2x2 system of linear equations...

[Chris*]

Here's the problem:

x+y = 50

(0.4x+0.15y)
------------- = 0.21
(x+y)

How do I isolate the variables for this?

I have a quiz tomorrow, so any help would be greatly appreciated.

ETA...

So I tried once more to work it out and I think I have it...

Do I multiply the second equation by (x+y)?
So that it becomes
0.4x + 0.15y = 0.21x + 0.21y and then
0.19x - 0.06y = 0

So that makes...

x+y = 50 (multiply this by 0.06 or 0.19)
0.19x - 0.06y = 0

And then isolate from there...

Is that right?

 

[Mrspi]

Here's the problem:

x+y = 50

(0.4x+0.15y)
------------- = 0.21
(x+y)

How do I isolate the variables for this?

I have a quiz tomorrow, so any help would be greatly appreciated.

ETA...

So I tried once more to work it out and I think I have it...

Do I multiply the second equation by (x+y)?
So that it becomes
0.4x + 0.15y = 0.21x + 0.21y and then
0.19x - 0.06y = 0

So that makes...

x+y = 50 (multiply this by 0.06 or 0.19)
0.19x - 0.06y = 0

And then isolate from there...

Is that right?

That would work.....OR, you could solve the first equation for either x or y. I'll solve for y:

y = 50 - x

Then, substitute (50 - x) for y in the second equation:

0.19x - 0.06(50 - x) = 0

Solve that for x, and substitute the value you get into y = 50 - x to find y.

 

[soroban]

Hello, Chris*!

[1] $\displaystyle \;x\,+\,y\: =\: 50 \;\;\;$ [2]$\displaystyle \;\frac{0.4x\,+\,0.15y}{x\,+\,y} \:= \:0.21$

[2] becomes: $\displaystyle \:0.4x\,+\,0.15y\:=\:0.21x\,+\,0.21y$

Multiply by \(\displaystyle 100:\;\;\;40x\,+\,15y\:=\:21x\,+\,21y\\)

. . . And we have:. .$\displaystyle 19x\,-\,6y\:=\:0$
Multiply [1] by 6: $\displaystyle \;\;6x\,+\,6y\:=\:300$

. . . . . . . . . Add: . . . . .$\displaystyle 25x \:=\:300$

. . . . . . . . .Then: . . . . . .$\displaystyle \fbox{x\:=\:12}$

Substitute into [1]: $\displaystyle \:12\,+\,y\:=\:50\;\;\Rightarrow\;\;\fbox{y\:=\:38}$