Solve the system of equations for x and y.
x+y=a
x−y=b

x = (a+b)/2

y = (a-b)/2

Step-by-step explanation:

x+y=a

x−y=b

Add this two equations together to eliminate y

x+y=a

x−y=b

2x = a+b

Divide by 2

2x/2 = (a+b)/2

x = (a+b)/2

Subtract the two equations to eliminate x

x+y=a

-x+y=-b

2y = (a-b)

Divide by 2

2y/2 = (a-b)/2

y = (a-b)/2

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.