Verified Answer :

  • Identify the common denominator as \(5a + 2\).
  • Rewrite the entire expression over the common denominator: \(\frac{3a(5a + 2) + 7}{5a + 2}\).
  • Distribute and combine the numerator terms to get \(15a^2 + 6a + 7\).
  • The simplest form of the expression is \(\boxed{\frac{15a^2 + 6a + 7}{5a + 2}}\).

Explanation

  1. Analysis of the Expression

The objective is to simplify the expression \(3a + \frac{7}{5a + 2}\) by combining it into a single fraction. We observe that this is an algebraic expression where the variable \(a\) appears in both the term \(3a\) and the denominator of the fraction \(\frac{7}{5a + 2}\). For this expression to be defined, the denominator must not be zero, so we require \(5a + 2 eq 0\), which means \(a eq -\frac{2}{5}\).

  1. Finding a Common Denominator

To combine the terms into a single fraction, we need to find a common denominator. The term \(3a\) can be written as \(\frac{3a}{1}\). Therefore, we multiply the numerator and denominator of this term by \(5a + 2\) to obtain a common denominator:
\[3a = \frac{3a(5a + 2)}{5a + 2}\]

  1. Combining and Simplifying the Numerator

Now that both terms have the same denominator, we can add the numerators together:
\[\frac{3a(5a + 2)}{5a + 2} + \frac{7}{5a + 2} = \frac{3a(5a + 2) + 7}{5a + 2}\]
Next, we expand the numerator by distributing \(3a\):
\[3a(5a + 2) + 7 = 15a^2 + 6a + 7\]
Thus, the expression becomes \(\frac{15a^2 + 6a + 7}{5a + 2}\).

  1. Final Verification

Finally, we check if the quadratic expression in the numerator, \(15a^2 + 6a + 7\), can be factored or simplified further. We calculate its discriminant \(D = b^2 - 4ac = 6^2 - 4(15)(7) = 36 - 420 = -384\). Since the discriminant is negative, the quadratic has no real roots and cannot be factored over the real numbers. Therefore, the expression is already in its simplest form. The final result is \(\frac{15a^2 + 6a + 7}{5a + 2}\).

Examples

This type of algebraic simplification is fundamental in many fields. For example, in electrical engineering, when analyzing circuits with multiple impedances in series or parallel, you often encounter expressions where resistances or reactances are added to fractional components. Simplifying these into a single fraction helps in determining the equivalent impedance of a circuit, making it easier to solve for current or voltage using Ohm's Law, \(\(V = I \times Z\)\), where \(\(Z\)\) is the simplified expression.

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