Question

Two linear equations are represented by using the tables below.

A 2-column table with 4 rows is titled Equation A. Column 1 is labeled x with entries negative 5, negative 2, 0, 1. Column 2 is labeled y with entries negative 4, negative 1, 1, 2.
A 2-column table with 4 rows is titled Equation B. Column 1 is labeled x with entries negative 6, negative 3, 3, 6. Column 2 is labeled y with entries negative 4, negative 2, 2, 4.

The data points for equation A are plotted on the coordinate plane below and are connected by using a straight line.

On a coordinate plane, a line goes through points (negative 5, negative 4), (negative 2, negative 1), (0, 1), (1, 2).

What is the solution to the system of equations?
(–6, –4)
(–5, –4)
(–3, –2)
(0, 1)

Answer 1

(-3,-2)

Explanation:

Answer 2

(–3, –2)

Explanation:

A linear equation is an equation in the form y = mx + b, where m is the slope of the line and b is the y intercept.

Equation A is represented by the points:

x: -5 -2 0 1

y: -4 -1 1 2

Equation A can be gotten by picking any two pair of points. Let us use (-5, -4) and (0, 1). We use this formula:

`y- y_1=(y_2-y_1)/(x_2-x_1)(x-x_1)\\\\y-(-4)=(1-(-4))/(0-(-5)) (x-(-5))\\\\y +4=x+5\\\\y=x+1`

Equation B is represented by the points:

x: -6 -3 3 6

y: -4 -2 2 4

Equation B can be gotten by picking any two pair of points. Let us use (-6, -4) and (3, 2). We use this formula:

`y- y_1=(y_2-y_1)/(x_2-x_1)(x-x_1)\\\\y-(-4)=(2-(-4))/(3-(-5)) (x-(-6))\\\\y +4=(2)/(3)( x+6)\\\\y+4=(2)/(3) x+4\\\\y=(2)/(3)x`

The solution to the system of equations is gotten by solving y = x + 1 and y = (2/3)x simultaneously.

Substitute y = (2/3)x in y = x + 1:

(2/3)x = x + 1

2x = 3x + 3

x = -3

Put x = -3 in y = (2/3)x

y = (2/3) (-3) = -2

Hence (-3, -2) is the solution