anonymous
  • anonymous
What is the solution to the system of equations? 4x+y=2 x-y=3
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
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SolomonZelman
  • SolomonZelman
add the equations to each other, this will help you eliminate one parameter (the y).
anonymous
  • anonymous
how do you add them?
SolomonZelman
  • SolomonZelman
\(\large\color{black}{ \displaystyle 4x \quad+y\quad=2 }\) \(\large\color{black}{ \displaystyle ^{\huge ^+} \quad x\quad-y\quad=3 }\) \(\LARGE \color{black}{ \displaystyle ^\text{_____________} }\)

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SolomonZelman
  • SolomonZelman
\(\large\color{black}{ \displaystyle 4x+x=\quad ? }\) \(\large\color{black}{ \displaystyle y+(-y)=\quad ? }\) \(\large\color{black}{ \displaystyle 2+3=\quad ? }\)
anonymous
  • anonymous
4x+x= 4?
SolomonZelman
  • SolomonZelman
Don't ask why I ask, just answer if you can, please.
SolomonZelman
  • SolomonZelman
4 + 1 = ?
anonymous
  • anonymous
5
SolomonZelman
  • SolomonZelman
Just as: 4 +1 =5 So is: 4x +1x =5x
SolomonZelman
  • SolomonZelman
(note that when you say "x" you really mean "1x") So, 4x+x=?
anonymous
  • anonymous
5x
SolomonZelman
  • SolomonZelman
Yes, and 3+2=?
anonymous
  • anonymous
5
SolomonZelman
  • SolomonZelman
Very nice...
SolomonZelman
  • SolomonZelman
y - y = ?
SolomonZelman
  • SolomonZelman
1 - 1 = 0 or, (Anything) - (Anything) = 0
anonymous
  • anonymous
y+(-y)=0
SolomonZelman
  • SolomonZelman
yes
SolomonZelman
  • SolomonZelman
\(\large\color{black}{ \displaystyle \color{red}{4x} \quad\color{green}{+y}\quad=\color{blue}{2} }\) \(\large\color{black}{ \displaystyle ^{\huge ^+} \quad \color{red}{x}\quad\color{green}{-y}\quad=\color{blue}{3} }\) \(\LARGE \color{black}{ \displaystyle ^\text{_____________} }\) \(\large\color{black}{ \displaystyle \color{red}{5x} \quad\color{green}{+0}\quad=\color{blue}{5} }\) <─ the result of adding these 2 equations
SolomonZelman
  • SolomonZelman
Does everything make sense so far?
anonymous
  • anonymous
yes
SolomonZelman
  • SolomonZelman
To say: 5x +0 = 5 Is the same as saying: 5x = 5 (Because adding zero makes no difference) And now you are able to solve for x (by dividng both sides by 5).
anonymous
  • anonymous
x=1
SolomonZelman
  • SolomonZelman
Very good!
SolomonZelman
  • SolomonZelman
x=1. You can substitute this solution for x into any of the original equations to solve for y.
SolomonZelman
  • SolomonZelman
Lets use the 2nd equation: x - y = 3 YOu are given that x=1, so you can re-write theis equation as: 1 - y =3
SolomonZelman
  • SolomonZelman
Does everything make sense so far ? Can you solve for y?
anonymous
  • anonymous
yes
anonymous
  • anonymous
1-y=3 y = -2
SolomonZelman
  • SolomonZelman
yes, x=1 and y=-2 Perfect!
SolomonZelman
  • SolomonZelman
You made it !
anonymous
  • anonymous
on the question i need hep with it is giving me cordintes that i have to fill in would that be (5x,5)
SolomonZelman
  • SolomonZelman
Your coordinates are in the form of (x,y)
SolomonZelman
  • SolomonZelman
Plug in the x-solution for x, and the y-solution for y.
anonymous
  • anonymous
(5,5)
anonymous
  • anonymous
is that correct
SolomonZelman
  • SolomonZelman
|dw:1447084085292:dw|
SolomonZelman
  • SolomonZelman
See ?
anonymous
  • anonymous
yes
SolomonZelman
  • SolomonZelman
Alright ... Good \(:~)\)
anonymous
  • anonymous
tyvm for your help
anonymous
  • anonymous
.....can you help e with 1 more
SolomonZelman
  • SolomonZelman
Perhaps, but in a new post please...
SolomonZelman
  • SolomonZelman
GOod LUCk!
anonymous
  • anonymous
ok

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