Algebraic manipulation of fractions

Jan 2015
108
2
London
Hi, could I have some help with these questions.
\(\displaystyle 1+\frac x{2} = \frac {x-3}{4}\)
do I times by the lowest common denominator, so 4?
but the answer is \(\displaystyle -7\) and I don't get how you get to that?

Also these questions:
\(\displaystyle x+1 = \frac2{x}\)

and

\(\displaystyle \frac3{x-1} - \frac8 {x+2} = 1\)
 
Last edited:

skeeter

Math Team
Jul 2011
3,363
1,854
Texas
Hi, could I have some help with these questions.
\(\displaystyle 1+\frac x{2} = \frac {x-3}{4}\)
do I times by the lowest common denominator, so 4?
multiply every term by 4 ...

$4 + 2x = x-3$

now solve for x

\(\displaystyle x+1 = \frac2{x}\)
multiply every term by x ...

\(\displaystyle x^2+x = 2\)

solve the quadratic for x ...

\(\displaystyle \frac3{x-1} - \frac8 {x+2} = 1\)
multiply every term by $(x-1)(x+2)$ ...

$3(x+2) - 8(x-1) = (x-1)(x+2)$

expand & solve the resulting quadratic ...
 
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Jan 2015
108
2
London
Thanks man know I fully understand so for the last example is it x=-8 or x=2 ?
 

skeeter

Math Team
Jul 2011
3,363
1,854
Texas
Thanks man know I fully understand so for the last example is it x=-8 or x=2 ?
sub in each solution into the original equation and see if each one works
 
Sep 2016
88
30
India
For the first question,

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

$4+2x=x-3$

$2x-x=-3-4\;\implies x= -7$

for the second equation,

$x+1=\dfrac{2}{x}$

$x^2+x=2$

write this as-

$x^2+x-2=0$

$x^2+2x-x-2=0$

$x(x+2)-1(x+2)$

$(x-1)(x+2)$

$x= 1, -2$

For third equation-

$\dfrac{3}{x-1}-\dfrac{8}{x+2}=1$

$\dfrac{3(x+2)-8(x-1)}{(x-1)(x+2)}=1$

$3x+6-8x+8=x^2+2x-x-2$

$-5x+14=x^2+x-2$

$x^2+x+5x+14-2=0$

$x^2+6x+12=0$

Similarly solve it for $x$

I hope it' ll help.
 
Jun 2014
945
191
Earth
for the second equation,

$x+1=\dfrac{2}{x} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $Note that x cannot equal 0.

$x^2+x=2$

write this as-

$x^2+x-2=0$

$x^2+2x-x-2=0$

$x(x+2)-1(x+2) \ \ \ \ \ \ \ $This is still supposed to be an equation.


$(x-1)(x+2) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $And so is this.

$x= 1, -2$

$x(x+2) \ - \ 1(x+2) \ = \ 0$

$(x-1)(x+2) \ = \ 0$


$x \ = \ 1, \ -2 \ \ \ \ \ \ \ \ \ \ \ $ Compare these to the restricted value.

For third equation-

$\dfrac{3}{x-1}-\dfrac{8}{x+2}=1$

$\dfrac{3(x+2)-8(x-1)}{(x-1)(x+2)}=1$

$3x+6-8x+8=x^2+2x-x-2$

$-5x+14=x^2+x-2$

$x^2+x+5x+14-2=0 \ \ \ \ \ \ \ \ \ \ \ \ \ \ $ This step is incorrect.

$x^2+6x+12=0 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $Then so is this step.

Similarly solve it for $x$
$-5x+14 \ = \ x^2+x-2$

$x^2+x+5x-14-2 \ = \ 0 $

$x^2+6x-16 \ = \ 0$

Similarly solve it for $x$.


.
 
Last edited:
Sep 2016
88
30
India
$x(x+2) \ - \ 1(x+2) \ = \ 0$

$(x-1)(x+2) \ = \ 0$


$x \ = \ 1, \ -2 \ \ \ \ \ \ \ \ \ \ \ $ Compare these to the restricted value.



$-5x+14 \ = \ x^2+x-2$

$x^2+x+5x-14-2 \ = \ 0 $

$x^2+6x-16 \ = \ 0$

Similarly solve it for $x$.


.
Yes! I know in second equation I have mistakenly typed.
and thanks for pointing out mistake in third step.:)