Question

Solve the following system of equations.

2x + y = 3
x = 2y - 1

Answer 1

simply put (2y-1) in the first equation in place of x

this gives you an equation in y only which is simple to solve

THEN put that value of y in the second equation - and you can solve x

Answer 2

2x + y = 3
x = 2y -1

so
2(x) + y = 3
2(2y -1) + y = 3
4y -2 + y = 3
5y -2 = 3
5y = 5
so y = 1

now use y = 1 to solve for x

Answer 3

im lost lmao cx

Answer 4

@Jack1
It is discouraged to give answers here.

See Code of Conduct

@nathshely If you are lost it is because you replied before you even had a chance to look at the options given to you.

Follow the instructions above - and try it for yourself.

Answer 5

i still read it and im still lost.
and we're here to help each other and i love him for giving me the answer :P

Answer 6

is y=1 the answer to your NEXT similar question?

Answer 7

sorry @nathshely , i didn't have a lot of time to got through it all with you yesterday, the method i typed above as an example it called substitution, and its used heaps for solving simultaneous equations
u can read more about it here tho:
http://www.basic-mathematics.com/substitution-method.html

basically, if u have 2 equations with the same variables (so x and y), u rearrange one and "sub"that into the new equation:

Answer 8

so like \(\large 2x + y = 10\) ==> equation 1
and \(\large y-x = 1\) ==> equation 2

so rearrange equation 2 by adding x to both sides
\(\large y-x = 1\)
\(\large y-x +x = 1 +x\)
\(\large y ~~\color{red}{\cancel {-x}}~~\color{red}{\cancel {+x}} = 1 +x\)
\(\large y = 1+x\) ==> equation 2a

now substitute equation 2a back into equation 1
\(\large 2x + y = 10\) ==> equation 1
\(\large 2x + (1+x) = 10\) (as y = 1+x)

so this can now be solved for x
\(\large 2x + (1+x) = 10\)
\(\large 2x +x + 1 = 10\)
\(\large 3x + 1 = 10\) (combine like terms)
\(\large 3x + 1-1 = 10-1\) (subtract 1 from both sides)
\(\large 3x = 9\)
\(\large \color{blue}{ x = \huge{\frac 93} \large = 3}\) (solved for x)

then use x = 3 to solve for y
\(\large y-x = 1\) ==> equation 2
\(\large y-3 = 1\)
\(\large y-3+3 = 1+3\)
\(\large\color{blue}{ y = 4}\) (solved for y)

solved in this example! c:

Answer 9

@MrNood it is discouraged to be unfriendly on this site

See Code of Conduct

The first line of the "pledge" is:
"Be Nice - I will stay \(\color{red}{positive}\), be friendly, and not mean"

Had I just told her only that that x=1 and y = 1, then sure, call me out bro, but instead, I demonstrated a method that I would use to solve the first \(\Large\color{red}{half}\) of the problem... and left the other half for her to solve...

tl;dr - check ur facts bro, and D.B.A.D (Dont Be A D...) and we'll get along just fine, k? ;)

Answer 10

DAMNN! lol thanks @Jack1 really helped :'D youre so nice =DD

Answer 11

I had already 'demonstrated the method'

You executed it right through to the answer, leaving the OP nothing to do for the key part of the question.

It is strongly stated that answers should not be given.
Some evidence of input from the OP should be encouraged before further illustrating/solving the issue.

However - for my part the discussion is now closed.

We'll get along just fine anyhow k?

Answer 12

no! its not nice from you.
why do you care if he gives me the answer or not?
does it affect you?
people break rules allll the time...
@MrNood :DD

Answer 13

no @MrNood, ur having an argument inside someone's question, that's not ok dude.

It is strongly stated that OpenStudy is an online learning "Community", learn to get along

"simply put (2y-1) in the first equation in place of x" => now who's executing it:
- without an explanation why
- or what the method's called
- or even if the maths is correct

Maybe next time give the OP a reason to trust u an ur method, and explain a bit about "what" and "why"ur doing things, not just "how"

However, for my part, this discussions' finally closed.

Follow the above simple rules and I'm sure we'll get along, k?