I'm trying to help my sister with her Pre-Calculus homework. But...I need help help too. Please help us.?

Question

It's been a while since I've done this and I wanna to know if my answers are correct. So I'll use whatever answers you provide and my reference. Thanks in advance. Here goes:

Solve the system. There are three different systems.
If the system has infinitely many solutions, express your answer in the form "X = X" and "Y" as a function of "X".

Your answer is: X=_____ and Y=_____.

1st System.
-x + y = 0,
4x - 3y = -3

2nd System.
3x + 2y = 6,
x - 2y = -6

3rd System.
2x - 6y = 2,
-3x + 9y = -3


My sister and I thank you all in advance for helping us. Mainly her. =)

Answer 1

1) x = y
Substituting:
4x - 3x = -3
x = -3
y = -3

2) 3x + 2y = 6
x = 2y - 6
Substituting:
3(2y - 6) + 2y = 6
6y - 18 + 2y = 6
8y = 24
y = 3
x = 2(3) - 6
x = 0
y = 3

3) 2x - 6y = 2
-3x + 9y = -3
If you divide the top by 2 and the bottom by -3 you get:
x - 3y = 1
x - 3y = 1

These are equivalent, so you can't determine a single solution.

The best you can do is simplify the equation to get a formula for y given x:
3y = x - 1
y = (1/3)x + (-1/3)
Because you can't isolate a single variable, you essentially have one equation and two unknowns. The solutions will lie along the line with a slope of 1/3 and a y-intercept of -1/3.

Answer 2

Hi!

A few tips: in each system, use algebra to isolate y or x, whichever looks easier to do. Then, use that new definition of y or x in the other equation. That equation will then only contain one variable and therefore it can be solved.

(Remember that the equals sign means that the two terms on either side of it are indeed equal and can be substituted for each other.)

The best thing is that you can check your answers! Try plugging in your values of x and y, both equations should come out true.
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In the first system, -x + y = 0. This is the same as saying y=x. (Essentially, we are saying y is the same as x.) Since we know this, we can change the second equation by substituting x any place we see y. We get

4x - 3x = -3 ---> x = -3.

The second system is a little tougher because it isn't quite as obvious what y or x is.

Therefore x = -3. To find y, substitute x = -3 any place you see x in either of the equations (typically you use the other equation). Since y is the same as x here, y is also -3.
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In the second system, it would be easiest to use

x - 2y = -6 -----> x = 2y - 6

Notice that I am now expressing x only in terms of y. Substitute that new x into the other equation.

3(2y-6) + 2y = 6 Now you should distribute the 3

6y - 18 + 2y = 6 Now gather terms

8y = 24; y = 3.
Now that we know what y is, use that new value of y in the other equation.

x - 2y = -6
x - 2(3) = -6 ----> x = 0
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Again, defining one variable in terms of the other. I'll use
2x - 6y = 2 (given)
2x = 6y + 2
x = 3y + 1 (x is now by itself!)

Now I'll use that in the other equation.
-3x + 9y = -3
-3(3y+1) + 9y = -3
-9y - 3 + 9y = -3
0 = 0 ????

Oh no! The y terms canceled out giving a useless (but true, and that's important) statement 0 = 0. Don't worry! This is because the two equations are multiples of each other and are effectively the same equation. And as we know a single equation can have infinite solutions. So per the directions, let's write y as a function of x.

2x - 6y = 2 (you can use either equation as usual)
-6y = 2 - 2x
6y = 2x - 2
y = (1/3)x - (1/3) and x = x.

HOPE THAT HELPED! :-)

Answer 3

1st X = -3, Y = 3

2nd X = 0, Y = 3

3rd Infinitely many solutions for both X and Y

Answer 4

1) x = y
so 4y - 3y = -3 --> y = -3 and x = -3

2) x = 2 - 2y/3
so 2 - 2y/3 - 2y = -6
8 = 8y/3 --> y = 3 and x = 0

3) x = (2 + 6y)/2 --> x = 3y +1
-3(3y + 1) + 9y = -3
-9y + 9y -3 = 3 --> INFINITE # OF SOLUTIONS (THESE 2 EQUATIONS ARE THE SAME LINE y = (x - 1)/3 )

so x = x and y = (x - 1)/3

Answer 5

1st system:
X = -3 Y = -3

2nd system:
X = 0 Y = 3

3rd system:
Infinitely many solutions
X = X Y = (X - 1)/3

Answer 6

1.-x+y=0 times 4
-4x+4y=0
4x-3y=-3
adding
y=-3
plugging in (1)
-3+x=0
x=3

2.3x+2y=6
x-2y=-6
adding
4x=0
x=0
substituting
0-2y=-6
y=3

3.2x-6y=2
-3x+9y=-3
(1)*3 and (2)*2
6x-18y=6
-6x+18y=-6
since both the equations are multiples/coincidental the solutions are infinite

Answer 7

1st sys:
X= -3, Y= -3

2nd sys:
X=0, Y=3

3rd sys:
X=x, Y=(x-1)/3

Answer 8

1st system
x = y = -3
2nd system
x=-3 , y = 3/2
3rd system
has no answer

Answer 9

sin(2arcsin(2x)) = 2sin(arcsin(2x))*cos(arcsin(2x)) = 4xcos(arcsin(2x)) enable z = arcsin(2x) => sin(z) = 2x and cos(z) = ?(a million - 4x^2) hence sin(2arcsin(2x)) = 4x?(a million - 4x^2)