Solving Two Equations: x1 = 0.8x2, x2 = 0.2x1 + h

I'm getting "re-familiar" with solving equations again.

I have two equations that I need to solve (are already solved). They read as follows:

x1 = 0.8x2
x2 = 0.2x1 + h

Therefore, x2 = 0.2 (0.8x2) +h
Or 0.84x2 = h
Or x2 = h/0.84

And x1 = 0.8x2

How do I arrive at the "0.84"? Any pointers for solving it... step by step?

Thanks,
odumath

odumath said:

Therefore, x2 = 0.2 (0.8x2) +h
Or 0.84x2 = h

How do I arrive at the "0.84"?

Click to expand...

Multiply 0.2 by 0.8. Subtract the 0.16x₂ from the 1x₂.

odumath said:

Any pointers for solving it... step by step?

Click to expand...

Didn't the book give you the step-by-step solution...?

Eliz.

Hello, odumath!

Those subscripts are annoying . . . I'll rewrite the problem.

\(\displaystyle \begin{array}{ccc}x &= & 0.8y \\
y & = & 0.2x\,+\,h\end{array}\)

Therefore: \(\displaystyle \:y \:= \:0.2(0.8y)\,+\,h\;\) **

Or:\(\displaystyle \:0.84y \:=\: h\)

Or:\(\displaystyle \:y\: =\:\frac{h}{0.84}\)

And: \(\displaystyle \:x\:=\:0.8y\)

How do I arrive at the "0.84"?

Click to expand...

You may kick yourself . . .

**
We have: \(\displaystyle y\:=\:0.16y\,+\,h\)

Then: \(\displaystyle \:y\,-\,0.16y\:=\:h\)

This is: \(\displaystyle \:1.00y\,-\,0.16y\:=\:h\)

Hence: \(\displaystyle \:0.84y\:=\:h\)

Thanks, I appreciate it.

odumath