Question

LESSON 14-1 PRACTICE 19. Create a table of values to determine the solution to the following system of equations. y=-x-2 +3 20. Determine which of the points |(1, -2).(-1, 2), (1, 2).(---1. - 20). if any, are solutions to the system of equations. 2x + y = -1 21. Graph each of the equations below and use your graph to determine at which value of x the values of y are the same. (2x-12y = -3 x + 4y = -4

Answer 1

19) x = -3, y = 1

20) None of the points given are solutions to this system of equations.

21) The graph of this system of equations is presented below

The point of intersection is (-3, -0.25). Hence, the solution of this system of equations is

x = -3

y = -0.25

Step-by-step explanation

19)

y = -x - 2

y = (2x/3) + 3

We will just solve this to obtain its solution

Equating the values of y to each other,

-x - 2 = (2x/3) + 3

Multiply through by 3 to simplify this

-3x - 6 = 2x + 9

-3x - 2x = 9 + 6

-5x = 15

Divide both sides by -5

(-5x/-5) = (15/-5)

x = -3

We can then solve for y

y = -x - 2

y = -(-3) - 2 = 3 - 2 = 1

x = -3. y = 1

20) We are asked to check if a set of values given are solutions of this system of equations

x = (4y/5) - 7

2x + y = -1

To do this, we will insert each of these given points into at least one of the equations.

Using the second equation because it is more relatable,

(1, -2)

x = 1, y = -2

2x + y = -1

2(1) + (-2) = -1

2 - 2 = -1

0 ≠ -1

(-1, 2)

2x + y = -1

2(-1) + 2 = -1

-2 + 2 = -1

0 ≠ -1

(1, 2)

2x + y = -1

2(1) + 2 = -1

2 + 2 = -1

4 ≠ -1

(-1, -2)

2x + y = -1

2(-1) + (-2) = -1

-2 - 2 = -1

-4 ≠ -1

None of the points given are solutions to this system of equations.

21) To do this, we will plot the two equations on the same graph, and the point of intersection is the solution.

To plot this, we will use intercepts to obtain two points on each line and draw the lines.

For the first equation,

2x - 12y = -3

when x = 0

2(0) - 12y = -3

-12y = -3

Divide both sides by -12

(-12y/-12) = (-3/-12)

y = ¼ = 0.25

First point on this line is (0, 0.25)

when y = 0

2x - 12(0) = -3

2x = -3

Divide both sides by 2

(2x/2) = (-3/2)

x = -(3/2) = -1.5

Second point on this line is (-1.5, 0)

For this second equation,

x + 4y = -4

when x = 0

0 + 4y = -4

4y = -4

Divide both sides by 4

(4y/4) = (-4/4)

y = -1

First point on this line is (0, -1)

when y = 0

x + 4(0) = -4

x = -4

Second point on the line is (-4, 0)

For the graph, the first equation is represented with the red line and the second equation is represented with the blue line.

The graph and solution is presented under 'Answer' above.

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Hope this Helps!!!