Answer:
Option B
Step-by-step explanation:
Hope this helps
Answer: The answer is x < 12 but i don't see that as an option.
Answer:
x = 7
y = 2
Step-by-step explanation:
In the above question, we are given 2 equations which are simultaneous. To solve this equation, we have to find the values of x and y
x + 3y = 13 ........ Equation 1
x - y = 5...........Equation 2
From Equation 2,
x = 5 + y
Substitute 5 + y for x in Equation 1
x + 3y = 13 ........ Equation 1
5 + y + 3y = 13
5 + 4y = 13
4y = 13 - 5
4y = 8
y = 8/4
y = 2
Since y = 2, substitute , 2 for y in Equation 2
x - y = 5...........Equation 2
x - 2 = 5
x = 5 + 2
x = 7
Therefore, x = 7 and y = 2
<h3>
Answer: False</h3>
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Explanation:
I'm assuming you meant to type out
(y-2)^2 = y^2-6y+4
This equation is not true for all real numbers because the left hand side expands out like so
(y-2)^2
(y-2)(y-2)
x(y-2) .... let x = y-2
xy-2x
y(x)-2(x)
y(y-2)-2(y-2) ... replace x with y-2
y^2-2y-2y+4
y^2-4y+4
So if the claim was (y-2)^2 = y^2-4y+4, then the claim would be true. However, the right hand side we're given doesn't match up with y^2-4y+4
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Another approach is to pick some y value such as y = 2 to find that
(y-2)^2 = y^2-6y+4
(2-2)^2 = 2^2 - 6(2) + 4 .... plug in y = 2
0^2 = 2^2 - 6(2) + 4
0 = 4 - 6(2) + 4
0 = 4 - 12 + 4
0 = -4
We get a false statement. This is one counterexample showing the given equation is not true for all values of y.
