Verify if (a²b-c)/(-c+ba²) = 1: Algebraic Fraction Analysis

Indicate whether true or false

a2bcc+ba2=1 \frac{a^2\cdot b-c}{-c+b\cdot a^2}=1

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations
00:00 Determine if the equation is correct
00:04 Let's use the commutative law and arrange the expression
00:31 Let's reduce what we can, when reducing the entire fraction we get 1
00:37 Let's compare the expressions
00:44 And this is the solution to the question

Step-by-step written solution

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1

Understand the problem

Indicate whether true or false

a2bcc+ba2=1 \frac{a^2\cdot b-c}{-c+b\cdot a^2}=1

2

Step-by-step solution

Let's first examine the problem:

a2bcc+ba2=?1 \frac{a^2 b-c}{-c+b a^2}\stackrel{?}{= }1 Looking at the expression on the left side, note that using the distributive property in addition (first) and multiplication (later) in the fraction's denominator (or alternatively in the fraction's numerator) we get:

a2bcc+ba2=a2bcba2+(c)=a2bcba2c=a2bca2bc=1 \frac{a^2 b-c}{-c+b a^2}=\\ \frac{a^2 b-c}{b a^2+(-c)}=\\ \frac{a^2 b-c}{b a^2-c}=\\ \frac{a^2 b-c}{a^2b -c}=\\ 1 where in the final step we used the fact that dividing any number by itself always yields 1,

Therefore the expressions on both sides of the equality (which is assumed to hold) are indeed equal, meaning:

a2bcc+ba2=a2bca2bc=!1 \frac{a^2 b-c}{-c+b a^2}=\frac{a^2 b-c}{a^2b -c}\stackrel{!}{= }1

(In other words, there is an identity equality- which is true for all possible values of the parameters a,b,c a,b,c )

Therefore the correct answer is answer A.

3

Final Answer

True

Practice Quiz

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Determine if the simplification below is correct:

\( \frac{4\cdot8}{4}=\frac{1}{8} \)

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