A system of equations and its solution are given below. System A Complete the sentences to explain what steps were followed to obtain the system of equations below. System B To get system B, the equation in system A was replaced by the sum of that equation and times the equation. The solution to system B the same as the solution to system A.

To get to system B the second equation in system A was replaced by the sum of that equation and the first equation multiplied by 4. The solution to system B is the same as the solution to system A.

Step-by-step explanation:

A system of equations and it’s solution are given below.

System A.

x-y=3

-2x+4y=-2

Solution: (5,2)

Complete the sentences to explain what steps were followed to option the system of equations below.

System B

x-y=3

2x=10

To get to system B the (first/second) equation in system A was replaced by the sum of that equation and the (first/second) equation multiplied by (42-3) . The solution to system B (is/ is not) the same as the solution to system A.

Solution:

To get system B:

First write the first equation in system A. x - y = 3

Multiply the second equation in system A by 4 and add to the first equation in system A to get:

2x = 10

x = 10/2 = 5

Put x = 5 in x - y = 3

5 - y = 3

y = 5 - 3 =2

The solution to system B is the sam as that of system A

To get to system B the second equation in system A was replaced by the sum of that equation and the first equation multiplied by 4. The solution to system B is the same as the solution to system A.

geometric series converge the common ratio r has magnitude less than one.

which is equivalent to

for the example we need

|2-x| < 1

-1 < 2-x   and 2-x < 1

x < 3 and x > 1

we can write that

answer: 1 < x < 3

1a. common ratio r=250/25=10 > 1     diverges

1b.   r=50/100=1/2 converges

when these converge they converge to

where a is the first term.

answer: 200

1c. r=-1   so it's not the case that |r| < 1.     diverges

2a. x is the common ratio so we converge when |x|< 1

2b.   r=-10x / -2 = 5x.

we converge when

|5x| < 1

|x| < 1/5

2c.   r=3-2x

-1 < 3-2x and 3-2x < 1

-4 < -2x and 2 < 2x

x < 2 and 1 < x

1 < x < 2

2d. [/tex] r=-48x^3/12 = 4x^3[/tex]

|4x^3| < 1

|x^3| < \frac 1 4

|x| < 4^{-1/3}

10000

step-by-step explanation: divide by 2,