Answer:
Please, see the explanation.
Step-by-step explanation:
DETERMINING THE EQUATION FOR THE FIRST TABLE
Given the first table
x y
-5 10
-1 2
0 0
11 -22
From the given equation, we can verify and determine that
[tex]y=-2x[/tex] is the required equation that satisfies the table values of the first table.
substituting the table values of all the ordered pairs to check
FOR (-5, 10)
y =-2x
10 = -2(-5)
10 = 10
L.H.S = R.H.S
FOR (-1, 2)
y =-2x
2 = -2(-1)
10 = 2
L.H.S = R.H.S
FOR (0, 0)
y =-2x
0 = -2(0)
0 = 0
L.H.S = R.H.S
FOR (11, -22)
y =-2x
-22 = -2(11)
-22 = -22
L.H.S = R.H.S
As all the ordered pairs of the first table satisfy the equation.
Hence, [tex]y=-2x[/tex] is the equation for the first table.
DETERMINING THE EQUATION FOR THE SECOND TABLE
Given the first table
x y
-8 -11
-2 -5
1 -2
7 4
From the given equation, we can verify and determine that
[tex]y=x-3[/tex] is the required equation that satisfies the table values of the second table.
substituting the table values of all the ordered pairs to check
FOR (-8, -11)
y =x-3
-11 = -8-3
-11 = -11
L.H.S = R.H.S
FOR (-2, -5)
y =x-3
-5= -2-3
-11 = -5
L.H.S = R.H.S
FOR (1, -2)
y = x-3
-2= 1-3
-2 = -2
L.H.S = R.H.S
FOR (7, 4)
y = x-3
4 = 7-3
4 = 4
L.H.S = R.H.S
As all the ordered pairs of the second table satisfy the equation.
Hence, [tex]y = x-3[/tex] is the equation for the second table.
Therefore, the equations are:
[tex]y=-2x[/tex]
[tex]y = x-3[/tex]
Now, let us solve to determine the solution
[tex]\begin{bmatrix}y=-2x\\ y=x-3\end{bmatrix}[/tex]
Arrange equation variables for elimination
[tex]\begin{bmatrix}y+2x=0\\ y-x=-3\end{bmatrix}[/tex]
[tex]y-x=-3[/tex]
[tex]-[/tex]
[tex]\underline{y+2x=0}[/tex]
[tex]-3x=-3[/tex]
[tex]\begin{bmatrix}y+2x=0\\ -3x=-3\end{bmatrix}[/tex]
solving
[tex]-3x=-3[/tex]
[tex]\frac{-3x}{-3}=\frac{-3}{-3}[/tex]
[tex]x=1[/tex]
For [tex]y+2x=0[/tex], plugin [tex]x = 1[/tex]
[tex]y+2\cdot \:1=0[/tex]
[tex]y+2=0[/tex]
[tex]y=-2[/tex]
Therefore, the solution to the system of equations are:
[tex]y=-2,\:x=1[/tex]
