Asystem of equations can be solved by elimination in the table below. does the system of equations on line 4 have the same solution set as the system of equations on line 1? why or why not?

Full question attached below.

Yes, the system of equations on line 4 has the same solution set as the system of equations on line 1

Step-by-step explanation:

There is no difference between the systems of line 1 and 4

1st system

5x + 3y = 14

4x - 2y = -2

4th system

10x + 6y = 28

12x - 6y = -6

In the fourth system, we factor number 2 from the first line, and number 3 from the second line

4th system

2[5x + 3y] = 2[14]

3[4x - 2y] = 3[-2]

If we simplify,

4th system

[5x + 3y] = [14]

[4x - 2y] = [-2]

We get the same system as the 1st

Therefore, the solution is the same.

(0,-3) see attached graph

step-by-step explanation:

a. to solve the system, graph the two equations.

x = y+3 becomes y=x-3 has m=1/1 and b=-3. start at (0,-3) on the y-axis and plot a point. then move up 1 unit and 1 unit to the right. plot a point and connect.

y=-4x+3 has m =-4/1 and b=-3. start at (0,-3) on the y-axis and plot a point. then move down four units and to the right 1 unit. plot a point and connect.

the solution is the point (x,y) where the two lines intersect. by the graph below it is (0,-3).

b. to solve by substitution, substitute y = -4x-3 into x= y+3

x= (-4x-3) +3

x=-4x

5x=0

x=0

now substitute x = 0 back in to find y.

y= -4(0)-3

y=0-3

y=-3

solution is (0,-3)

c. use elimination by adding the equations together.

x-y=3

4x+y=-3

5x=0

x=0

then substitute into one equation to find y.

y=-4(0)-3

y=-3