What is the solution of the system of equations? -4x+3y=-12 -2x+3y=-18 *Note: I believe it has no solution, but I would like to make sure. Thank you all for your help.

Step-by-step explanation:

I've enclosed a graph to show you that there is one solution.

-4x + 3y = - 12

- 2x + 3y = - 18           Subtract

-2x = 6                        Divide by - 2

-2x/-2 = 6/-2

x = - 3

- 4x + 3y = - 12

-4(-3) + 3y = - 12

12 + 3y = - 12

3y = - 12 - 12

3y = - 24

3y/3 = - 24/3

y = - 8

Which is exactly what the graph shows.

You can tell immediately that this has 1 or more solutions just by looking at the number in front of x and y in each each equation.

It's always a good idea to graph a question like this one. Desmos is a pretty good tool to use if you don't have a graphing calculator.

The y's are the same so you are 1/2 way home in saying what you did.

The x's are different so that means you have at least 1 solution.

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}

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step-by-step explanation:

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