algebra please anyone help?

1st problem:
Value of x:
3x - 4 = - 2x + 1
5x = 5
x = 1

Value of y:
= 3(1) - 4
= 3 - 4
= - 1 OR

= - 2(1) + 1
= - 2 + 1
= - 1

Answer: x = 1, y = - 1

2nd problem:
Value of x:
4/3x - 2 = 2/3x
4x - 6 = 2x
2x = 6
x = 3

Value of y:
= 4/3(3) - 2
= 4 - 2
= 2 OR

= 2/3(3)
= 2

Answer: x = 3, y = 2

3rd problem:
Value of x:
1/4x - 1 = - 2x - 10
x - 4 = - 8x - 40
9x = - 36
x = - 4

Value of y:
= 1/4(- 4) - 1
= - 1 - 1
= - 2 OR

= - 2(- 4) - 10
= 8 - 10
= - 2

Answer: x = - 4, y = - 2

1. 3x-4=-2x+1
5x=5
x=1

y=3-4
y=-1

2. 2/3x=4/3x-2
-2/3x=-2
x=3

y=2/3(3)
y=2

3. 1/4x-1=-2x-10
9/4x=-9
x=-4

y=-2(-4)-10
y=8-10
y=-2

Question 1
3x - 4 = - 2x + 1
5x = 5
x = 1
y = - 1

Question 2
(4/3) x - 2 = (2/3) x
16x - 24 = 8x
8x = 24
x = 3

y = (2/3) 3
y = 2

x = 3 , y = 2

Question 3
- 2x - 10 = (1/4) x - 1
- 8x - 40 = x - 4
- 36 = 9x
x = - 4

y = - 1 - 1
y = - 2

x = - 4 , y = - 2

1)
y = 3x - 4 (solve by using elimination)
y = -2x + 1

...y = 3x - 4
‒) y = -2x + 1 (subtraction)
-------------------------
0 = 5x - 5

0 = 5x - 5
5x = 5
x = 5/5
x = 1

y = 3x - 4
y = 3(1) - 4
y = 3 - 4
y = -1

∴ x = 1 , y = -1

= = = = = = = =

2)
y = 4/3(x) - 2 (solve by using substitution)
y = 2/3(x)

y = 2/3(x)
y = 2x/3

y = 4/3(x) - 2
2x/3 = 4x/3 - 2
2x/3 - 4x/3 = -2
-2x/3 = -2
-2x = 3(-2)
-2x = -6
x = -6/-2
x = 3

y = 2/3(x)
y = 2/3(3)
y = 6/3
y = 2

∴ x = 3 , y = 2

= = = = = = = =

3)
y = 1/4(x) - 1 (solve by using substitution)
y = -2x - 10

y = 1/4(x) - 1
y = x/4 - 1

y = -2x - 10
x/4 - 1 = -2x - 10
x/4 + 2x = 1 - 10
4(x/4 + 2x) = 4(-9)
x + 8x = -36
9x = -36
x = -36/9
x = -4

y = -2x - 10
y = -2(-4) - 10
y = 8 - 10
y = -2

∴ x = -4 , y = -2

1) (1, -1)

2) (3, 2)

3) (-4, -2)

For each one just input one value of y for the other. You'll get the value of x and then substitute that value in to get y.

1) y = 3x - 4
....y = -2x + 1

3x - 4 = -2x + 1
3x + 2x = 1 + 4
5x = 5
x = 1

y = 3x - 4
y = 3(1) - 4
y = -1

Answer is (1,-1)

2) y = (4x/3) - 2
....y = 2x/3

(4x/3) - 2 = 2x/3
(4x/3) - (2x/3) = 2
(2x/3) = 2
2x = 6
x = 3

y = (4x/3) - 2
y = (4(3)/3) - 2
y = 4-2
y = 2

Answer is (3,2)

3.) y = (x/4) - 1
.....y = -2x - 10

(x/4) - 1 = -2x - 10
(x/4) + 2x = -10 + 1
(x/4) + (8x/4) = -9
9x/4 = -9
9x = -36
x = -4

y = (x/4) - 1
y = (-4/4) - 1
y = -2

Answer is (-4,-2)

traditional ordered pairs use 0, 1, -1, 2 for x values.
just start plugging in the values for x in each equation to get the y value and then you have an ordered pair. I think this is what you are looking for, the question is a bit vague. For example, in #1
(0,-4), (1,-1), (-1,-7), (2,2)
(0,1), (1,-1), (-1,3), (2,-3)

I hope this is what you are looking for.

