Solve the system of linear equations using elimination. 10x + 7y = 12
8x + 7y = 18
A)(3, 6)
B)(3, −6)
C)(−3, 6)
D)(−3, −6)​

C. (-3,6)

Step-by-step explanation:

Take 10x + 7y = 12 as first equation and 8x + 7y = 18 as second equation.

Subtracting both equations, you will get

10x + 7y - 8x - 7y = 12 - 18

2x = -6

x = -3

Substitute value of 'x' in any of the given equation :-

10x + 7y = 12

10 * -3 + 7y = 12

7y = 12 + 30

y = 42/7

y = 6

C) (−3, 6)

Step-by-step explanation:

10x + 7y = 12 ___ (−3, 6)

10×(-3) + 7×6 = -30 + 42 = 42 - 30 = 12

8x + 7y = 18 ___ (−3, 6)

8×(-3) + 7×6 = -24 + 42 = 42 - 24 = 18

37

step-by-step explanation:

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}