Question
SOLVING EQUATIONS Solve the equation.$$ rac{1}{2} x-1=-1$$
SOLVING EQUATIONS Solve the equation. $$ \frac{1}{2} x-1=-1 $$
Answers
SOLVING EQUATIONS Solve the equation. $$ \frac{1}{2} x-1=-1 $$
We have to solve the linear equation. X multiplied with one plus two x is equal to two X minus one multiplied that X minus two. So we'll use the distributive property on both sides to simplify the comes first, so we get X Plus two. X squared is equal to two x squared minus X minus two X plus two that IHS two x squared plus X will be equal toe two X squared minus three X plus two. Once they have rearranged. Now we'll bring the light terms onto one side. So we get two X squared minus two X square plus X plus three X is equal to two, which means for X is equal to two. So now dividing both sides of the equation by four will get X is equal to two divided by four. That is one divided by two, which means exit will do their appoint fight. Therefore, the solution is that excess it will do 0.5
This question asks us to solve the given equation. What we know that the first thing we're going to do is we're going to get all the exes on the right hand side one over X minus two over access negative one over X, which means our X has to be negative, too.
You have a divided by Negative, too, is equal. The one opposite of division is multiplying. Multiply both science by the same value. Multiplying cancels. Dividing is equal to negative, too.
Okay for this problem were given wound about about X minus one plus one over X plus one is equal to two over X squared, minus one. Okay. Looking at our denominators, we can see the X cannot equal one. Oh, are negative one. We're in a multiple of our at least common multiple of our denominators, which is X squared minus one. That's gonna get us a new equation of X plus one plus X minus. One is equal to two. Combining like terms will give us two X is equal to two and therefore X is equal to one. But that's in conflict with what we said before. So X equals one is not a solution. So in that case, there is no solution to this problem to this equation. Thank you very much.
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