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

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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

Mathematics
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5x + 0 = 15 x = 3 plug to the first equation 2.3 + y = 6 y = 0
so i would just add the numbers then divide
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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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 ......
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
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.
Now they asked you to use addition method which means. some how you have to add equation 1 and equation 2
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.
give me another problem, i'll explain in more details because this is problem is very simple.
ok 4a-5b=7 4a+5b=17
alright. this is also simple. Let's change to 4a + 5b = 7 2a + 3b = 9 instead.
now what do you do? I want to know how you do it, then I'll explain
i add 4a to 2a right
continue
then i add5b and3b
ok. but the new equation doesn't get rid of any variables right? so your stuck
you have to modify the original equations somehow so when adding you will get either 0x or 0y
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.
did you learn how to modify the equations yet? or just purely adding stuff?
no did learn to modify the equations
okay then please try to modify this one: 4a + 5b = 7 2a + 3b = 9
i cnt i dnt understand how to modify im sorry
let's try to modify so when adding I will get 0a ok?
i see first equation have 4a. the 2nd have 2a. ... I need to have some how to have 2nd equation to have -4a. Right?
ummm yes
you can do this by multiply 2nd equation by -2
what's the modified equation now?
ok the modified (2nd equation) becomes: -4a -6b = -18
now you can add the first and the modified.
so then id add the 4a and 5b
no no. it's gonna be like this: 4a + 5b = 7 + -4a -6b = -18 ----------------------- 0a -b = -11
you only add the a and a... b and b not a and b.
o
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
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
you have to modify both equations for b.
4a+2a right
yes what about b
5b+3b?
so you will get 6a + 8b = 16?
it doesn't get rid of b does it?
so i subtract
you need to get help from professor about this. But those problems you gave me. you can just add them directly without modifying anything.
you are not clear about the concept of adding.
one thing I can tell you that when adding you must get rid of one of the variables.
no im not im sorry thnks for tryin to help tho
i have to go home. sorry

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