Question

MEDAL AND FAN! PLEASE HELP!

how do i find the solution to

3Y = 15 + x
x + 7Y = 5

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Answer 1

the first step is to isolate one of the variables

let's say we pick on the second equation and solve for x


x + 7y = 5
x+7y = 5 - 7y ... subtract 7y from both sides
x = 5 - 7y


so after solving `x + 7y = 5` for x, we end up with `x = 5 - 7y`

Answer 2

now move onto the other equation

we will replace every `x` with `5-7y`

so...

\[\Large 3y = 15 + x\]

\[\Large 3y = 15 + \color{red}{x}\]

\[\Large 3y = 15 + \color{red}{5-7y}\]

\[\Large 3y = 15 + 5-7y\]

do you know how to solve for y from here?

Answer 3

no can you explain please

Answer 4

does everything make sense so far up to this point?

Answer 5

yes it makes sense. Once we get the X we substitute every X in the other equation with 5-7Y

Answer 6

ok on the right side of the equation `3y = 15 + 5-7y` we can combine like terms, right?

Answer 7

so that will make it 3y= 20-7y?

Answer 8

yes, now we must move the y terms all to one side

Answer 9

or will that make 10y = 20?

Answer 10

good, you added 7y to both sides

Answer 11

10y means 10 times y

so 10 times ______ = 20

what must go in the blank? you can guess and check, or you can divide both sides by 10 to isolate y

Answer 12

divided by 10 to both sides will give me Y = 2 ?

Answer 13

yep y = 2

now use this to find x

Answer 14

x = 5 - 7y

x = 5 - 7*2 ... replace y with 2

x = ??

Answer 15

-9 ?

Answer 16

good

Answer 17

so the solution is (x,y) = (-9, 2)

let's check that answer

we do this by plugging it back into the original equations. It has to make ALL of the equations of the original system true

Answer 18

3y = 15 + x

3y = 15 + (-9) ... replace x with -9

3*2 = 15 + (-9) ... replace y with 2

6 = 6 ... equation is true

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x + 7y = 5

-9 + 7y = 5 ... replace x with -9

-9 + 7*2 = 5 ... replace y with 2

-9 + 14 = 5

5 = 5... equation is true


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both original equations are true when (x,y) = (-9, 2) so the solution has been confirmed

Answer 19

this check is highly recommended for any equation you solve because later on in algebra, you'll learn of cases where you'll get to an "answer" but it won't be the solution because it doesn't make the original equation true

Answer 20

is this the only way to solve this equation? isn't there a way that goes like this.

3y = 15 + x
-7y = -5 - x
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? where you can cancel out x?

sorry if this confuses you but i saw my teacher do this

Answer 21

that's another way and that method is called the addition method or the elimination method

Answer 22

notice on the right side of each equation, you can add them up and the +x and the -x combine to 0x, so the x terms go away completely

Answer 23

can you show me the way to do it with this method? sorry if i'm bothering you. I have a test on this tomorrow

Answer 24

it's no bother, so you're fine

Answer 25

once you get it to this form

3y = 15 + x
-7y = -5 - x

you just add straight down

Answer 26

3y plus -7y = -4y

15 plus -5 = 10

x plus -x = 0x

Answer 27

-4y = 10 ?

Answer 28

hmm something went wrong

Answer 29

how did you get the -5? it should be +5 or just 5

Answer 30

the system should be

3y = 15 + x
-7y = 5 - x

Answer 31

oh wait I'm misreading, let me try again

Answer 32

ok

Answer 33

let's start with x + 7Y = 5

and subtract x from both sides, so we have just the y term on the left side

so we now have this new system


3Y = 15 + x
7Y = 5 - x

Answer 34

now add straight down

Answer 35

10y = 20

but to get the bottom x to be a negative dont we have to times everything by -1?

Answer 36

when you subtract x from both sides, it becomes negative on the right side