Urgent pls help me out.!!!!!!!!!!?

Question

Solve each of the following systems of linear equations.
(a)

5x - y = 11
3x + 2y = 4

(b)
3x + y = 2
y = -3x -5

Answer 1

(a) use substitution
change 5x-y=11 to y =5x-11
then substitue that into the other equation
3x+ 2(5x-11)=4
3x+ 10x-22=4
13x-22=4
13x=26
x=2
plug back in
5x-y=11
5(2)-y=11
10-y=11
-y=1
y=-1

(b) do the same thing and subsitute, this time you do not have to change an equation
3x+ (-3x-5) =2
3x -3x -5=2
-5=2
this equation does not work and has no solution (since both sides are not equal)

hope I helped!
email me if you have any more questions...it should be on my profile
~SmS~


ADDED

Elimination process-

5x-y=11
3x+2y=4

multiply both equations to make one of the variables additive inverses

2(5x-y)=2(11)
1(3x+2y)=1(4)

10x-2y=22
3x+2y=4

then eliminate the y
13x=26
then solve,
x=2

plug back in...

5x-y=11
5(2)-y=11
10-y=11
-y=1
y=-1

The solution to (a) is (2,-1)
Using this, you can probably do (b). You just have to change the 2nd equation to a+b=c format
it would be 3x+y=-5

Answer 2

Since the equations are in two variables, two seperate non-reducing propositions are necessary for the equations to be solved simultaneously as given here. Then,

a) 5x - y =11 --------(1)
3x + 2y = 4 -------(2)
Now multiply equation (1) by a scalar 2, note that 2 is the coefficient of the variable y in equation (2)
(1) *2 gives, 10x - 2y = 22 ------(3)
Let this new relation(which is reducible to (1)) be (3)
then,
(2) + (3) gives,
13x = 26 => x = 2
Notice how we take out one variable so that we may work with the other.
Since we know now that x=2, plug this value to any of the equations (1), (2) or (3) and they will each give you a single value for y. Consider (1) then,
x =2 => 5(2) - y =11
then, y = 10 -11 = -1

so x =2 , y=-1

Try b) 3x + y = 2 -----(1)
y = -3x -5 ------(2)
Now consider (2), if you *** 3x to both sides you get,
3x+y = -5
but the first equation says 3x+y = 2
both relations cannot hold at the same time so the claims are false having been presented simultaneously.

If however this was a typography error, the first example will allow you to solve a new two variable system of linear equations.

Answer 3

Question B cannot be performed as you have written down.
I don't konw how advanced you are so I will be very thorough.
Question A can be as follows:

5x-y=11 - equation 1
3x+2y=4 - equation 2

Multiply one of these equations by an amount so that it makes a pronumeral (a letter, x or y) the same quantity as the other equation.

I chose to multiply equation 1 by 2

this turns equation 1 into

10x-2y=22 - new equation 1
3x+2y=4 - equation 2

Add the 2 equations together. This forms equation 3. Notice the y's cancel out.

13x=26

which means

x=2

sub back into equation 2 to check what y equals.

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

So x=2, y=-1 which can be written as (2, -1) Good luck with the rest.

Answer 4

(a)
5x - y = 11
3x + 2y = 4
Multiply the 1st equation by 2
10x - 2y = 22
3x + 2y = 4
Add:
13x = 26
Solve for x, substitue into either equation, and solve for y.

(b)
This pair is nonsensical. They cannot be solved as written. You end up with -5 = 2, which is untrue.

Answer 5

Don't you just want to use substitution or subtraction??

a) y = -11 +5x (sub into 3x + 2y = 4)
3x + 2(-11 +5x) = 4
x = 2 y = -1

b) y= -3x - 5 (sub into 3x + y =2)
3x + (-3x -5) = 2
(Cant solve this last one, did you write it down wrong??)

Answer 6

for 1st eq
5x-y=11 *2
3x+2y=4

10x-2y=22
3x+2y=4

13x=26
x = 2
5(2)-y=11
y= -1


2nd eq maybe is typed wrong checkout