Answer:
x = (-15)
y = 6
z = 3
Step-by-step explanation:
x + y + z = -6______(1)
2x - y - 5z = -21____(2)
-x - y -6z = -9______(3)
(1) + (2)
![x + y + z + ( - x - y - 6z) = - 6 + ( - 9) \\ x + y + z - x - y - 6z = - 15 \\ - 5z = - 15 \\ z = \frac{ - 15}{ - 5} \\ z = 3 \\](https://tex.z-dn.net/?f=x%20%2B%20y%20%2B%20z%20%20%2B%20%28%20-%20x%20-%20y%20-%206z%29%20%3D%20%20-%206%20%2B%20%28%20-%209%29%20%5C%5C%20x%20%2B%20y%20%2B%20z%20%20-%20%20x%20%20-%20%20y%20-%206z%20%3D%20%20-%2015%20%5C%5C%20%20-%205z%20%3D%20%20-%2015%20%5C%5C%20z%20%3D%20%20%5Cfrac%7B%20-%2015%7D%7B%20-%205%7D%20%20%5C%5C%20z%20%3D%203%20%5C%5C%20)
z = 3, (1) + (2)
![x + y + 3 + (2x - y - 5z) = - 6 + ( - 21) \\ x + y + 3 +( 2x - y - 5( - 3) = - 27 \\ x + y + 3 + (2x - y + 15) = - 27 \\ x + y + 3 + 2x - y + 15 = - 27 \\ 3x + 18 = - 27 \\ 3x = - 27 - 18 \\ 3x = - 45 \\ x = \frac{ - 45}{3} \\ x = ( - 15) \\](https://tex.z-dn.net/?f=x%20%2B%20y%20%2B%203%20%2B%20%282x%20-%20y%20-%205z%29%20%3D%20%20-%206%20%2B%20%20%28%20-%2021%29%20%5C%5C%20x%20%2B%20y%20%2B%203%20%2B%28%202x%20-%20y%20-%205%28%20-%203%29%20%3D%20%20-%2027%20%5C%5C%20x%20%2B%20y%20%2B%203%20%2B%20%282x%20-%20y%20%2B%2015%29%20%3D%20%20-%2027%20%5C%5C%20x%20%2B%20y%20%2B%203%20%2B%202x%20-%20y%20%2B%2015%20%3D%20%20-%2027%20%5C%5C%203x%20%2B%2018%20%3D%20%20-%2027%20%5C%5C%203x%20%3D%20%20-%2027%20%20-%2018%20%5C%5C%203x%20%3D%20%20-%2045%20%5C%5C%20x%20%3D%20%20%5Cfrac%7B%20-%2045%7D%7B3%7D%20%20%5C%5C%20x%20%3D%20%28%20-%2015%29%20%5C%5C%20)
x = (-15), (1)
![x + y + z = - 6 \\ ( - 15) + y + 3 = - 6 \\ - 12 + y = - 6 \\ y = - 6 + 12 \\ y = 6 \\](https://tex.z-dn.net/?f=x%20%2B%20y%20%2B%20z%20%3D%20%20-%206%20%5C%5C%20%28%20-%2015%29%20%2B%20y%20%2B%203%20%3D%20%20-%206%20%5C%5C%20%20-%2012%20%2B%20y%20%3D%20%20-%206%20%5C%5C%20y%20%3D%20%20-%206%20%2B%2012%20%5C%5C%20y%20%3D%206%20%5C%5C%20)
Your equation would be 7-3x=40. Subtract 7, divide by -3, and your answer is -11
Answer: 1. y = 2(x + 4)² - 3
![\bold{2.\quad y=-\dfrac{1}{3}(x+8)^2-7}](https://tex.z-dn.net/?f=%5Cbold%7B2.%5Cquad%20y%3D-%5Cdfrac%7B1%7D%7B3%7D%28x%2B8%29%5E2-7%7D)
![\bold{3.\quad y=-\dfrac{1}{2}(x-7)^2+1}](https://tex.z-dn.net/?f=%5Cbold%7B3.%5Cquad%20y%3D-%5Cdfrac%7B1%7D%7B2%7D%28x-7%29%5E2%2B1%7D)
<u>Step-by-step explanation:</u>
Notes: The vertex form of a parabola is y = a(x - h)² + k
- (h, k) is the vertex
- p is the distance from the vertex to the focus
![\bullet\quad a=\dfrac{1}{4p}](https://tex.z-dn.net/?f=%5Cbullet%5Cquad%20a%3D%5Cdfrac%7B1%7D%7B4p%7D)
1)
