Answer:
(x, y) = (6, -2)
Step-by-step explanation:
Sometimes this solution method is called "addition" or "elimination." You want to multiply one or both equations by a value or values that make the coefficients of one of the variables be opposites. Then when you add the equations, the result is an equation without that variable. (One variable has been eliminated.)
Multiply the first equation by -2 to make its x-coefficient the opposite of that in the second equation. Then add the two equations:
-2(3x +3y) = -2(12)
-6x -6y = -24
Add the second equation ...
(-6x -6y) +(6x +11y) = (-24) +(14)
5y = -10 . . . . . simplify
y = -2 . . . . . . . divide by 5
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3x +3(-2) = 12 . . . . . substitute into the first equation
3x = 18 . . . . . . . . . . add 6
x = 6 . . . . . . . . . . . . divide by 3
The solution is (x, y) = (6, -2).
Answer:
- inverse is not a function
- unless the domain is restricted to |x| ≥ 1.2 (approximately)
Step-by-step explanation:
The test to see if the inverse function is also a function is called the "horizontal line test." The test passes if any horizontal line intersects the graph in only one place.
Here, a horizontal line can intersect the graph in 1, 2, or 3 places, so the test fails. The function does not have an inverse that is a function.
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If the domain of the inverse relation is restricted to |x| > 1.2, then that inverse will map any x to only a single value of y. Then it will be a function.
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The graph shows the original function (dashed red line) and the inverse relation (blue). The green shading marks values of x for which there is a single value of y, so the inverse relation is a function in those regions.
(We could be more specific as to the limits on the domain of f^-1(x), but the given graph seems to have an unknown vertical scale factor.)
Answer:
Step-by-step explanation: