Question

A by x minus A + B by x minus b equal to 2 C by x minus C

Answer 1

Answer:

x = 0 or x =  (2ab - ac - bc)/(a + b - 2c).

Step-by-step explanation:

Hi,

Given a/(x - a) + b/(x - b) = 2c/( x - c)

On adding 2 on both sides, we get

a/(x - a) + 1 + b/(x - b) + 1 = 2c/(x - c) + 2

On simplifying , we get

x/( x - a) + x/( x - b) = 2x/(x - c)

x/( x - a) + x/( x - b) - 2x/(x - c) = 0

Taking x common out we get

x[ 1/ x-a + 1/x-b - 2/x-c] = 0

Hence either x = 0 or

1/ x-a + 1/x-b - 2/x-c = 0

1/ x-a + 1/x-b =  2/x-c

(2x - (a+b))(x -c) = 2(x-a)(x-b)

2x² -x(2c + a + b) + c(a + b) = 2(x² -x(a + b) + ab)

(a + b - 2c)x = 2ab - ac - bc

x = (2ab - ac - bc)/(a + b - 2c).

Hence, solutions are x = 0 or x =  (2ab - ac - bc)/(a + b - 2c).

Hope, it helps !

Answer 2

Answer:heyyo!!! Just a diiferent method ..

change the question like.... a/(x-a)+b/(x-c)= c/(x-c) +c/(x-c)

Shift a few places like... a/(x-a)-c/(x-c) = c/(x-c) - b /(x-b)

Step-by-step explanation:

Take lcm on both sides..

ax-ac -cx +ac = -bx +bc +cx -bc

__________ ____________

X^2-ax-cx+ac x^2-bx-cx+bc

...after cancling a few variable that are in negetive n positive on both sides.....

cross multipy... shift denominators to multiply on both sides with numerators

(ax-cx)(x^2-bx-cx+bc) = (bx-cx)(x^2-ax-cx+ac)

x(a-c) (x^2-bx-cx+bc) = x(b-c) (x^2-ax-cx+ac)

Cancle x as it is common on both sides......

Now, open the brackets,

ax^2 - abx -acx +abc -cx^2 +bcx +c^2x -bc^2=

-bx^2 + abx +bcx -abc +cx^2 -acx -c^2x +ac^2

Cancle certain terms that can be cancled....

ax^2 -abx +abc -cx^2 -bc^2 +c^2x=

-bx^2 +abx -abc +cx^2 +ac^2 -c^2x

finally put all terms with 2 on one side like this:

ax^2 + bx^2 - bc^2 -ac^2 = 2(abx+abc + cx^2 -c^2x)

Take commons out on both side and factorize.....

you will get,

(x^2-c^2)(a+b) = 2(ab+cx)(x-c)

Split using identity in the lhs

(x+c)(x-c) (a+b) = 2 (ab+cx) (x-c)

Hence, (x+c)(a+b)= 2ab +2cx

multiply lhs

ax +ac +bx +bc = 2ab + 2cx

keep all x terms in rhs and take common i.e. x out

2ab +ac +bc = ax +bx -2cx

2ab +ac+bc= x( a+b-2c)

Therefore,

X= 2ab+ ac+ bc /(a+b-2c)

And there was a lot to type so i couldn't mention some steps.

hope you understand,

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