1.) y = 3x - 4
y = -2x + 1

so 3x-4 = -2x + 1 --->x = 1 and y = 3(1)-4 = -1
the ordered pair (1,-1)

similarly for 2 or 3/

I know Im about to complicate this but...whatever here goes! What you have in front of you is a square 2x2 matrix and what you are trying to do is find the solution of the Vector space with the basis vectors. Equation 1 can be rewritten as

3x-y=4
-2x-y=-1 so 2x+y=1

The system becomes:

3x-y=4
2x+y=1

Therefore

[3 -1 [x = [4
2 1] * y] 1]

Let [3 -1
2 1] be Matrix A and and [x is Matrix B hence [4 is Matrix C
y] 1]

So we have A*B=C
MATRIX MULTIPLICATION IS NOT COMMUTATIVE...

AB IS NOT BA order of multiplication matters as we are dealing with matrices and the determination of the dot products will differ if you multiply AB or BA.....

Dimension of a Vector Space is the number of linearly independent basis vectors that span the set...in this case we have 2 vectors that are linearly independent so the dimension of the space is TWO (2) and the basis spans the set because if we remove one of the vectors, we no longer have the same vector space!

If we remove the second equation from our set of 2, we no longer can find an ordered pair! Makes sense, huh?

We need to find A^-1 (Inverse of A) such that B=(A^-1)*C It is important to note that you can only find inverses of square matrices. This is a 2x2 matrix (2 equations in 2 variables) so it has an inverse.
Also (A)(A^-1) = I the Identity Matrix [1 0
0 1]

so then the solution becomes I*B=C*A^-1 or [1 0 [x = (A^-1)*C
0 1] * y]
[x = (A^-1)*C
y]

Notice that the inverse of A is also of dimension 2 this will always be the case even for n dimensional matrices....

Matrix multiplication is essentially finding dot products between the rows and columns of matrices noting that the resultant values are the solution of the matrix product in corresponding order.

A nxw matrix when multiplied by a nxl matrix will yield a wxl matrix
In our case a (2x2) matrix times a (2x1) yields a (2x1) matrix

This is how we know that the result of (A^-1)*C will be a 2x1 matrix which will correspond with the [x matrix! These are your solutions!
y]
To find the inverse we shall use the following method:

Find the determinant of Matrix A, to find the determinant:

[a b
c d] det = (ad-bc) so in our case det A = (3*1-2*(-1))

det A= (3+2) = 5

Then rewrite [a b as adjoint matrix [d -b in our case [3 -1 is then [1 1
c d] -c a] 2 1] -2 3]

Multiply 1 / det A * [d -b which would be (1/5) * [1 1 = [1/5 1/5
-c a] -2 3] -2/5 3/5]

This is your inverse matrix! :-D Now....find the dot products

[x = [1/5 1/5 [4 = [4*(1/5) + 1*(1/5) = [(4/5)+(1/5) = [5/5
y] -2/5 3/5] * 1] 4*(-2/5) + 1*(3/5)] (-8/5) +(3/5) -5/5]

Which is....OF COURSE!

[x = [1
y] -1]

Why did I do all this? You presented a vector space of dimension 2 (two equations in two unknowns)....but it could have easily been 3, 4, 5...n equations in n unknowns! This would make the problem VERY difficult to solve using conventional methods...

This method will obtain the solutions for systems with n equations in n unknowns....pretty powerful wouldnt you say! ;-)

There are slight differences though....as the dimension of the vector space increases, its determinant (which is corresponds to the area of parallelogram inside the vector space!) becomes increasingly hard to calculate and DOES NOT follow the method I showed you above. That only works for 2x2 matrices and I chose it out of convenience! I will not go into detail as to how to do it for larger matrices it becomes NEEDLESSLY COMPLICATED! HAHA lol :-D

This is REAL ALGEBRA....dont worry, I know what you are doing is brainless operating but viewing it from a deeper mathematical perspective is fascinating! That's why MOST people don't like math...they have never been exposed to the deeper concepts of it and find it "worthless" hahaha I PITTY those people! :-) If they only knew that this IS the way computers communicate with each other and interpret information, through SYSTEMS OF NUMBERS...If it were BS as many claim it to be, I wouldn't be typing the answer to your question! :-) Enjoy your math and DON'T let it frustrate you, if you do, step away from it for a while, do something fun and come back to it. Eventually, you WILL get it! ;-)