X-2y= 15 2x + 4y = -18
OA. x= 1, y= -6
OB. x= 1, y=-7
OC. x= 3, y= -6
D. x = 3, y=-7

Option C: x=3 and y= -6 is the correct answer.

Step-by-step explanation:

Given that:

x - 2y = 15     Eqn 1

2x+4y = -18    Eqn 2

Multiplying Eqn 1 by 2

2(x-2y=15)

2x-4y=30    Eqn 3

Adding Eqn 2 and Eqn 3,

(2x+4y)+(2x-4y)=-18+30

2x+4y+2x-4y=12

4x=12

Dividing both sides by 4

Putting x=3 in Eqn 2

2(3)+4y=-18

6+4y=-18

4y= -18-6

4y= -24

Dividing both sides by 4

Hence,

Option C: x=3 and y= -6 is the correct answer.

option a, c,d are correct.

step-by-step explanation:

from the given figure, it is given that z is equidistant from the sides of the triangle rst, then from triangle tzb and triangle szb, we have

tz=sz(given)

bz=zb(common)

therefore, by rhs rule,δtzb ≅δszb

by cpctc, sz≅tz

also, from δctz and δasz,

tz=sz(given)

∠tcz=∠saz(90°)

by rhs rule, δctz ≅ δasz, therefore by cpctc, ∠ctz≅∠asz

also,from δasz and δzsb,

zs=sz(common)

∠zbs=∠saz=90°

by rhs rule, δasz ≅δzsb, therefore, by cpctc, ∠asz≅∠zsb

hence, option a, c,d are correct.