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There are several different answers one could make and an exam question would just require one for each.
1) Propanal - from the ending "al" this is an aldehyde.
- Add Tollens' reagent. Positive result: Silver mirror formed.
- Add Fehling's / Benedict Solution. Positive result: Red precipitate formed.
- Acidified potassium dichromate. Positive result: Orange to green (due to oxidation).
2) Benzoic acid - You probably have not come across this in the lab at school. Nevertheless from the name you can work out that it is a carboxylic acid.
- Add Sodium carbonate/ Sodium hydrogen carbonate. Positive result: Effervescence is observed.see more
Let the first equation be equation 1 and the second equation be equation 2.
Firstly, you must substitute equation 1 into equation 2. This is because there are currently two unknown values in each equation (x and y) and therefore you must eliminate one of them so that you just have x's or y's in a singe equation.
Therefore, lets sub in equation 1 into equation 2. But first we must square:
From equation 1: y2 = (2x-3)2 = (2x-3)(2x-3) = 4x2 -12x +9
Then substituting this into equation 2: x2 +y2 = 2
x2 +4x2-12x+9 =2
5x2 -12x+9 =2
Subtracting 2 from both sides of the equation:
5x2 -12x + 7 = 0
Now factorising the quadratic equation:
(5x -7)(x-1) = 0
Now to solve for x: either of the brackets need to equal 0 so that the left hand side of the equation equals 0.
Therefore, 5x-7 = 0, which with some simple rearrangement gives rise to x = 7/5
x-1 = 0, which with some simple rearrangement gives rise to x = 1
Please note we have two values for x; this is because we have a quadratic equation (one which has powers of 2 in it). As we have two values for x it means we have two corresponding values for y. To find these values substitute the values of x you have just found back into equation 1:
From equation 1: When x = 7/5, y = -1/5
When x = 1, y = -1
And that's the final answer!see more