Plz help me only 7 min left or I get in trouble
1 answer:
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Answer:
We conclude that (4, 2) is NOT a solution to the system of equations.
Step-by-step explanation:
Given the system of equations
Important Tip:
- In order to determine whether (4, 2) is a solution to the system of equations or not, we need to solve the system of equations.
Let us solve the system of equations using the elimination method.
Arrange equation variables for elimination
Subtract the equations
Now, solve -2x = 6 for x
Divide both sides by -2
Simplify
For y - x = -2 plug in x = -3
Subtract 3 from both sides
Simplify
The solution to the system of equations is:
(x, y) = (-3, -5)
Checking the graph
From the graph, it is also clear that (4, 2) is NOT a solution to the system of equations because (-3, -5) is the only solution as we have found earlier.
Therefore, we conclude that (4, 2) is NOT a solution to the system of equations.
Remark
This question likely should be done before the other one. What you are trying to do is give C a value. So you need to remember that C is always part of an indefinite integral.
y =
y = sin(x) - cos(x) + C
y(π) = sin(π) - cos(π) + C = 0
y(π) = 0 -(-1) + C = 0
y(π) = 1 + C = 0
C = - 1
y = sin(x) - cos(x) - 1 <<<<< Answer
Problem Two
Remember that
y( - e^3 ) = ln(|x|) + C = 0
y(-e^3) = ln(|-e^3|) + C = 0
y(-e^3) = 3 + C = 0
3 + C = 0
C = - 3
y = ln(|x|) - 3 <<<< Answer
This question likely should be done before the other one. What you are trying to do is give C a value. So you need to remember that C is always part of an indefinite integral.
y =
y = sin(x) - cos(x) + C
y(π) = sin(π) - cos(π) + C = 0
y(π) = 0 -(-1) + C = 0
y(π) = 1 + C = 0
C = - 1
y = sin(x) - cos(x) - 1 <<<<< Answer
Problem Two
Remember that
y( - e^3 ) = ln(|x|) + C = 0
y(-e^3) = ln(|-e^3|) + C = 0
y(-e^3) = 3 + C = 0
3 + C = 0
C = - 3
y = ln(|x|) - 3 <<<< Answer