Which first step for solving the given system using substitution results in an equation without fractions? Solve for x in the first equation. Solve for y in the first equation. Solve for x in the second equation. Solve for y in the second equation.

solve for x in the second equation is correct

Step-by-step explanation:

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Which first step for solving the given system using substitution results in an equation without fractions?

(2x+6y=7

(6x + 18y = 24 2

Solve for x in the first equation,

Solve for y in the first equation,

Solve for x in the second equation

The best equation to she will be equation 2 since all the coefficients are end numbers

Let us make x t subject f the formula from equation 2;

From 2: 6x+18y = 24

Subtract 18y from both sides

6x+18y-18y = 24-18y

6x = 24-18y

Divide through by 6

6x/6 = 24/6 - 18y/6

x = 4-3y.

You can see that the resulting expression of x is not a fraction. Hence solve for x in the second equation is correct

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}

answer: what are the answer choices?

step-by-step explanation: