Answer:
8 1/8
Step-by-step explanation:
1.
The description of set K as <span> {(x, y) | x - y = 5}, means the following:
the elements of set K are pairs (x,y)
such that: x-y=5, that is we can write(x, y) = (x, x-5).
2.
A set of (ordered) pairs is a function, if each of the first coordinates is paired to only 1 specific value, not 2 or more.
for example: {(1,2), (3, 5), (3, 8)} is not a function because 3 is not paired to only one second value, we have (3, 5) but also (3, 8).
whereas, {(-2, 4), (3, 5), (8.1, 17)} is a function, because each first coordinate is unique, we don't see it again in another pair.
3.
Back in our set K, the description of pairs (x,y) as </span>(x, x-5)
makes sure that each x, produces a specific y, for example in K we have:
(5, 0), and we cannot have (5, a value ≠0), because it would not fit the description (x, x-5)
Answer: yes
Answer:
25. (x, y) = (5, 11)
26. (x, y) = (-1, 1)
Step-by-step explanation:
Both equations are of the form y=( ), so you can set the expressions for y equal to each other. Or, you can subtract the equation with the smaller y-coefficient from the other one.
25.
x +6 = y = 2x +1 . . . . . equate expressions for y
5 = x . . . . . . . . . . . subtract x+1
y = 5+6 = 11 . . . . . using the first equation to find y
(x, y) = (5, 11)
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26.
(y) -(y) = (3x +4) -(x+2) . . . . subtract the first equation from the second
0 = 2x +2 . . . . . . . . . . . . . . simplify
0 = x + 1 . . . . . . . . . . . . . . . . divide by the x-coefficient
x = -1 . . . . . . . . . . . . . . subtract the constant
y = -1 +2 = 1 . . . . . . . . . use the first equation to find y
(x, y) = (-1, 1)
_____
Of course, when we say "subtract ..." or "divide ..." we mean that you should do the same operation to both sides of the equation. That way the equal sign remains valid. You can always use an expression or variable in place of its equal (this is the substitution property of equality).
The expression (x+1) that we subtract in problem 25 is the smaller x-term plus the constant on the opposite side of the equal sign. That way, we eliminate both the unwanted x-term and the unwanted constant. You can do these operations one at a time (and you were probably taught to do it that way). That is, subtract x; subtract 1.
For 26, the method of solution that puts both the variable and the constant on the same side of the equation and 0 on the other side has certain advantages. Subtracting one side of the equation from both sides (to make an expression equal to zero) will always work, regardless of the expressions involved. After simplification, you can divide by the coefficient of the variable to get the form x+constant=0, and the answer is always x = -constant. These simple instructions require no judgment. You may find it easier to choose to subtract the side with the smaller coefficient, so the result has a positive coefficient. That's not necessary, but it can reduce anxiety and errors.
