What is the solution for the system of equations?
y= 2x - 3
4x - 3y = 31

Use Substitution.

Plug in 2x-3 in place of y in the second equation.

4x - 3(2x-3) = 31

4x - 6x + 9 =31

            -9     -9

-2x=22 so x=-11

Then, y=2(-11)-3, so y=-25

So the solution would be (-11,-25).

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.

idk lol

step-by-step explanation: