Mathematics, 21.11.2020 21:00 Lizzyloves8910

A system of linear equations is given by the tables.

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Answer from: landon5718

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

y=-2x 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, y=-2x 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

y=x-3 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, y = x-3 is the equation for the second table.

Therefore, the equations are:

y=-2x

y = x-3

Now, let us solve to determine the solution

$\begin{bmatrix}y=-2x\\ y=x-3\end{bmatrix}$

Arrange equation variables for elimination

$\begin{bmatrix}y+2x=0\\ y-x=-3\end{bmatrix}$

y-x=-3

$\underline{y+2x=0}$

-3x=-3

$\begin{bmatrix}y+2x=0\\ -3x=-3\end{bmatrix}$

solving

-3x=-3

$\frac{-3x}{-3}=\frac{-3}{-3}$

x=1

For y+2x=0 , plugin x = 1

$y+2\cdot \:1=0$

y+2=0

y=-2

Therefore, the solution to the system of equations are:

$y=-2,\:x=1$

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A system of linear equations is given by the tables.
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