Lost my basics

[iryshsteam]

Working on a question for some reason cannot get my head around it.
Any help is appreciated.
original question:
(3x^2+10x+3)/(x+3)=(x^2-4x+3)/(x-1) SOLVE FOR X

STEP 1 X-1(3X^2+10X+3) = X+3(X^2-4X+3)

STEP 2 X-3X^2-10X-3 = X+3X^2-12X+9

STEP 3 X-10X-3X^2-3 = X-12X+3X^2+9

STEP 4 -9X-3X^2-3+3 = -11X+3X^2+9+3

STEP 5 -9X-3X^2 = -11X+3X^2+12

STEP 6 -9X-3X^2+11X = -11X+11X+3X^2+12

STEP 7 2X-3X^2 = 3X^2+12


I know that the answer is x=-2 but I am lost
please rescue an old man thanks

 

[Deleted member 4993]

Working on a question for some reason cannot get my head around it.
Any help is appreciated.
original question:

(3x^2+10x+3)/(x+3)=(x^2-4x+3)/(x-1) SOLVE FOR X

(x+3)(3x+1)/(x+3) = (x-3)(x-1)/(x-1)

3x + 1 = x - 3

2x = -4

x = -2

STEP 1 X-1(3X^2+10X+3) = X+3(X^2-4X+3)

STEP 2 X-3X^2-10X-3 = X+3X^2-12X+9

STEP 3 X-10X-3X^2-3 = X-12X+3X^2+9

STEP 4 -9X-3X^2-3+3 = -11X+3X^2+9+3

STEP 5 -9X-3X^2 = -11X+3X^2+12

STEP 6 -9X-3X^2+11X = -11X+11X+3X^2+12

STEP 7 2X-3X^2 = 3X^2+12


I know that the answer is x=-2 but I am lost
please rescue an old man thanks

 

[iryshsteam]

Thank you Very much

 

[soroban]

Hello, iryshsteam!

Subhotosh had the best approach.

Your method should have worked, but you multiplied incorrectly.

\(\displaystyle \text{Solve for }x\!:\quad \frac{3x^2+10x+3}{x+3} \;=\; \frac{x^2-4x+3}{x-1}\)

\(\displaystyle \text{We have: }\;(x-1)(3x^2 + 10x + 3) \;=\;(x+3)(x^2-4x+3)\)


\(\displaystyle \text{Multiply: }\;3x^2 + 10x^2 + 3x - 3x^2 - 10x - 3 \;=\;x^3 - 4x^2 + 3x + 3x^2 - 12x + 9\)

. . \(\displaystyle \text{and we have: }\;2x^3 + 8x^2 + 2x - 12 \:=\:0\)


\(\displaystyle \text{Factor: }\;2(x-1)(x+2)(x+3) \;=\;0\)

. . \(\displaystyle \text{and we have: }\;x \:=\:1,\:-2,\:-3\)


\(\displaystyle \text{But when we check }x = 1\text{ or }x = -3\text{, we get the indeterminate form: }\tfrac{0}{0}\)


\(\displaystyle \text{Therefore, the only solution is: }\:x = -2\)