Solving a system of equations problem

Question

I'm ytrying to solve the following for $x$ as an expression of the other variable (as few as possible)

$$a+bx+(n-1)de=0$$ $$f+gx+(n-1)he=0$$

I'm not sure how to solve this.

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I tried equating both, but I cant get rid of $(n-1)$ or $e$. $$ a+bx+(n-1)de=f+gx+(n-1)he $$

I think the answer is $$x=\frac{fh-ad}{bd-gh}$$

Is this correct?

Answer 1

If your other variable is $$y=(n-1)e$$ as it seems to be from your own attempts and the following quote:

I tried equating both, but I cant get rid of $(n−1)$ or $e$.

then the solution for $x$ is: $$x=\frac{d f-a h}{b h-d g}$$

And the other variable is:

$$y=(n-1)e=\frac{a g-b f}{b h-d g}$$

Answer 2

a+bx= (1-n)(de)

f+gx= (1-n)(he)

so

(a+bx)/(f+gx)= (1-n)(de)/(1-n)(he)

for n not equal to 1 and e not equal to 0, we can simplify further

(a+bx)/(f+gx)= d/h

ah+bhx= df+dgx

(bh-dg)x= df-ah

x= (df-ah)/(bh-dg)

Answer 3

Hint: eliminate $\,n\,$ between the equations, then solve $(1)$ for $\,x\,$:

$$ \begin{align} {a+bx+(n-1)de = 0 \quad \mid \cdot\,h \\ f+gx+(n-1)he = 0 \quad \mid \cdot\,d} \end{align} \quad\Bigg| \,-\, \Bigg| \\[5px] $$ $$\require{cancel} ah+bhx+\cancel{(n-1)deh} - \big(\,fd + gdx + \cancel{(n-1)hed}\,\big) = 0 \tag{1} $$