Question

Enter the values for the highlighted variables to complete the steps to the sum (3x)/(2x - 6) + 9/(6 - 2x) = (3x)/(2x - 6) + 9/(a(2x - 6)) = 3x 2x-6 + 6 2x-6 = 3x-c 2x-6 = d(x-x) f(x-3) =g

Answer 1

Answer:

a= -1

b= -9

c=9

d=3

e=3

f=2

$g=\frac32$

Step-by-step explanation:

Rule of sign:

1. (-)×(+)=(-), (-)÷(+)=(-)
2. (+)×(-)=(-) , (+)÷(-)=(-)
3. (+)×(+)=(+), (+)÷(+)=+
4. (-)×(-)=(+), (-)÷(-)=(+)

Given that,

$\frac{3x}{2x-6}+\frac{9}{6-2x}$

We can rewrite 6-2x as 2x-6, taking (-1) as common factor of (6-2x)

$=\frac{3x}{(2x-6)}+\frac{9}{-1(2x-6)}$

So, a= -1

$\frac9{-1}=-9$

$=\frac{3x}{(2x-6)}+\frac{-9}{(2x-6)}$

So, b= -9

The L.C.M of (2x-6) and (2x-6) is (2x-6)

and (2x-6)÷(2x-6)=1

$=\frac{1 \times 3x+1\times (-9)}{(2x-6)}$

$=\frac{( 3x-9)}{(2x-6)}$

∴c= 9

(3x-9) has a common factor 3 and (2x-6) has a common factor 2.

(3x-9)=3(x-3)

(2x-6)=2(x-3)

$=\frac{3(x-3)}{2(x-3)}$

∴d=3, e=3 and f=2

Since the denominator and numerator are the product of two polynomial. So, if there is any common element, then can cancel the common factor.

Here the common factor is (x-3). So cancel  out (x-3).

$=\frac32$

$\therefore g=\frac32$