![\text{Vertex}=(-4,-3)\qquad \text{Directrix}:y=-\dfrac{25}{8}\\\\\text{Given}:(h, k)=(-4, 3)\\\\p=\dfrac{-24}{8}-\dfrac{-25}{8}=\dfrac{1}{8}\\\\\\a=\dfrac{1}{4p}=\dfrac{1}{4(\frac{1}{8})}=\dfrac{1}{\frac{1}{2}}=2](https://tex.z-dn.net/?f=%5Ctext%7BVertex%7D%3D%28-4%2C-3%29%5Cqquad%20%5Ctext%7BDirectrix%7D%3Ay%3D-%5Cdfrac%7B25%7D%7B8%7D%5C%5C%5C%5C%5Ctext%7BGiven%7D%3A%28h%2C%20k%29%3D%28-4%2C%203%29%5C%5C%5C%5Cp%3D%5Cdfrac%7B-24%7D%7B8%7D-%5Cdfrac%7B-25%7D%7B8%7D%3D%5Cdfrac%7B1%7D%7B8%7D%5C%5C%5C%5C%5C%5Ca%3D%5Cdfrac%7B1%7D%7B4p%7D%3D%5Cdfrac%7B1%7D%7B4%28%5Cfrac%7B1%7D%7B8%7D%29%7D%3D%5Cdfrac%7B1%7D%7B%5Cfrac%7B1%7D%7B2%7D%7D%3D2)
Now input a = 2 and (h, k) = (-4, -3) into the equation y = a(x - h)² + k
y = 2(x + 4)² - 3
******************************************************************************************
2)
![\text{Vertex}=(-8,-7)\qquad \text{Directrix}:y=-\dfrac{-25}{4}\\\\\text{Given}:(h, k)=(-8, -7)\\\\p=\dfrac{-28}{4}-\dfrac{-25}{4}=\dfrac{-3}{4}}\\\\\\a=\dfrac{1}{4p}=\dfrac{1}{4(\frac{-3}{4})}=\dfrac{1}{-3}=-\dfrac{1}{3}](https://tex.z-dn.net/?f=%5Ctext%7BVertex%7D%3D%28-8%2C-7%29%5Cqquad%20%5Ctext%7BDirectrix%7D%3Ay%3D-%5Cdfrac%7B-25%7D%7B4%7D%5C%5C%5C%5C%5Ctext%7BGiven%7D%3A%28h%2C%20k%29%3D%28-8%2C%20-7%29%5C%5C%5C%5Cp%3D%5Cdfrac%7B-28%7D%7B4%7D-%5Cdfrac%7B-25%7D%7B4%7D%3D%5Cdfrac%7B-3%7D%7B4%7D%7D%5C%5C%5C%5C%5C%5Ca%3D%5Cdfrac%7B1%7D%7B4p%7D%3D%5Cdfrac%7B1%7D%7B4%28%5Cfrac%7B-3%7D%7B4%7D%29%7D%3D%5Cdfrac%7B1%7D%7B-3%7D%3D-%5Cdfrac%7B1%7D%7B3%7D)
Now input a = -1/3 and (h, k) = (-8, -7) into the equation y = a(x - h)² + k
![\bold{y=-\dfrac{1}{3}(x+8)^2-7}](https://tex.z-dn.net/?f=%5Cbold%7By%3D-%5Cdfrac%7B1%7D%7B3%7D%28x%2B8%29%5E2-7%7D)
******************************************************************************************
3)
![\text{Focus}=\bigg(7,\dfrac{1}{2}\bigg)\qquad \text{Directrix}:y=\dfrac{3}{2}](https://tex.z-dn.net/?f=%5Ctext%7BFocus%7D%3D%5Cbigg%287%2C%5Cdfrac%7B1%7D%7B2%7D%5Cbigg%29%5Cqquad%20%5Ctext%7BDirectrix%7D%3Ay%3D%5Cdfrac%7B3%7D%7B2%7D)
The midpoint of the focus and directrix is the y-coordinate of the vertex:
![\dfrac{focus+directrix}{2}=\dfrac{\frac{1}{2}+\frac{3}{2}}{2}=\dfrac{\frac{4}{2}}{2}=\dfrac{2}{2}=1](https://tex.z-dn.net/?f=%5Cdfrac%7Bfocus%2Bdirectrix%7D%7B2%7D%3D%5Cdfrac%7B%5Cfrac%7B1%7D%7B2%7D%2B%5Cfrac%7B3%7D%7B2%7D%7D%7B2%7D%3D%5Cdfrac%7B%5Cfrac%7B4%7D%7B2%7D%7D%7B2%7D%3D%5Cdfrac%7B2%7D%7B2%7D%3D1)
The x-coordinate of the vertex is given in the focus as 7
(h, k) = (7, 1)
Now let's find the a-value:
![p=\dfrac{2}{2}-\dfrac{3}{2}=\dfrac{-1}{2}}\\\\\\a=\dfrac{1}{4p}=\dfrac{1}{4(\frac{-1}{2})}=\dfrac{1}{-2}=-\dfrac{1}{2}](https://tex.z-dn.net/?f=p%3D%5Cdfrac%7B2%7D%7B2%7D-%5Cdfrac%7B3%7D%7B2%7D%3D%5Cdfrac%7B-1%7D%7B2%7D%7D%5C%5C%5C%5C%5C%5Ca%3D%5Cdfrac%7B1%7D%7B4p%7D%3D%5Cdfrac%7B1%7D%7B4%28%5Cfrac%7B-1%7D%7B2%7D%29%7D%3D%5Cdfrac%7B1%7D%7B-2%7D%3D-%5Cdfrac%7B1%7D%7B2%7D)
Now input a = -1/2 and (h, k) = (7, 1) into the equation y = a(x - h)² + k
![\bold{y=-\dfrac{1}{2}(x-7)^2+1}](https://tex.z-dn.net/?f=%5Cbold%7By%3D-%5Cdfrac%7B1%7D%7B2%7D%28x-7%29%5E2%2B1%7D)
Answer:
The equation that says P is equidistant from F and the y-axis is
.
