solving by the addition and subtracting method?

Question

2x+y=18
2x-y=22

12a-5b=18
15a-5b=30

How do I tell when to use the addition and subtraction method?? and if you could walk me through one of the problems that would be great....

Answer 1

2X+Y=18
2X-Y=22
------------
add the two equations together
4X+Y -Y =40, the +&- Y cancel giving you X = 10
plug it into either equation solves for the system, Y = -2

use it the same for the other equation
12a - 5b = 18
15a - 5b = 30
since both equations have a -5b, you can no straight add them,...
therefore do this
15a - 5b = 30 15a -5b = 30
-(12a -5b = 18) or -12a+5b =-18
-----------------
now its an addition 3a = 12 problem with a =4

you can also multiply a value by an entire equation to help it work out.
say if 15a -5b = 30 were instead 3a -b = 6
you could multiply by 5(3a-b = 6) to get the equation in a form you could utilize.

hope that helps.

Answer 2

A way to spot the need to use the A/S method is to see if the x's and y's line up when you place on equation on top of the other. In both cases, this has already been done for you.

The method involves eliminating one variable so you can solve ONE equation with ONE variable.

So...

2x + y = 18
2x - y = 22
--------------
4x = 40

Adding the two equations eliminates the variable y.

Solve that equation for the variable...

4x = 40 -> x = 10.

Plug x = 10 in for y in one of your original equations. Either one will do.

2(10) + y = 18
20 + y = 18
y = -2

It is preferred that you check your answer in the other equation to make sure it checks out.

2(10) - (-2) = 20 + 2 = 22.
It checks.

So, x = 10 and y = -2

Answer 3

Find the variable in which coeficients in both equations have the same absolute value. In the first pair of equations, that would be y.

If the sign is the same, subtract. If the sign is opposite, add.

Take the first pair:

2x + y = 18
2x - y = 22

Add the the two and get:
4x + 0y = 40
x = 10

Substitute x=10 into either of the equations and solve for y:

2 x 10 - y = 22
y = -2

Answer 4

You can almost always use EITHER the substitution method or the addition and subtraction method. It really doesn't matter. You should know how to do both, though.

2x+y = 18
2x-y = 22
add the two equations to get
4x = 40
x = 10
then y = 18 - 2x or -2

YOu can do the second one, but I'll give you a hint: SUBTRACT the second equation from the first instead of adding it.

Answer 5

If you mean systems of inequalities
Elimination:
2x+y=18
+2x -y=22
-------------
4x=40
divide both sides by 4 to get x=10
Go back to one of the original equations and plug 10 in for x
2(10)-y=22
20-y=22
subtract 20 from both sides -y=2
divide both sides by -1 to get y=-2


or Substitiution:
2x+y=18
subtract 2x from both sides to get y=-2x+18
put that into the other equation 2x-y=22 for y
2x-(-2x+18)=22
distribute the -1 to make it 2x+2x-18=22
combine like terms 4x-18=22
add 18 to both sides
4x=40
divide by 4, x=10
plug 10 into the equation y=-2x+18
y=-2*(10)+18
y=-20+18
y=-2


It depends on what the directions say as to what you'll use. If it does not specify which to use, chose the one you are most comfortable.

Answer 6

1)You use the addtion method:
2x+y=18
2x-y=22
======
4x=40
x=10

2(10)+y=18
20+y=18
y=-2

The solution set is (10,-2)

Check:
2(10)-2=18
20-2=18
18=18

2(10)-(-2)=22
20+2=22
22=22

2)You would use the subtraction method and you can choose which equation you want to multiply by a negative, I am choosing to multiply the first equation by a negative:
-(12a-5b=18)
15a-5b=30
==========
-12a+5b=-18
15a-5b=30
==========
3a=12
a=4

15(4)-5b=30
60-5b=30
-5b=-30
b=6

The solution set is (4,6)

Check:
12(4)-5(6)=18
48-30=18
18=18

15(4)-5(6)=30
60-30=30
30=30

Answer 7

You can add two equations in order to get a third, simpler equation.

For example, in the first case you can add both sides separately, to get:
(2x + y) + (2x - y) = 18 + 22
2x + y + 2x - y = 4 x = 40

What you do is look for the easiest way to get rid of one unknown, without getting rid of the other one (or without creating a strange event -- like dividing by zero).

Again using the first case, we could have used subtration:

(2x + y) - (2x - y ) = 18 - 22

2x + y -2x -(-y) = y + y = 2 y = -4

In the second case, you will have to subtract.

Other methods include multiplying one of the equations (or both) before adding or subtracting.

for example, if you have 2x + y = 18 and 6x - 2y = 64, you can multiply both sides of the first equation by 3 to get
6x + 3y = 54
Then you can subtract to get rid of the x:

6x + 3y -(6x -2y) = 54 - 64 = -10
6x + 3y -6x + 2y = 5y = -10 (and from there, you find y = -2)

Answer 8

As Tim said, you choose an operation that gets rid of one of your variables... you must look at the sign and the coefficients both variables have...

let's take the first two equations for an instance...

to solve for x and y, you will just add equation 1 (2x + y = 18) to equation 2 (2x - y = 22)... why? because once we add two equations, that gets rid of y (same coefficients but different signs) and you can easily evaluate for x...

4x = 40
x = 10

and then by substitution in either of the two equations, you can find y...

y = -2

You could also hav possibly substracted 2nd equation from 1st equation... but since it is a bit "lengthy" (just one more step in this case), we choose the easy way... in this case, the addition operation is the key...

now let's take the second set of equations... which variable is easy to take out first??? we can easily say that it is b because in both equations coefficients of b is same... now we just need to decide on which operation we need to perform... can we perform addition and be content? No because despite performing this operation, we still have two variables as it is and our purpose is it to get rid of one variable, so we can not perform addition... we must do substraction... substract 2nd equation from the 1st one (or substract 1st equation from 2nd equation, it will result in same solution set)... because that gets rid of variable b (the substraction changes the sign of coefficient to exactly opposite as the other one) and we can easily find a...

-15a + 5b = -30... eq 2
12a - 5b = 18... eq 1
after performing operation, we get the following...
-3a = -12
a = 4

insert a in any of the two equations and solve for b,

b = 6

for 1st set of equations... y has same coefficients and opposite signs so ideal operation is addition...

for 2nd set of equations... while b has same coefficients, it has same signs so we must change sign of one of them...

in some set of equations, you may not have same coefficients for both variables... in that case, you will need to find least common multiple and then perform the appropriate operation (addition or substraction)... to choose that operation, you must look at the signs... the 2nd set of equations can be solved this way if you want... but it is a lengthy way... therefore, we avoid it... ;)

I hope I am not confusing you with all this... :)

Answer 9

you choose the one that makes one of your variables go away.

2x + y = 18
2x - y = 22

You can do this one either way. Let's do addition first:

2x+y = 18
2x-y = 22

Adding, you get 4x = 40, which solves to x = 10
With x = 10, 2(10)-y=22; 20 - y = 22; y = -2

Subtracting, you get 2y = -4, or y = -2
with y = -2, 2x - (-2) = 22, 2x = 20, x = 10

So either way will work; just choose the one that simplifies your system.

Hope it helps.

Answer 10

Use the ELIMINATION method

2x+y=18
2x-y=22
------------(subtract)
2y=-4
y=-2

2x-2=18
2x=20
x=10

and the second like that too!!

12a-5b=18
15a-5b=30
--------------(subtract)
3a=-12
a=-4

-60a-5b=30
-5b=90
b=-18


hope this usefull!!