Solve this system of equations using the addition method.
3x-y=9
2x+y=6 Solve this system of equations using the addition method.
3x-y=9
2x+y=6 @Mathematics

- anonymous

- jamiebookeater

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- anonymous

5x + 0 = 15
x = 3
plug to the first equation
2.3 + y = 6
y = 0

- anonymous

so i would just add the numbers then divide

- anonymous

first add the equation 1 and 2 so that it will get rid of one of the variable
In this example, you got it easy because in equation 1 you have -y, and in equation 2 you have y, so the sum of these equations will get rid of y

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- anonymous

in case you want to get rid of x first. then you would have to do the following:
multiply 1st equation by 2
multiply 2nd equation by -3 so that it would appear like this
6x ....
-6x
-------------
0x ......

- anonymous

im sorry im very confused..see im takin algebra online and dnt have a teacher so i have to take tests online and have no idea how to do them

- anonymous

ok so you have two equations and two variables which is x, y. They ask you to find those values of x, y so that when you plug them in to any equation, they have to be true. In this case if you plug x = 0 and y = 3. Both equations hold. That's how it goes anyways.

- anonymous

Now they asked you to use addition method which means. some how you have to add equation 1 and equation 2

- anonymous

3x-y=9
+
2x+y=6
----------
5x + 0 = 15
the new equation only has x now.
So you know what x is, just plug into any of the equations they gave you, you will get y value.

- anonymous

give me another problem, i'll explain in more details because this is problem is very simple.

- anonymous

ok 4a-5b=7
4a+5b=17

- anonymous

alright. this is also simple. Let's change to 4a + 5b = 7
2a + 3b = 9 instead.

- anonymous

now what do you do? I want to know how you do it, then I'll explain

- anonymous

i add 4a to 2a right

- anonymous

continue

- anonymous

then i add5b and3b

- anonymous

ok. but the new equation doesn't get rid of any variables right? so your stuck

- anonymous

you have to modify the original equations somehow so when adding you will get either 0x or 0y

- anonymous

look at this problem you gave me:
4a-5b=7
4a+5b=17
notice the -5b (1st) and 5b (2nd equation). when adding the result is 0b. that's why i said they gave you a simple problem. Just adding.

- anonymous

did you learn how to modify the equations yet? or just purely adding stuff?

- anonymous

no did learn to modify the equations

- anonymous

okay then please try to modify this one:
4a + 5b = 7
2a + 3b = 9

- anonymous

i cnt i dnt understand how to modify im sorry

- anonymous

let's try to modify so when adding I will get 0a ok?

- anonymous

i see first equation have 4a. the 2nd have 2a. ... I need to have some how to have 2nd equation to have -4a. Right?

- anonymous

ummm yes

- anonymous

you can do this by multiply 2nd equation by -2

- anonymous

what's the modified equation now?

- anonymous

ok the modified (2nd equation) becomes:
-4a -6b = -18

- anonymous

now you can add the first and the modified.

- anonymous

so then id add the 4a and 5b

- anonymous

no no. it's gonna be like this:
4a + 5b = 7
+
-4a -6b = -18
-----------------------
0a -b = -11

- anonymous

you only add the a and a... b and b not a and b.

- anonymous

o

- anonymous

so now you can solve for the b value. Once you have one variable solved, plug to any of the original equation, then bam. That's it

- anonymous

can you modify to get rid of b instead of a this time? it's a bit harder, but same method.
4a + 5b = 7
2a + 3b = 9

- anonymous

you have to modify both equations for b.

- anonymous

4a+2a right

- anonymous

yes what about b

- anonymous

5b+3b?

- anonymous

so you will get 6a + 8b = 16?

- anonymous

it doesn't get rid of b does it?

- anonymous

so i subtract

- anonymous

you need to get help from professor about this. But those problems you gave me. you can just add them directly without modifying anything.

- anonymous

you are not clear about the concept of adding.

- anonymous

one thing I can tell you that when adding you must get rid of one of the variables.

- anonymous

no im not im sorry thnks for tryin to help tho

- anonymous

i have to go home. sorry

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