help plse:
(4a-3) = a+33/a+1
solution is

Question

anonymous · Answer

help plse:
(4a-3) = a+33/a+1
solution is

zarkon · Answer

is the equation
$$4a-3=\frac{a+33}{a+1}$$?

anonymous · Answer

yes where 4a-3 in parenthese

zarkon · Answer

multiply by a+1

$$(4a-3)(a+1)=a+33$$

multiply out

$$4a^2+4a-3a-3=a+33$$

$$4a^2+a-3=a+33$$



$$4a^2-3=33$$

$$4a^2=36$$

$$a^2=9$$


$$a=\pm3$$

anonymous · Answer

so the answer would be 3

zarkon · Answer

3 and -3

anonymous · Answer

Zarkon can ou help with this one
2-a-4/a+3 = a^2-2/a+3

zarkon · Answer

you need to work on formatting your questions.  The way you type them is very ambiguous.

is this your equation either of these?

$$2-a-\frac{4}{a+3}=\frac{a^2-2}{a+3}$$

$$2-\frac{a-4}{a+3}=\frac{a^2-2}{a+3}$$

anonymous · Answer

it is the 2nd one I'm sorry for the formatting I do not know how to use the equation format too goo so I did the best I could sorry for the confusion.l

zarkon · Answer

ok

$$2-\frac{a-4}{a+3}=\frac{a^2-2}{a+3}$$

$$2=\frac{a^2-2}{a+3}+\frac{a-4}{a+3}$$

$$2=\frac{a^2-2+a-4}{a+3}$$

$$2(a+3)=a^2-2+a-4$$

$$2a+6=a^2-2+a-4$$

$$0=a^2-2+a-4-2a-6$$

$$0=a^2-a-12$$

$$0=(a-4)(a+3)$$

so a=4

-3 is not a solution since it does not satisfy the original equation (can't divide by 0)