# Algebraic manipulation of fractions

#### jamesbrown

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

$$\displaystyle \frac3{x-1} - \frac8 {x+2} = 1$$

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#### skeeter

Math Team
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 ...

• 2 people

#### jamesbrown

Thanks man know I fully understand so for the last example is it x=-8 or x=2 ?

#### skeeter

Math Team
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

#### jamesbrown

sub in each solution into the original equation and see if each one works
Yeah it works thanks man

#### deesuwalka

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.

#### Math Message Board tutor

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 $x$.

.

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#### deesuwalka

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

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

$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. 