how do i solve this linear equation?

Question

2y-6=2(x+2)
1.) I have to find the slope
2.) A line parallel to the given equation 2y-6=2(x+2) passes through point (2,1). Find the equation of the parallel line.
3.) A line perpendicular to the given equation 2y-6=2(x+2) passes through point (2,1). Find the equation of the perpendicular line.

Someone PLEASE help I am HORRIBLE at math!

I have summer school..........

Answer 1

rewrite it:
2y- 6 = 2(x+2)
2y - 6 = 2x + 4
2y = 2x + 10
y = 1x + 5

slope is 1. lines parallel have the same slope (which means the equations start out y = 1x + something). plug in the point you know to get "something".
y = x + b
1 = 2 + b
b = -1
so the parallel line is y = x - 1

perpendicular lines have "negative reciprocal" slopes, so slopes of lines perpendicular to y = x + b look like y = -x + c. again plug in point.
y = -x + c
1 = -(2) + c
3 = c
so the perpendicular line is y = -x + 3.

Answer 2

You need to convert the answer into the standard form Y=mX + b. Where m is equal to the slope and b is the point at which your line crosses the Y axis over point 0 for x (the center of the graph, or point (0,b) . Isolating Y is fairly simple:

1. Add 6 to both sides. equation is now 2y=2(x+2) + 6
2. Change 2(x=2) to 2x + 4 on the right side : 2y=2x+4+6
3. resolve addition on the right side 2y=2x+10
4. Divide both sides by 2 to isolate y: y=x+5

Now you have y alone, and x alone. Your slope is 1 because you only have ONE x (get it?), rather than some number times x.

For part 2 you simply need to draw out your first line, that goes through point (0,5) you got this number from the five at the end of the equation, a.k.a. "b", and has a slope of 1. This means it goes through point (5,0) as well. So draw a dot at both points and connect with a ruler. Now find point (2,1) Draw a line with a slope of 1 through this point, a slope of 1 will make it parallel to the other line. *hint* a slope of 1 will make this line pass through (0,-1) as well as (2,1). The equation of this line is Y=x-1, because the slope is 1 and the Y intercept is -1.

The third question is similar to the second except the line slope will be negative one and the y intercept will change. This line goes in the opposite direction of the second and makes a big X on your graph. The equation of this line is Y=-x+3.

Now write all of this down and show your teacher. Draw out the lines and you should have a line on the left side of your graph going from bottom left to upper right. Another line like that to the right side of your graph, and the last one crosses this line into a big X.

But, be prepared to explain that:

1. A positive slope tells me how many times a line goes up and over for each square on the grid. In your case once up and once over each square.
2. Parallel lines look like two tall guys next to each other, or like 2 bulbs in a florescent light fixture.
3. Perpendicular lines look like the two lines on an X, but only when they cross at EXACTLY 90 degree angles.

Answer 3

The slope intercept form of a equation is given by
y = mx + b
where m is the slope and b is the y-intercept
if you can write the given equation in the form of y = mx + b
you can easily identify the m value.
we need to solve for y. Solving for y means we need to undo whatever is done to y
2y - 6 = 2(x+2)
you can see 2 is multiplied with y... so to undo it.. we need to perform the reverse operation of multiplication. Reverse operation of multiplication is division. so
divide 2 on both side of the equation .. that will give you
2y/2 - 6/2 = 2(x+2)/2
y - 3 = x + 2
Now 3 is subtracted from y.. so we need to add 3 on both sides
y - 3 +3 = x + 2 +3
y = x + 5
y = mx + b
comparing them you can see m = 1
so answer for you first question is slope = 1
--------------------------------------------------------
2) We need to find the equation of line which is parallel to the line y = x + 5 and passing through (2, 1).
Note: Slope of parallel lines are equal.
so we need to find equation of line whose slope, m = 1 and passing through the point (2, 1)
When a point and slope of a line is given.. we can use point slope formula to construct the equation of line which is...
y - y1 = m(x - x1) where (x1, y1) is the point and m is the slope
so y - 1 = 1(x - 2) is the line passing through point (2,1) and
have the slope 1
-----------------------------
The slope of line is negative reciprocal to the slope of the line perpendicular to it..
y = x - 1 has slope 1
the slope of the line perpendicular to it will be 1/-1 = -1
so we need to find the line passing through (2, 1) and having the slope m = -1
Point slope formula : y - y1 = m (x - x1)
plugging what we know: y - 1 = -1(x - 2)
Hope this helps.
you can check the site below.. it will give you a good insight which help you to do these kind of problems.

Answer 4

1) to find the slope, you need to change the form of the equation into y=mx+c, where
m= gradient/ slope, c= y-intercept
2y-6=2(x+2)
2y-6=2x+4
2y=2x+10
divide both sides by 2, we got
y=x+5
m=1
The slope is 1
2) The slope of two parallel lines are always equal,
so the slope of the line you're looking for is also 1.
since it pass through (2,1), you can use the formula
y-y1=m(x-x1)
y-1=1(x-2)
y-1=x-2
y=x-1
the equation is y=x-1
3) The slope of two perpendicular lines are defined as
m1xm2=-1, where
m1= slope of first line
m2= slope of the second line
so the slope of the line you're looking for is -1
using the formula y-y1=m(x-x1), just like before, we got
y-1=-1(x-2)
y-1=-x+2
y=-x+3
the line equation is y=-x+3

Answer 5

Start by reducing the equation to y= something.
2y-6=2(x+2)
2y-6 = 2x +4
so, 2y= 2x + 10
so, y= x+5
So the slope is 45 degrees and the line passes through the y axis at +5. Draw it out on graph paper and then astound yourself by getting the rest of the answers for yourself!

Answer 6

1)Solve for y

2y-6 = 2(x+2)
2y-6 = 2x+4
2y-6+6 = 2x +4 +6
2y = 2x +10
y = x + 5

so the slope is +1 (coefficeient of x) and the y intercept is +5

2) y = x-1

3) y = -x +3

Answer 7

2y-6=2(x+2)
it can be written as
y=x+5
comparing with
y=mx+c
m=1
slope is 1

parallel lines have equal slopes and perpendicular lines have the product of their slopes = -1

m1=1 m2= -1
use
y2-y1=m(x2-x1)

Answer 8

isnt it summer vacation? um. yeah im horrible at math too. sorry my answers no help.