How do you do this problem? simple and concise explanation, !

Step-by-step

This is a problem of partial fractions.

Step 1:

factor all denominators on the left-hand side (LHS).

LHS = a/[(x+1)(x-1)] + x/[(x-3)(x+1)]

Note the common factor (x+1) in both denominators, which makes the combined/common denominator [(x+1)(x-1)(x-3)]

Step 2:

multiply each term, top and bottom, by the factor "missing" from the common denominator.

The first term is missing (x-3), the second term is missing (x-1)

a(x-3)/[(x+1)(x-1)(x-3)] + x(x-1)/[(x+1)(x-1)(x-3)]

which can be simplified to:

[a(x-3)+x(x-1)]/[(x+1)(x-1)(x-3)]

Expand numerator:

[x^2 + ax-x -3a]/[(x+1)(x-1)(x-3)]

Step 3:

For the two expressions on each side of the equal sign (LHS and RHS) to be equivalent (for ALL values of x), the numerators and denominators must be identical when expanded, so

For denominator, we have factors [(x+1)(x-1)(x-3)], or b,c,d = -3, -1, 1 (in ascending order).

For the numerator, we need to have LHS = RHS

x^2 + (a-1)x -3a  = x^2 + 2x -9

putting a=3 gives the

LHS = x^2 + (3-1)x -3(3) = x^2 + 2x - 9 which makes equality with the RHS.

So we have solved for all values required.

given that triangle abc is transformed two times to get triangle

a"b"c"

first abc is transformed in to a'b'c' as follows:

the coordinates of a are (1,-1) and transformed into a'(-1,1)

similarly b (4,-2) became b'(-4,2) and

c(7,-2) became c'(-7,2)

i.e. (x,y) becomes (-x,-y)

this is nothing but reflection about a point here origin.

thus first transformation is reflection on the origin.

next is exactly shifting 3 units down

source:

answer: okw

step-by-step explanation: