Answer:
The solution to the system of equations is (-6, 3)
Step-by-step explanation:
Plot both lines on a graph, find where they intersect IF they even intersect at all like in this case, and boom there's your answer. Hope that helps and have a good rest of your day.
When you distribute and combine like terms, the result is x^3-27
Answer:
We know that the equation of the circle in standard form is equal to <em>(x-h)² + (y-k)² = r²</em> where (h,k) is the center of the circle and r is the radius of the circle.
We have x² + y² + 8x + 22y + 37 = 0, let's get to the standard form :
1 - We first group terms with the same variable :
(x²+8x) + (y²+22y) + 37 = 0
2 - We then move the constant to the opposite side of the equation (don't forget to change the sign !)
(x²+8x) + (y²+22y) = - 37
3 - Do you recall the quadratic identities ? (a+b)² = a² + 2ab + b². Now that's what we are trying to find. We call this process <u><em>"completing the square"</em></u>.
x²+8x = (x²+8x + 4²) - 4² = (x+4)² - 4²
y²+22y = (y²+22y+11²)-11² = (y+11)²-11²
4 - We plug the new values inside our equation :
(x+4)² - 4² + (y+22)² - 11² = -37
(x+4)² + (y+22)² = -37+4²+11²
(x+4)²+(y+22)² = 100
5 - We re-write in standard form :
(x-(-4)²)² + (y - (-22))² = 10²
And now it is easy to identify h and k, h = -4 and k = - 22 and the radius r equal 10. You can now complete the sentence :)
Answer:
Step-by-step explanation:
Hope this helps.
(1) y² + x² = 53
(2) y - x = 5 ⇒ y = x + 5
subtitute (2) to (1)
(x + 5)² + x² = 53 |use (a + b)² = a² + 2ab + b²
x² + 2x·5 + 5² + x² = 53
2x² + 10x + 25 = 53 |subtract 53 from both sides
2x² + 10x - 28 =0 |divide both sides by 2
x² + 5x - 14 = 0
x² - 2x+ 7x - 14 = 0
x(x - 2) + 7(x - 2) = 0
(x - 2)(x + 7) = 0 ⇔ x - 2 = 0 or x + 7 = 0 ⇔ x = 2 or x = -7
subtitute the values of y to (2)
for x = 2, y = 5 + 2 = 7
for x = -7, y = 5 + (-7) = 5 - 2 = 3
Answer: x = 2 and y = 7 or x = -7 and y = 3
