Question

The tables represent two linear functions in a system
X
-4
-2
0
2
Y
26
18
10
2
X
-4
-2
0
2
Y
14
8
2
-4
What is the solution to this system?
(1,0)
(1,6)
(8,26)
(8,-22)

Answer 1

Answer:

D.22

Step-by-step explanation:

Answer 2

The solution to the system of equations is (-14, -54)

Table A

The equation of table A is calculated using:

$y = \frac{y_2 - y_1}{x_2-x_1} \times (x -x_1) + y_1$

This gives

$y = \frac{14- 2}{3-0} \times (x -0) +2$

$y = \frac{12}{3} \times x +2$

$y = 4 \times x +2$

$y = 4x +2$

Table B

The equation of table B is calculated using:

$y = \frac{y_2 - y_1}{x_2-x_1} \times (x -x_1) + y_1$

This gives

$y = \frac{-3+ 12}{3-0} \times (x -0) -12$

$y = \frac{9}{3} \times x -12$

$y = 3x -12$

Equate both equations

$4x + 2 = 3x -12$

Collect like terms

$4x -3x=- 2 -12$

$x =-14$

Substitute -14 for x in $y = 4x +2$

$y = 4 \times -14 + 2$

$y = -54$

Hence, the solution to the system of equations is (-14, -54)

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