Iwill award

solve the equations for x:

1) a−(a+b)x=(b−a)x−(c+bx), if b≠ 0

x = if b ≠

2) 2(3x−5a)+9(2a−7b)+3(5a−2x)=0;

if a = 63b/23 then x =
if a ≠ 63b/23 then x =

1) x = (a+c)/b

2) The equation is correct no matter what value x takes.

Step-by-step explanation:

1) a−(a+b)x=(b−a)x−(c+bx)

<=> a - ax - bx = bx - ax - c - bx

<=> a - ax - bx - (bx - ax - c - bx) = 0

<=> a - ax - bx - bx + ax + c + bx = 0

<=> a - ax + ax - bx - bx + bx + c = 0

<=> a - bx + c = 0

<=> bx = a + c

If b ≠ 0, x = (a+c)/b

2) 2(3x−5a)+9(2a−7b)+3(5a−2x)=0

<=> 2×3x + 2×(-5a) + 9×2a + 9× (-7b) + 3×5a + 3×(-2x) = 0

<=> 6x -10a + 18a - 63b + 15a - 6x = 0

<=> (6x - 6x) + (15a - 10a + 18a) - 63b = 0

<=> 23a - 63b = 0

<=> 23a = 63b

=> a = 63b/23

with all values of x, the equation is correct.

1) x= (a+c)/b if b ≠ 0

2) if a =  63b23  then x = all real numbers

if a ≠  63b23  then x = no solutions

Step-by-step explanation:

dont think so

step-by-step explanation:

first one because it can't have a fraction in it

step-by-step explanation: