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I am just trying to solve through some problems in my book. The problem was

Let $a,b,c,d$ and $e$ be constants in the system of equations

$ax+by = d$

$ax+cy = e$

suppose $b$ and $c$ are not equal and $a$ is not $0$. Must the system of equations have exactly one solution $(x,y)$?

My attempt:

I added equation $1$ and equation $2$ so to find

$y$ = $\frac{d-e}{b-c}$

I plugged this in equation $1$ to find the value of $x$. I noticed the solution for $x$ depended on the constant $a$, which means that there are infinite solutions for $x$.

Is my logic correct? I am not sure if I am approaching this problem the corret way

Bing Chilling
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1 Answers1

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There are infinite solutions for $x$, but only one for each constant $a$ since you can determine $y$ and hence $x$ from $a,b,c,d,e$. You have shown that there is a unique solution $(x,y)$ for the given constants and conditions.

(Formally, we can verify this through the determinant of the system, which is $ac-ab=a(c-b)$. Since $a\ne0$ and $b\ne c$, the determinant cannot be $0$.)

Parcly Taxel
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