Find the ages of the people and tree

callum is five years younger than eilidh caitlin is 15 years older than eilidh. the sum of their ages is 31. ewan has a tree in his garden which is seven times old as he is their ages add up to 352

show working out plz

1. Callum = 2; Eilidh = 17; Caitlin's =22.

2. Ewan = 44; tree = 308.

Explanation:

Q1. Callum, Eilidh, and Caitlin

Let x = Callum's age

     y = Eilidh's age

     z = Caitlin's age

We have three conditions.

(1)  x = y - 5

(2) z = y + 15

(3) x + y + z = 31

Step 1. Eliminate one of the variables in two of the equations

Subtract (1) from (2):                      z - x = 20

Rearrange:                 (4)             -x + z = 20

Solve (2) for y:           (5)                     y = z - 15

Substitute into (3):         x +  z - 15  + z = 31

                                 (6) x + 2z             = 46

Step 2. Set up two new equations in two variables

(5)                          -x +   z = 20

(6)                           x + 2z = 46

Add (5) and (6):           3z = 66

Divide each side by 2: z = 22

Step 3. Substitute z into (5)

-x + 22 = 20

 x          = 2

Step 4. Substitute x into (1)

                                    2 = y - 5

Add 5 to each side     y = 7

Callum's age is 2; Eilidh's age is 7; Caitlin's age is 22.

Q2. Ewan and the tree

Let x = Ewan's age

      y = tree's age

We have two conditions.

                                     (1)          y = 7x

                                     (2)   x + y = 352

Rearrange (1)                (3) -7x + y =     0

                                     (2)    x + y = 352

Subtract (3) from (2)          8x       = 352

Divide each side by 8         x       =    44

Substitute y into (1)                    y = 7 × 44

                                                  y =   308

Ewan's age is 44; the tree's age is 308.

there is no picture

step-by-step explanation:

put up a picture so people can you

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.