Question

SOLVE
1-2x/3-1/3(1-3x)=-1/3(2+3x)​

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Simplify:

Step-by-step explanation:

$\ Given \ value:\\\\1-\frac{2x}{3}-\frac{1}{3(1-3x)}=-\frac{1}{3(2+3x)} \\\\\ Solution:\\\\1-\frac{2x}{3}-\frac{1}{3(1-3x)}=-\frac{1}{3(2+3x)} \\\\\rightarrow \frac{3(1-3x)-2x(1-3x)-1}{3(1-3x)}=-\frac{1}{3(2+3x)} \\\\\rightarrow \frac{3-9x-2x+6x-1}{3-9x}=-\frac{1}{6+9x} \\\\\rightarrow \frac{2-5x}{3-9x}=-\frac{1}{6+9x} \\\\\rightarrow (2-5x)(6+9x)=-(3-9x) \\\\\rightarrow 12+18x-30x-45x^2= -3+9x \\\\\rightarrow -45x^2+12-12x=-3+9x \\\\\rightarrow -3+9x+45x^2-12+12x=0 \\\\\rightarrow45x^2+21x-15=0$

$\rightarrow 3(15x^2+7x-5)=0 \\\\\rightarrow 15x^2+7x-5=0$

$\ compare \ with \ ax^2+bx+c= 0 \\\\a= 15\\b= 7\\c= -5 \\\ formula: \\\\x= \frac{-b\± \sqrt{b^2-4ac}}{2a} \\\\x= \frac{-(7)\± \sqrt{7^2-4\times 15\times -5}}{2\times 15} \\\\x= \frac{-7\± \sqrt{49+300}}{30} \\\\x= \frac{-7\± \sqrt{349}}{30}$

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