algebra 2 help?

Question

Graph each system of equations and state its solution.
y=(3/2)x + 19/2
x - y = -7

A (0, -2)
B (-3, 2)
C (-5, 2)
D (2, 5)
E (-2, -4)
F No Solution

Answer 1

We can't really graph them for you, so you're on your own there.

Draw each one on the same graph and see where they intersect. (The answer is C)

Answer 2

[C] is the correct answer:

As user (Mathematica) said, we can't make a graph here but what you need to do is just to draw a cartesian coordinates with two axes of x and y and give two arbitrary points in each equation to draw the lines.

For the first equation: (x_1, y_1) & (x_2, y_2)
and for the second equation: (x_3, y_3) & (x_4, y_4)

For instance you may plug (x_1 = 0) in the first equation and you will find (y_1 = 9.5) and you may plug (x_2 = - 4) and you get (y_2 = 3.5) then you draw the first line.

For the second line, you may plug (x_3 = 0) and you get (y_3 = 7) and you may plug (x_4 = -4) and you get (y_4 = 3). Now you draw the second line.

Then what you need is a ruler to extend both lines in a way to intersect each other and you will see the intersection point.

A quick solution is just to plug in the answers one by one into the second equation (x - y) = -7 which is simpler equation.

A: when you plug x = 0 and y = -2,
0 - (-2) = 2
It does not satisfy the equation! It should be -7 if the point is on the line in the second equation!
So [A] is incorrect!

B: x = -3 and y = 2 so again
-3 - 2 = -5 ! Not (-7)
So [B] is incorrect!

C: x = -5 and y = 2 so
-5 - 2 = -7 !
So [C] can be the correct answer!

D: x = 2 and y = 5 so again
2 - 5 = -3 ! Not (-7)
So [D] is incorrect!

E: x = -2 and y = -4 so again
-2 - (-4) = 2 ! Not (-7)
So [E] is incorrect!

Now you plug the only possible correct answer (-5, 2) into the first equation to see if it satisfy the equation:
x = -5 and y = 2 so
2 = (3/2) * (-5) + (19/2) = (-15/2) + (19/2) = (-15 + 19)/2 = 4/2 = 2!

So certainly [C] is correct answer!

For an analytical solution to this, you may do this:

You have these two equations:
y = (3/2)x + 19/2 (eq - 1)
x - y = -7 (eq - 2)

(eq - 2) can be written as: y = x + 7
So to intersect we need to equal the "y" in both equations:So you have:

x + 7 = (3/2)x + 19/2

then:
(3/2)x - x = 7 - (19/2) which gives you:
((3-2)/2)x = ((14-19)/2) so:
(1/2)x = (-5/2) so

x = -5

Then you insert (x = -5) in either equations (eq-1 or eq-2) and you get the "y" for the intersection:

For instance you insert (x = -5) in (eq-2) which is simpler and you get:

(-5) - y = -7 which gives you: y = (-5) + 7 and that finally gives you "y" as:

y = 2

So the intersection point is: ( -5 , 2) which is the answer in [C]

Answer 3

2y = 3x + 19

3x - 2y = - 19
x - y = - 7

3x - 2y = - 19
- 2x + 2y = 14---ADD
x = - 5

10 + 2y = 14
2y = 4
y = 2

x = - 5 , y = 2
ANSWER C