System of equations plus some problems?

Solve:
1) x² + 6x - 27 = 0 factorise this
(x+9)(x-3)=0>>>>x+9=0 and x-3 =0
x=-9; x=3
2) 2x²-5x + 3=0 factorise this
(2x-3)(x-1)=0>>>>2x-3=0 and x-1=0
2x=3, x=3/2 and x=1

3) 5/2y + 1/12 = 3/y (to rid fraction, multiply all by 12)
30/y +1=36/y ( to get the y from denominator to numerator, multiply all by y)
30 +y= 36>>>>y=6

4. System of equations: refers to a stack of adding or subtracting equations; also where substituiton is used to solve for variables
3a + b = 7
b= a -1, so a-b=1
3a + b = 7
1a- b = 1
---------
4a=8>>>a=2
substitute 2 in any equation
3a + b = 7>>>6 +b=7>>> b=1

5. 4x + y = -1 (mutiply this row by 3, then add to 2nd row)

12x +3y=-3
x - 3y = 16
----------
13x=13>>>>x=1
substitute 1 in either equation and find y
x - 3y = 16>>>1-3y=16>>>>>-3y=15
y=-5

1)
x^2 + 6x - 27 = 0
(x + 9)(x - 3) = 0

x + 9 = 0
x = -9

x - 3 = 0
x = 3

∴ x = -9 , 3

= = = = = = = =

2)
2x^2 - 5x + 3 = 0
(2x - 3)(x - 1) = 0

2x - 3 = 0
2x = 3
x = 3/2 (1.5)

x - 1 = 0
x = 1

∴ x = 3/2 (1.5) , 1

= = = = = = = =

3)
5/2y + 1/12 = 3/y
5(6)/2y(6) + 1(y)/12(y) = 12(3)/12(y)
30/12y + y/12y = 36/12y
30 + y = 36
y = 36 - 30
y = 6

= = = = = = = =

4)
3a + b = 7
3a = 7 - b
a = (7 - b)/3

b = a - 1
b = (7 - b)/3 - 1
3(b) = 3[(7 - b)/3 - 1]
3b = 7 - b - 3
3b + b = 7 - 3
4b = 4
b = 4/4
b = 1

3a + b = 7
3a + 1 = 7
3a = 7 - 1
a = 6/3
a = 2

∴ a = 2 , b = 1

= = = = = = = =

5)
4x + y = -1
4x = -1 - y
x = (-1 - y)/4

x - 3y = 16
(-1 - y)/4 - 3y = 16
4[(-1 - y)/4 - 3y] = 4(16)
-1 - y - 12y = 64
-y - 12y = 64 + 1
-13y = 65
y = 65/-13
y = -5

4x + y = -1
4x + (-5) = -1
4x = -1 + 5
x = 4/4
x = 1

∴ x = 1 , y = -5

(4x+y=-1) multiply by 3
12x+3y=-3
x-3y=16 add them
13x=13
x=1 12+3y=-3 3y=-3-12=-15 y=-15/3=-5


b=a-1 a-b=1
3a+b=7 add them 4a=8 a=8/4=2
2-b=1 2-1=b=1

Let me first answer your question re: "systems of equations":

One equation by itself can only have at most one variable in it... otherwise you can't fully solve. However, if you have two or more equations working together, so to speak, you can solve for more than one variable. This group of equations is called a 'system of equations'.

For example:

2x + y = 5

can't be solved because the values of x and y rely on one another. If x=0 then y=5, but if x=1 then y=3 .. ugh!

However, the 'system of equations'

2x + y = 5
4x - 2y = 6

CAN be solved because the two equations offer information about eachother.
(btw ~ x = 2, y = 1).

Now, for your solutions:

1)
x^2+6x-27 = 0
(x+9)(x-3) = 0
x =3, -9

2)
2x^2-5x + 3=0
(2x+1)(x-3)=0
x= 3, -1/2

3)
5/2y + 1/12 = 3/y
5+2y/12 = 6 <--- Multiply through by 2y
60+2y =72 <--- Multiply through by 12
2y=12
y=6

Now, for the systems of equations:

3a + b = 7
b= a -1

turn the two equations into one by
plugging the second equation into the first:

3a + b = 7
3a + (a-1) = 7
4a = 8
a=2

Now you know what 'a' is, plug it into one of the two original equations:

b = a -1
b = 2 - 1
b = 1

so your solution is a=2, b=1

next system:

4x + y = -1
x - 3y = 16

multiply the first one by 3 to get the 'y' s matched up.

12x + 3y = -3
x - 3y = 16

adding the two equations will cancel out the 'y' s:

13x =13

which means
x=1

Now you know what x is, plug it into one of the two original equations:

12x + 3y = -3
12(1) + 3y = -3
3y = -15
y=-5

so your solution is x=1, y=-5

1)x^2+6x=27 - transposing 27

thats the ans you stop solving as u have got unlike terms(different variables)

2)2x^2-5x=-3
3)5/2 y-3/y=1/12

5-6/2y=1/12

-1/2y=1/12
y=1/12 *-2
y= -1/6

1) x^2 + 6x - 27 = 0
(x+9)(x-3)=0
x = -9 or 3

2) similarly, factorise it and solve for x to get 3/2 and 1

3) multiply the whole eqn by y and solve for y

system of eqns just means a few equations together.
3a + b = 7 ... (1)
b= a -1 ... (2)
Susbt (2) into (1):
3a + a-1 = 7
=> a = 2
subst back into (2) to get b=1

Do the same for the next set of eqns.

System of equations means as it says ie a system involving a number of equations.
Question 1
(x + 9)(x - 3) = 0
x = - 9 , x = 3

Question 2
(2x - 3)(x - 1) = 0
x = 3/2 , x = 1

Question 3
Question is not clear.
Do you mean (5/2) y or 5 / (2y) ??
Should not have to guess but will try 5 / (2 y) :-
5 / (2y) + 1/12 = 3 / y
30 + y = 36
y = 6

Question 4
3a + (a - 1) = 7
4a = 8
a = 2
b = 1

Question 5
12x + 3y = - 3
x - 3y = 16-------ADD

13x = 13
x = 1

4 + y = - 1
y = - 5

x = 1 , y = - 5

PS
Hope this helps but remember to use brackets !