Solve the system.
y= -2x + 1

y= 2x - 3

Equation 1: y = -2x + 1

Equation 2: y = 2x - 3

Since both equations already have y isolated, we are able to simply set the right side of both equations equal to each other. Since we know that the value of y must be the same, we can do this.

-2x + 1 = 2x - 3

1 = 4x - 3

4 = 4x

x = 1

Then, we need to plug our value of x back into either of the original two equations and solve for y. I will be plugging x back into equation 2 above.

y = 2x - 3

y = 2(1) - 3

y = 2 - 3

y = -1

Hope this helps!! :)

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.