solve the systems pf equations alegbraically: y+4=3x y=x^2+5x-39

y+ 4 = 3x...........(1)

y=
x^2 + 5x -39 .............(2)

We will use the substitution
method to solve.

We will substitute y into equation
(1).

==> x^2 + 5x -39 + 4 =
3x

Now we will combine all
terms.

==> x^2 + 2x - 35 =
0

Now we will
factor.

==> (x+7) (x-5) =
0

==> x1= -7 ==> y1= 3x-4 = 3*-7 -4=
-25

==> x2= 5 ==> y2 = 3*5 -4 =
11

Then we have two sets of solution for the
system.

The solution is the pair ( -7, -25)
and ( 5, 11)