\[ R(x) = rac{(x - 1)(x - 3)}{x - 1} = x - 3. \]

["# Simplifying R(x): Proving That ( R(x) = x - 3 )", "When faced with a rational expression like\n[\nR(x) = \frac{(x - 1)(x - 3)}{x - 1},\n]\nit’s natural to wonder whether this seemingly complex fraction simplifies to a linear function:\n[\nR(x) = x - 3.\n]\nThis article explores how simplification works, evaluates the domain considerations, and confirms the equivalence of ( R(x) ) to ( x - 3 ) for all valid inputs.", "## Understanding the Expression", "The original function is\n[\nR(x) = \frac{(x - 1)(x - 3)}{x - 1}.\n]\nAt first glance, the presence of ( (x - 1) ) in both numerator and denominator suggests a removable discontinuity at ( x = 1 )—a common simplification in algebra. However, understanding why this simplification is valid requires careful step-by-step analysis.", "### Step 1: Identify Domain Restrictions\nBefore simplifying, it’s critical to identify where the expression is undefined. Since division by zero is undefined, the denominator ( x - 1 ) must not be zero, meaning:\n[\nx <br/>\ne 1.\n]\nSo, ( R(x) ) is defined for all real numbers except ( x = 1 ).", "### Step 2: Simplify the Expression", "For all values where ( x <br/>\ne 1 ), provided ( x - 1 <br/>\ne 0 ), we can safely cancel the common factor ( (x - 1) ):\n[\nR(x) = \frac{(x - 1)(x - 3)}{x - 1} = x - 3,\n]\nas long as ( x <br/>\ne 1 ).", "So, while algebraically ( R(x) = x - 3 ) for all ( x <br/>\ne 1 ), this simplified expression—it’s equivalent to the original function on its entire domain.", "### Step 3: Graphical Interpretation", "The graph of ( R(x) ) matches the straight line ( y = x - 3 ), except for a hole at ( x = 1 ), where the original function is undefined. This visual insight reinforces that for practical purposes,\n[\nR(x) = x - 3 \quad \ ext{for all } x <br/>\ne 1.\n]", "### Step 4: Mathematical Justification", "By algebraic rules, when ( a <br/>\ne 0 ), the expression\n[\n\frac{a \cdot b}{a} = b \quad \ ext{for any } b \n]\nholds true, as long as ( a <br/>\ne 0 ). Here, ( a = (x - 1) ), which is zero only at ( x = 1 )—not in the domain we consider for simplification. Thus, the simplification is valid across the function’s domain.", "### Why This Matters: Simplification in Algebra and Calculus", "Simplifying rational expressions not only clarifies analysis but also simplifies integration, differentiation, and solving equations. Knowing that ( R(x) = x - 3 ) (where defined) allows easier computation across calculus operations. It also highlights key concepts like removable discontinuities and domain restrictions—essential for mastery in precalculus and calculus.", "---", "### Conclusion", "The rational function\n[\nR(x) = \frac{(x - 1)(x - 3)}{x - 1}\n]\nsimplifies algebraically to\n[\nR(x) = x - 3\n]\nfor all ( x <br/>\ne 1 ). This equivalence reflects the function’s behavior across its domain, with a single removable discontinuity at ( x = 1 ). Simplifying functions in this way enhances clarity, reveals hidden structure, and supports advanced mathematical problem solving.", "Whether you're teaching algebra, preparing calculus exams, or analyzing real-world models, understanding how rational functions simplify helps turn complexity into insight.", "---", "Keywords: ( R(x) = \dfrac{(x - 1)(x - 3)}{x - 1} ), simplification, rational functions, domain restrictions, removable discontinuity, ( R(x) = x - 3 ), algebraic identity, calculus preparation.", "---", "Meta Description:\nDiscover how ( R(x) = \dfrac{(x - 1)(x - 3)}{x - 1} ) simplifies algebraically to ( R(x) = x - 3 ), valid for ( x <br/>\ne 1 ). Learn why simplifying rational expressions enhances understanding and problem solving in algebra and calculus."]