A system of linear equations includes the line that is created by the equation y = 0.5 x minus 1 and the line through the points (3, 1) and (–5, –7), shown below. On a coordinate plane, points are at (negative 5, negative 7) and (3, 1). What is the solution to the system of equations? (–6, –4) (0, –1) (0, –2) (2, 0) Mark this and return

htrhyjk

step-by-step explanation:

oiluiylkyuk4515kt

u65ku45kuyfr

4y - 6 + y^2 - 9 = y^2 + 4y - 15

y^2 term

terms

- terms

total of 3 terms when simplified

step-by-step explanation:

place r′ so that qr′ is parallel to the tangent.

now we get that ∠r′py=∠pr′q, where y is on the tangent, on the opposite side of p from x.

also, because of the symmetry, ∠r′py=∠qpx.

finally, note that no matter where we place r, as long as it's on the correct arc, we have ∠prq=∠pr′q.

stringing all these equalities together, we end up with ∠prq=∠qpx.

this picture will show how its going to look like.

hope this by jack monhen