What is the solution to the system of equations? 4x+y=2 x-y=3
add the equations to each other, this will help you eliminate one parameter (the y).
how do you add them?
\(\large\color{black}{ \displaystyle 4x \quad+y\quad=2 }\) \(\large\color{black}{ \displaystyle ^{\huge ^+} \quad x\quad-y\quad=3 }\) \(\LARGE \color{black}{ \displaystyle ^\text{_____________} }\)
\(\large\color{black}{ \displaystyle 4x+x=\quad ? }\) \(\large\color{black}{ \displaystyle y+(-y)=\quad ? }\) \(\large\color{black}{ \displaystyle 2+3=\quad ? }\)
4x+x= 4?
Don't ask why I ask, just answer if you can, please.
4 + 1 = ?
5
Just as: 4 +1 =5 So is: 4x +1x =5x
(note that when you say "x" you really mean "1x") So, 4x+x=?
5x
Yes, and 3+2=?
5
Very nice...
y - y = ?
1 - 1 = 0 or, (Anything) - (Anything) = 0
y+(-y)=0
yes
\(\large\color{black}{ \displaystyle \color{red}{4x} \quad\color{green}{+y}\quad=\color{blue}{2} }\) \(\large\color{black}{ \displaystyle ^{\huge ^+} \quad \color{red}{x}\quad\color{green}{-y}\quad=\color{blue}{3} }\) \(\LARGE \color{black}{ \displaystyle ^\text{_____________} }\) \(\large\color{black}{ \displaystyle \color{red}{5x} \quad\color{green}{+0}\quad=\color{blue}{5} }\) <─ the result of adding these 2 equations
Does everything make sense so far?
yes
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).
x=1
Very good!
x=1. You can substitute this solution for x into any of the original equations to solve for y.
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
Does everything make sense so far ? Can you solve for y?
yes
1-y=3 y = -2
yes, x=1 and y=-2 Perfect!
You made it !
on the question i need hep with it is giving me cordintes that i have to fill in would that be (5x,5)
Your coordinates are in the form of (x,y)
Plug in the x-solution for x, and the y-solution for y.
(5,5)
is that correct
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