Prove the following by using a contrapositive argument. "Let $f(x)=7-3x$. If $x_1$ $\neq$ $x_2$, then $f(x_1)$ $\neq$ $f(x_2)$"

Question

This is supposed to be in proof form. I am completely lost as to where to start but this is what I have. Prove the following by using a contrapositive argument.

Let $$f\left(x\right)=7-3x.$$ Prove If\ $$f\left(x_1\right)=f\left(x_2\right)$$\ then $$x_1=x_2$$.

$$f(x)=7-3x$$.\ Given $$f(x_1)=f(x_2)$$ Given $$7-3x_1=7-3x_2$$ Given $$7-3x_1+7=7-3x_2+7$$ additive inverse property $$3x_1=3x_2$$
$$\frac{{3x}_1}{3}=\frac{{3x}_2}{3}$$ multiplicative inverse property $$x_1=x_2$$

Answer 1

Here's an interesting solution.

For sake of contradiction, assume that there exists two non equal real numbers $x_1$ and $x_2$, for which $f(x_1)=f(x_2)$. Since, $f$ is an affine function, its graph will be strictly monotonic, and this forces $x_1=x_2$ which contradicts our assumption.