Answer:
An extraneous solution is a root of a transformed equation that is not a root of the original equation because it was excluded from the domain of the original equation. Example 1: Solve for x , 1x − 2+1x + 2=4(x − 2)(x + 2) .Set the equation to equal zero. (this ends up being √x+4−x+2=0 )
Plug this into the y= button on your TI-83/84 calculator.
Find the value of each of your solutions (go to 2nd->Calc->Value and enter your solution for x )
You should get zero as an answer for each of them.
Step-by-step explanation:
Extraneous solutions are not solutions at all. They arise from outside the problem, from the method of solution. They are extraneous because they are not solutions of the original problem. ... To tell if a "solution" is extraneous you need to go back to the original problem and check to see if it is actually a solution.
Standard form is ax + by = c. Begin our transformation by multiplying both sides by 2 to get rid of the fraction. Notice that ax + by = c does not allow for fractions. Doing that multiplication gives us 2y=-x + 3. Now move the x over by the y by addition. x + 2y = 3. And there you go!
Answer:
68%
Step-by-step explanation:
Answer:
19d + 11
Step-by-step explanation:
7(d + 2) - 3 (1- 4d)
Expand the brackets.
7d + 14 - 3 + 12d
Add or subtract like terms.
7d + 12d + 14 - 3
19d + 11
