linear equations again sorry i just dont understand them?

Question

1)x+y=5
x-y+7
2)x-2y=8
-x+3y+-5
3)x-4y=14
-x+3y=-11
4)2x-y=-3
-5x+y=9
5)3x+y=-3
x-4y+11
6)-2x+3y=14
-3x+4y=9
7)2x-3y=-16
x+3y=10

Answer 1

There are several ways to solve linear equations.

The easiest one is often to add or subtract one equation from the other.

Almost all of these are pretty easy to do that way.

For instance:

x + y = 5
+x - y = 7
------------
2x + 0y = 12

Therefore x = 6, and y = 5 - x = 5 - 6 = -1

x - 2y = 8
-x + 3y = -5
---------------
0x + 1y = 3

Therefore y = 3, and x = 8 + 2(3) = 14

get it?

5 and 6 are a little trickier, but you can still do them this way.

For number 5:

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

You can't cancel either x or y immediately by adding or subtracting. BUT, you can multiply the first equation by 4, so that it will work:

4(3x + y) = 4(-3)
12x + 4y = -12
x - 4y = 11
-----------------
13x + 0y = -1
x = -1/13
3(-1/13) + y = -3
y = -3 + 3/13 = -39/13 + 3/13 = -36/13

Answer 2

These are systems of linear equatione. It's like a game. Line up the x and y terms on teh left, equal to a constant.

x + y = 5 and is the other x-y equals 7?
x - y = 7 Now add them together and the y terms will disappear

2x = 12
x = 6 Now put 6 in for x in one of the equations and find y.

6 + y = 5
y= -1


2)
x - 2y = 8 and do you mean equals -5. Check your typing.
-x + 3y =-5

add them

y= 3 put it in an equation (I choose the first)

x-2(3)=8
x-6=8
x=14

You should always check you answers in teh equation you didn't use so check these values in the second equation.

3)
x - 4y = 14
-x + 3y = -11 add

-1y=3
y=-3
find x

4 and 7 can be done the same way but 5 and 6 cannot.

5)
3x+y =-3 is the second supposed to be = 11?
x -4y = 11

You have a choice here to eliminate either the x or the y. I choose they because the signs in fromt of them are oppposite. Now I have to get the numbers the same so I will multiply teh first equation by 4 and get
12x + 4y = -12 copy the second again
x - 4y = 11 now add
13x = -1

x = -1/13

solve for y

6)
-2x + 3y = 14
-3x + 4y = 9

Again a choice. I want to eliminate the y

I must have one coefficient = 12 and the other = -12 if I want to add ( and I always do...I make mistakes when I subtract) so I will multiply the first by -4 and the other by 3

8x - 12y = -56
-9x + 12y = 27
-1x = -29
x=29

solve for y

Good Luck

Answer 3

To solve linear equations: [question 3 as example]

1) Rearrange one of the equations for a variable
x - 4y = 14
rearrange for x
x = 4y + 14

2) Substitute the rearranged equation into the other equation
Sub x = 4y + 14 into -x + 3y = -11
-(4y + 14) + 3y = -11
-4y - 14 + 3y = -11
-y = 3
y = 3

3) Substitute the found variable into a different equation
Sub y = 3 into x = 4y + 14
x = 4(3) + 14
x = 26

Answer 4

What don't you understand?

Since you've given two equations for each problem you must be trying to solve the two equations "simultaneously". Here you try to add or subtract the equations so that either x or y cancels. Then you solve for the variable you have left.

Once you have one variable you substitute it back into either of the two original equations to get the other variable.

1. If you add the two equations together the "y" terms will cancel.

This gives you 2x = 12 so x = 6.
If you substitute "6" for x in either equation you will find what y is.

using the first equation: 6 + y = 5, so y = -1

#2,3,4, and 7 work the same way.

In #5 you have to multiply the top equation (multiply each term) by 4 because the second equation has "4y".

In #6 you have to multiply the top equation by -3 (don't forget the change the sign of EACH term) and multiply the bottom equation by +2.

Answer 5

the simpliest way how to solve this kind of equations is:
(I'll show you on the last equation...)

1) you need to get one variable from the first equation to be only on one side and alone:
2x - 3y = -16 / +3y
2x = -16 + 3y / +16
3y = 2x + 16 / /3
y = 2/3 * x + 16/3

2) now you just need to put this value of the variable on the left side into the second equation:

x + 3y = 10
and because we know y from the first equation:
x + 3*(2/3 * x + 16/3) = 10

3) we just need to solve this simple equation:

we put away the ():
x + 2 * x + 16 = 10 / - 16
3 * x = -6 / /3

x = -2

4) the last thing we need to do is to put x into one of the two equations you have in the start:

2x - 3y = -16

putting x = -2

2* (-2) - 3y = -16 / +3y
-4 = -16 + 3y / +16
3y = 12 / /3
y = 4

and this is it!

Answer 6

So you have a system of equations. I'll work through #1 as an example.

x+y=5
x-y=7

Solve the top equation for y:

y=5-x (Call this Equation 1)

Now using this expression for y, substitute into the second equation:

x-(5-x)=7

Now simplify:

x-(5-x)=7
x-5+x=7
2x-5=7
2x=12
x=6

Once you solve for x, take the value and plug it into Equation 1.

y=5-x
y=5-(6)
y= -1

So the answer is x=6, y= -1. You can employ the same strategy for all the other problems.

Answer 7

How to find the equations is by( Y=mx+b ) ..

1) -y=x+7
here behind the Y is - so divide it by -1 to let only Y so when you divide Y you divide all the equation x and 7 then u will have the answer .. and how to find the slope (m) in the eqaution is by mines y2-y1 divide by x2-x1..

and if you get - result the linear equations is not appear and if you get normal number without - then the linear equation is street line.

Try to solve the others because it is easy . and it is not hard , Try your best ;)

Answer 8

For the first one, your couples are basically (a million,3) (2,6) (3,9) and so on, so that you receives a immediately line with an excellent slope of three. For the 2d one, it will be a refection of the first interior the x axis, (a million, -3) (2, -6) (3,-9) and so on with a slope of -3. The third is a immediately horizontal line parallel to the x axis intersecting the y axis at 5.

Answer 9

Two methods:
1) Elimination - eliminating one unknown and subsequent substitution
2) Substitution - solving one unknown first and then substituting
I prefer the method of elimination rather than "outright" substitution, that is, solving for x or y right away. You have to cope with a "larger" expression.
Follow Jim Burnell's examples.

Answer 10

all you have to do is use one equation and solve for one variable then plug it back into the other equation

Example

1)x+y=5
2)x-y=7

y=-x+5
then subsitute into other equation
x-(-x+5)=7
2x-5=7
2x=12
x=6
then plug in value of x into equation 1 and solve for y
x+y=5
6+y=5
y=-1