(1, 0), (1, 3/2) and (1,6) are three points that are equdistant from F and the y-axis.
Step-by-step explanation:
Let
and
, where
is a point that is equidistant from F and the y-axis. The following vectorial expression must be satisfied to get the location of that point:
![F(x,y)-P(x,y) = P(x,y)-R(x,y)](https://tex.z-dn.net/?f=F%28x%2Cy%29-P%28x%2Cy%29%20%3D%20P%28x%2Cy%29-R%28x%2Cy%29)
![2\cdot P(x,y) = F(x,y)+R(x,y)](https://tex.z-dn.net/?f=2%5Ccdot%20P%28x%2Cy%29%20%3D%20F%28x%2Cy%29%2BR%28x%2Cy%29)
(1)
If we know that
and
, then the resulting vectorial equation is:
![P(x,y) = \left(1,\frac{3}{2} \right)+\left(0, \frac{y'}{2}\right)](https://tex.z-dn.net/?f=P%28x%2Cy%29%20%3D%20%5Cleft%281%2C%5Cfrac%7B3%7D%7B2%7D%20%5Cright%29%2B%5Cleft%280%2C%20%5Cfrac%7By%27%7D%7B2%7D%5Cright%29)
![P(x,y) =\left(1,\frac{3+y'}{2} \right)](https://tex.z-dn.net/?f=P%28x%2Cy%29%20%3D%5Cleft%281%2C%5Cfrac%7B3%2By%27%7D%7B2%7D%20%5Cright%29)
The equation that says P is equidistant from F and the y-axis is
.
If we know that
,
and
, then the coordinates for three points that are equidistant from F and the y-axis:
![P_{1}(x,y) = \left(1,\frac{3+y_{1}'}{2} \right)](https://tex.z-dn.net/?f=P_%7B1%7D%28x%2Cy%29%20%3D%20%5Cleft%281%2C%5Cfrac%7B3%2By_%7B1%7D%27%7D%7B2%7D%20%5Cright%29)
![P_{1}(x,y) = (1,0)](https://tex.z-dn.net/?f=P_%7B1%7D%28x%2Cy%29%20%3D%20%281%2C0%29)
![P_{2}(x,y) = \left(1,\frac{3+y_{2}'}{2} \right)](https://tex.z-dn.net/?f=P_%7B2%7D%28x%2Cy%29%20%3D%20%5Cleft%281%2C%5Cfrac%7B3%2By_%7B2%7D%27%7D%7B2%7D%20%5Cright%29)
![P_{2}(x,y) = \left(1,\frac{3}{2} \right)](https://tex.z-dn.net/?f=P_%7B2%7D%28x%2Cy%29%20%3D%20%5Cleft%281%2C%5Cfrac%7B3%7D%7B2%7D%20%5Cright%29)
![P_{3}(x,y) = \left(1,\frac{3+y_{3}'}{2} \right)](https://tex.z-dn.net/?f=P_%7B3%7D%28x%2Cy%29%20%3D%20%5Cleft%281%2C%5Cfrac%7B3%2By_%7B3%7D%27%7D%7B2%7D%20%5Cright%29)
![P_{3}(x,y) = \left(1,6 \right)](https://tex.z-dn.net/?f=P_%7B3%7D%28x%2Cy%29%20%3D%20%5Cleft%281%2C6%20%5Cright%29)
(1, 0), (1, 3/2) and (1,6) are three points that are equdistant from F and the y-axis.
I think its the first one