Hello, how would you solve x and y? Include the steps please.​

y=2 and x= 90

Step-by-step explanation:

13y+x+32y=180° [opposite angles of a cyclic quad]

Adding the like terms we get 45y+x=180°

Moving all the terms except x to the right side, we get x=180-45y. We will use this as the value of x(not an exact value)

2x-45y+x=180[opposite angles of a cyclic quad]

Adding the like terms we get 3x-45y=180°

Substitute the value of x into the equation 3x-45y=180°. You'll get something like 3(180°-45y)-45y=180°

Multiply the brackets out, you'll get 540°-135y-45y=180°. Adding the like terms we get 540-180y=180°

Moving all the other terms to one side and leaving - 180y we get: - 180y=180°-540°

Solving the right side we get - 180y= -360°

Dividing both sides of the equation by - 180, so that we are left with one variable(y), we get y=2

Substitute y=2 into the equation 3x-45y=180°. You'll get something like this: 3x-45(2)=180°

Multiplying out the brackets we get 3x-90°=180°

Moving all the terms to the other side so that we are left with one variable we get: 3x=180°+90°

Solving the right side we get 3x=270°.

Dividing out both sides of the equation by 3 we get x=90°

step-by-step explanation:

given that there are two triangles j'k'l' and jkl>

the former is got by dilating the second one.

when drawn on a grid, (graph paper) coordinates of vertices of first triangle is

(6,12)j, (9,6)k and (12,9)l

side jk has length = distance between j and k

=

side j'k'=distance between j' and k'

=

ie j'k'l' has sides 1/3 of that of corresponding sides of jkl.

hence scale factor of 1/3 is used to get j'k'l'from jkl