x) = +1

WE AREN'T FINISHED YET PEOPLE....
SUBSTITUTE IN ORIGINAL EQUATION

2 * 1 + y = 1
2 * 1 = 2 + y = 1
therefore y = 1 -2
y = -1

AM I RIGHT? DID I TWIST THE RULES,
or alternatively..
x²+x-2x²+3x-1+2x = 4
-x²+6x-1 = 4
x²-6x+5 = 0
(x-5)(x-1) = 0
x=5 x=1

and so on...

Cxx
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thewebmaster
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#26
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#26
x(-x + 6) = 5
TAKE ACROSS THE 6 <<<------------------- HERE. WHAT THE HELL?!
Ok so up until then i was doing quiet well? average? good 4 me though? lol

ok so i can't and am not allowed to say that
x(-x+6) = 5

x= 5
x+6 = 5 therefore x = 5-6 x = -1
??

But if it were

x(-x+6) = 0

I could say that x = 0 and x = +6?
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DaveManUK
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#27
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#27
(Original post by thewebmaster)
Ok so up until then i was doing quiet well? average? good 4 me though? lol

ok so i can't and am not allowed to say that
x(-x+6) = 5

x= 5
x+6 = 5 therefore x = 5-6 x = -1
??
do you know how to expand brackets?

But if it were

x(-x+6) = 0

I could say that x = 0 and x = +6?
yes.
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DaveManUK
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#28
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#28
also this is a pointless step:
FACTORISE
x(x + 1 - 2x + 3 + 2) = 5
you take x out, only in a later step to multiply back the brackets, wrongly.
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thewebmaster
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#29
Report Thread starter 13 years ago
#29
i realised that you didn't factorise to leave x(something here)
it was a quadratic so i could have done (x blah) (x blah)

I neglect factorising quadratics and must start doing it more often when its possible instead of trying to avoid it like i did before.

multiply by -1
x² -6x +5 = 0
factorise to give
(x-1)(x-5)

that was the bit that got me muddled up, i do not like negative constants in from of the x² term. got to pay more attention to multiplying by (- the constant) to get rid of it.
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thewebmaster
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#30
Report Thread starter 13 years ago
#30
slightly harder

x² + 2xy = 3
3x² - y² = 26

solve for x and y both values.

let y = (3-x²)/2x

3x²-(3-x²/2x)²=26

3x²-((3-x)(3-x)/(2x)(2x)) = 26

3x²- (9 + 3x - 3x + x²)/4x² = 26

12x^4 - 9 + 3x - 3x + x² = 26 + 4x²

STUCK NOW!

12^4 - 9 +x² = 26 + 4x²
12^4 +x² = 35 + 4x²

FROM HERE GUESSING
12^4 + x² - 4x² = 35
12^4 + x² - 4x² - 35 = 0
??
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thewebmaster
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#31
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#31
11y² - 13y + 2 = 0
Factorise for

(y-1)(11y-2)

y = -1
y = 2/11

substitute into... x+2y = 3

x + (-1 *2) = 3
x - 2 = 3
x = 3-2
x = 1

(1,-1)

x + (2/11 *2) = 3
x + 4/11 = 3
x = 3 - 4/11
x = 7/11

?
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Chewwy
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#32
Report 13 years ago
#32
plotting a graph tells me you're wrong. give me a minute.
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