Question

how to solve 4x-5y+16=0and 2x+y-6=0 by using substitution method??​

Answer 1

We have system of equation:

A:4x−5y+16=0−−−(1)

B:2x+y−6=0−−−−(2)

For A

x0−41 

y16/504 

For B

x031

y604

From solving equation (1) & (2), we get only one point (1,4) as

−7y+28=0⇒y=44x−5y+16=04x+2y−12=0−−+,x=26−y=22−4=1

So, the vertices are (−4,0),(1,4) & (3,0)