Question

Show that if \(a, b, and c\)are real numbers and\(a \ne 0\), then there is a unique solution of the equation\(ax + b = c\).

Short answer

Solution to \(ax + b = c\)is unique solution and is \(\frac{{c - b}}{a}\).

Step-by-step solution

Introduction

In a set of linear simultaneous equations, a unique solutions exists if and only if,

(a) The number of unknowns and the number of equations are equal,

(b) All equations are consistent, and

(c) Any two or more equations have no linear relationship, i.e., all equations are independent.

Assumption and calculation for proof

Let a, b and c be real numbers and \(a \ne 0\).

Consider\(ax + b = c\)

Add \( - b\)on both sides

So,

\(\begin{aligned}{}{\bf{ax + b - b = c - b}}\\\,\,\,\,\,\,\,\,{\bf{ax = c - b}}\\\,\,\,\,\,\,\,\,\,\,\,{\bf{x = }}\frac{{{\bf{c - b}}}}{{\bf{a}}}\end{aligned}\)

(As \(a \ne 0\))

Let \({x_1}\)and \({x_2}\)be two solutions to \(ax + b = c\).

So,

\(\begin{aligned}{}a{x_1} + b = c\\a{x_1} = c - b\end{aligned}\)

And,

\(\begin{aligned}{}a{x_2} + b = c\\a{x_2} = c - b\end{aligned}\)

Thus,

\(\begin{aligned}{}a{x_1} = a{x_2}\\{x_1} = {x_2}\end{aligned}\) (As\(a \ne 0\))

Therefore, solution to \(ax + b = c\) is unique solution and is \(\frac{{c - b}}{a}\).