$\frac{x}{x^2-5x+6}+\frac{x}{x+3}\\\\Step 1: \frac{x}{(x-2)(x-3)}+\frac{x}{x+3}\\Step2: \frac{x}{(x-2)(x-3)}+\frac{x(x-2)}{(x-2)(x+3)}\\Step3:\frac{x(x+3)+x(x-2)(x+3)}{(x-2)(x-3)(x+3)}$

Lets consider the steps given in the question:

$Step2: \frac{x}{(x-2)(x+3)}+\frac{3(x-2)}{(x-2)(x+3)}$

$Step3: \frac{x+3(x-2)}{(x-2)(x+3)}\\$

Lets compare both solutions:

In the original solution, the denominator of first term is (x-2)(x-3)

In the solution given in the question, the denominator of first term is (x-2)(x+3).

So the mistake she did in step 2 was that she change the sign of 3 in (x-3) from negative to positive, due to which she gets the wrong common denominator shown in Step 3.

7 0

3 years ago

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What is 100,000,000,000,000,000, divided by 100

s344n2d4d5 [400]

Answer:

1e+14

Step-by-step explanation:

5 0

2 years ago

If y=12x-7 were changed to y=12x+1, how would the graph of the new function compare to the original?

Likurg_2 [28]

When we have the equation of a line y=mx+b, the "b" value tells you how high or low the line is. your b value when from -7 to 1. with the -7, each point on the line would have been lowered 7 units from its original position. changing the -7 to a 1 will raise the line y=12x+1 8 units, because 1-(-7) = 1+7 = 8.

4 0

3 years ago

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