Question 56003
Hi nearlyhairless, lets see what we can do to:
Are the following lines parallel, perpendicular, or neither?   L1 with equation x – 6y = 30   L2 with equation 6x + y = 6
A)	Parallel
B)	Perpendicular
C)	Neither
Put the equations in slope intercept form:  {{{y=mx+b}}}, m=slope and (0,b)= y-intercept.  Parallel lines have equal slopes and perpendicular lines have slopes that are negative reciprocals (opposite signs and upside down).
L1) {{{x-6y=30}}}
{{{-x+x-6y=-x+30}}}
{{{-6y=-x+30}}}
{{{-6y/-6=-x/-6+30/-6}}}
{{{y=highlight(1/6)x-5}}}
 slope=m=+1/6
L2)  {{{6x+y=6}}}
{{{-6x+6x+y=-6x+6}}}
{{{y=highlight(-6)x+6}}}
m=-6/1
(B) Perpendicular
:
Given 
{{{f(x)=x^2-x+7}}}
find f(0).
Let x=0
{{{f(0)=(0)^2-(0)+7}}}
{{{f(0)=0-0+7}}}
{{{highlight(f(0)=7)}}}
:
Given g(x) = 4x – 3, find g(3a).
Let x=3a
g(3a)=4(3a)-3
g(3a)=12a-3
:
Given f(x) = 4x + 4, find f(0)
let x=0
f(0)=4(0)+4
f(0)=0+4
{{{highlight(f(0)=4)}}}
:	
Solve the system by addition.	
L1) 5x – 3y = 13	
L2) 4x – 3y = 11
Multiply L2 by -1 and add L1 and L2
-1(4x-3y)=-1(11)
-4x+3y=-11
-----------
5x-3y=13
-4x+3y=-11
____________
x+0y=2
x=2
Substitute x=2 into L1 and solve for x.
5(2)-3y=13
10-3y=13
-10+10-3y=-10+13
-3y=3
-3y/-3=3/-3
y=-1
The solution is (x,y)=(2,-1)
Happy Calculating!!!
funmath and a littleweirdmath