# Thread: A few Algebra 2 problems I need help with.

1. ## A few Algebra 2 problems I need help with.

I got all these wrong on my test and need help doing them for the remake.

1.

write an equation for each line in slope-intercept form

-passing through (-1,3) & (4,2).
-passing through (-2,3) and perpendicular to the line whos equation is y = -1/2x -4.
-passing through (6,-4) and parrallel to the line whose equation is x + 2y + 5.

2.

Solve by graphing

x + y = 6
4x - y = 4

This is one question.

3.

solve each system by subsitution or elimanation method

4x = 2y + 6
y = 2x - 3

2. Originally Posted by garbles
I got all these wrong on my test and need help doing them for the remake.

1.
write an equation for each line in slope-intercept form

-passing through (-1,3) & (4,2).
-passing through (-2,3) and perpendicular to the line whos equation is y = -1/2x -4.
-passing through (6,-4) and parrallel to the line whose equation is x + 2y + 5.

2.

Solve by graphing

x + y = 6
4x - y = 4

This is one question.

3.

solve each system by subsitution or elimanation method

(1) 4x = 2y + 6
(2) y = 2x - 3
Hi garbles,

It's my theory that if you post only 1 question per thread, you'll get your answers quicker. Several 'helpers' can be working on them at once.

I'll get you started with #3.

The second equation is already solved for y so we'll substitute that back into equation 1.

$4x=2(2x-3)+6$

$4x=4x-6+6$

$0=0$

Whoops! Looks like these two equations are the same. Therefore the lines they graph coincide. We have an infinite number of solutions. Closer inspection of the original equations reveal that equation (1) is two times equation (2).

3. thanks for answering number 3 for me.

4. Originally Posted by garbles
I got all these wrong on my test and need help doing them for the remake.

2.

Solve by graphing

(1) x + y = 6
(2) 4x - y = 4

This is one question.
For this one, solve each equation for y.

(1) y = -x + 6

(2) y = 4x - 4

Now set up a table of arbitrary x-values for each equation and find the corresponding y-value. Then graph the points, draw a straight line through the points and see if you can determine where the two lines intersect. From the tables, we can see...

Code:

(1) y = -x + 6                    (2) y = 4x - 4
________                            _________
x   |   y                            x  |   y
-----------                         ------------
0   |   6                            0  |  -4
1   |   5                            1  |   0
2   |   4                            2  |   4
3   |   3                            3  |   8
Now you can see from the table that the third entries in each table are identical. Guess what that means?

5. Originally Posted by garbles
I got all these wrong on my test and need help doing them for the remake.

1.
write an equation for each line in slope-intercept form

(a) -passing through (-1,3) & (4,2).
(b) -passing through (-2,3) and perpendicular to the line whos equation is y = -1/2x -4.
(c) -passing through (6,-4) and parrallel to the line whose equation is x + 2y + 5.
Ok, garbles, here are some hints for the first one.

(a) First, find the slope.

$m=\frac{y_2-y_1}{x_2-x_1}=\frac{2-3}{4-^-1}=\frac{-1}{5}$

Next, use the slope-intercept form of the linear equation to find the y-intercept.

$y=mx+b$

Use either point to substitute for x and y. m will be the slope we just found. Let's use point (4, 2).

$2=\frac{-1}{5}\left(4\right)+b$

Solve this thing for b. Then, use the slope-intercept form of the equation substituting for m and b and you're done.

6. Ok thanks alt

7. Originally Posted by garbles
I got all these wrong on my test and need help doing them for the remake.

1.
write an equation for each line in slope-intercept form

-passing through (-1,3) & (4,2).
using the general form $y = mx+c$ with $m = \frac{y_2-y_1}{x_2-x_1}$

you will get $m = \frac{2-3}{4-(-1)} = \frac{-1}{5}$

sub into original equation you get $y = \frac{-1}{5}x+c$ now use one of your points above to find c.

All good?