Please help due today
2 answers:
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Answer:
The system has no real solution.
Step-by-step explanation:
Hi there!
We have the following system of equations:
x² + y² = 1
5x + 6y = 12
The solutions of the system are those (x,y) pairs that satisfy both equations.
Let´s take the second equation and solve it for y:
5x + 6y =12
subtract 5x to both sides of the equation
6y = 12 - 5x
divide by 6 both sides
y = (12- 5x)/6
Now, let´s replace the y in the first equation:
x² + y² = 1
x² + [(12- 5x)/6]² = 1
Apply distributive property
x² + (12 - 5x)²/36 = 1
x² + (12 - 5x)(12 - 5x)/36 = 1
Apply distributive property
x² + (144 - 120x + 25x²)/36 = 1
Apply distributive property
x² + 4 - 10/3 · x + 25/36 · x² = 1
subtract 1 to both sides
61/36 · x² - 10/3 · x + 3 = 0
Using the quadratic formula, let´s find the solutions to this quadratic equation:
a = 61/36
b = -10/3
c = 3
[-b ± √(b² - 4ac)]/2a
[10/3 ± √(100/9 - 4(61/36)(3))]/2(61/36)
(10/3 ± √-83/9)/61/18
This expression has no real solution because √-83/9 is not defined in the set of real numbers. This means that both curves don´t intersect (see attached figure).
Have a nice day!
I am somehow confused with the given matrix. Because usually, when the system of equations is given as:
ax + by = c
dx + ey = f
the matrix would be written in this manner:
![\left[\begin{array}{ccc}a&b&c\\d&e&f\end{array}\right]](https://tex.z-dn.net/?f=%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Da%26b%26c%5C%5Cd%26e%26f%5Cend%7Barray%7D%5Cright%5D%20)
Let me just assume that the arrangement of the given corresponds to this system of equations:
4x + 8y = 0
2x + 5y = 6
So you write it as
![\left[\begin{array}{ccc}4&8&0\\2&5&6\end{array}\right]](https://tex.z-dn.net/?f=%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D4%268%260%5C%5C2%265%266%5Cend%7Barray%7D%5Cright%5D%20)
In order to solve this, the matrix should look like this where x and y are the answers
![\left[\begin{array}{ccc}1&0&x\\0&1&y\end{array}\right]](https://tex.z-dn.net/?f=%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%260%26x%5C%5C0%261%26y%5Cend%7Barray%7D%5Cright%5D%20)
So, the first thing to do is apply this pattern: Row 2 = Row 1 - 2*Row 2. The result will be
![\left[\begin{array}{ccc}4&8&0\\0&-2&-12\end{array}\right]](https://tex.z-dn.net/?f=%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D4%268%260%5C%5C0%26-2%26-12%5Cend%7Barray%7D%5Cright%5D%20)
Next, Row 1 = (Row 1/4) + R2:
![\left[\begin{array}{ccc}1&0&-12\\0&-2&-12\end{array}\right]](https://tex.z-dn.net/?f=%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%260%26-12%5C%5C0%26-2%26-12%5Cend%7Barray%7D%5Cright%5D%20)
Lastly, Row 2 = Row 2 / -2:
![\left[\begin{array}{ccc}1&0&-12\\0&1&6\end{array}\right]](https://tex.z-dn.net/?f=%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%260%26-12%5C%5C0%261%266%5Cend%7Barray%7D%5Cright%5D%20)
Thus, the answer is x=-12 and y=6.
ax + by = c
dx + ey = f
the matrix would be written in this manner:
Let me just assume that the arrangement of the given corresponds to this system of equations:
4x + 8y = 0
2x + 5y = 6
So you write it as
In order to solve this, the matrix should look like this where x and y are the answers
So, the first thing to do is apply this pattern: Row 2 = Row 1 - 2*Row 2. The result will be
Next, Row 1 = (Row 1/4) + R2:
Lastly, Row 2 = Row 2 / -2:
Thus, the answer is x=-12 and y=6.