Question

I WILL AWARD BRAINLIEST!! PLEASE HELP!!

Answer 1

Answer:

$1)\\\large\boxed{x=\dfrac{a+c}{b},\ if\ b\neq0}\\\\2)\\\boxed{if\ a=\dfrac{63b}{23}\ then\ x\ is\ any\ real \ number}\\\\\boxed{if\ a\neq\dfrac{63b}{23}\ then\ x\ not\ exist\ (no\ solution)}$

Step-by-step explanation:

$1)\\a-(a+b)x=(b-a)x-(c+bx)\qquad\text{use the distributive property}\ a(b+c)=ab+ac\\\\a-ax-bx=bx-ax-c-bx\qquad\text{cancel}\ bx\ \text{on the right side}\\\\a-ax-bx=-ax-c\qquad\text{add}\ ax\ \text{to both sides}\\\\a-bx=-c\qquad\text{subtract}\ a\ \text{from both sides}\\\\-bx=-a-c\qquad\text{divide both sides by}\ -b\neq0\\\\x=\dfrac{-a-c}{-b}\\\\x=\dfrac{a+c}{b}$

$2)\\2(3x-5a)+9(2a-7b)+3(5a-2x)=0\\\\\text{use the distributive property}\ a(b+c)=ab+ac\\\\(2)(3x)+(2)(-5a)+(9)(2a)+(9)(-7b)+(3)(5a)+(3)(-2x)=0\\\\6x-10a+18a-63b+15a-6x=0\qquad\text{combine like terms}\\\\(6x-6x)+(-10a+18a+15a)-63b=0\\\\23a-63b=0\qquad\text{add}\ 63b\ \text{to both sides}\\\\23a=63b\qquad\text{divide both sides by 23}\\\\a=\dfrac{63b}{23}\\\\if\ a=\dfrac{63b}{23}\ then\ x\ is\ any\ real \ number\\\\if\ a\neq\dfrac{63b}{23}\ then\ x\ not\ exist\ (no\ solution)$