Suppose you want to solve the equation 2a b=2a where a and b are nonZero real numbers describe the solution to this equation justify you description
To solve this equation we
can first assume that both a and b are nonzero real numbers. Hence,
A = 1 b = 1
1.
2 (1) + 1 =
2(1)
2.
2 + 1 = 2: now this a false equation since
there is not equality, the equation cannot retain the equal sign but will
become 2 + 1 > 2. Leaving the relationship unequal.
However, the alternative to
this problem is to be b = 0. To oversee the rule in order to solve the equation
retaining it as an “equation”. Further, there is no other solution for this
equation.
A = 1 b = 0
1.
Which
becomes 2(1) + 0 = 2(1)
2.
2 + 0 = 2 :
3. 2 = 2. Here we can observe the
equality.
We have the following equation:
\(2a+b=2a\)
By solving this equation we have that:
\(2a+b=2a \\ \\ \therefore (2a-2a)+b=0 \\ \\ \therefore \boxed{b=0}\)
So, the only solution to this problem is \(b=0\) for any real value of \(a\)
Then, the conclusion is:
\(a \ is \ a \ nonzero \ real \ number \\ \\ b \ \mathbf{must} \ be \ zero\)