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.