Question

Is it true that (x^2-3x+2)/(x-1) = x – 2, for all values of x? Justify your answer.

Answer 1

Answer:

It is true for all values of x

Step-by-step explanation:

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

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

First lets simplify Left hand side

LHS = $\frac{(x^2-3x + 2)}{x-1}$

Solving the numerator of LHS

x² - 2x - x +2

Using factorization to simplify

x(x-2) - 1(x-2)

(x-1)(x-2)

LHS becomes

$\frac{(x-1)(x-2)}{x-1}$

(x-1) will cancel out

LHS = (x-2)

LHS = RHS

x-2 = x-2

It is evident from the above expression that $\frac{(x^2-3x + 2)}{x-1} = x - 2$ is true for all values of x

Answer 2

Answer:

Yes, it is true for all values of x.

Step-by-step explanation:

We have given the equation:

(x²-3x+2)/(x-1) = x-2

First we solve the nominator L.H.S by factorization method.

x²-3x+2=0

x²-x-2x+2=0

x( x-1)-2( x-1)= 0

(x-1)(x-2)=0

Putting the value of nominator of L.H.S in the equation we get,

(x-1)(x-2)/(x-1)= (x-2)

Simplifying it we get,

(x-2) = (x-2)

Hence, L.H.S = R.H.S

So, it is true for all values of x.