Solve for $m$.
$\frac{5}{3} m+\frac{7}{4}=3$
Write your answer as an integer or simplified fraction.
$m=$
We are given the linear equation \(\frac{5}{3} m + \frac{7}{4} = 3\). Our objective is to determine the value of the variable \(m\). This is a standard algebraic problem that requires isolating the variable \(m\) using properties of equality. First, we need to eliminate the constant term on the left side of the equation. Then, we will address the coefficient of \(m\).
To isolate the term with \(m\), we subtract \(\frac{7}{4}\) from both sides of the equation:
\[\frac{5}{3} m = 3 - \frac{7}{4}\]
To subtract the fractions, we need a common denominator, which is 4. We rewrite 3 as \(\frac{12}{4}\):
\[\frac{5}{3} m = \frac{12}{4} - \frac{7}{4} = \frac{5}{4}\]
Now we have the equation \(\frac{5}{3} m = \frac{5}{4}\). To solve for \(m\), we multiply both sides of the equation by the reciprocal of the coefficient \(\frac{5}{3}\), which is \(\frac{3}{5}\):
\[m = \frac{5}{4} \cdot \frac{3}{5}\]
By simplifying the expression on the right-hand side, we notice that the 5s in the numerator and denominator cancel out:
\[m = \frac{3}{4}\]
Thus, the value of \(m\) is \(\frac{3}{4}\).
In real-life scenarios, solving linear equations like this is essential for scaling recipes or planning budgets. For instance, if a recipe for a certain dish requires \(\frac{5}{3}\) cups of flour plus an extra \(\frac{7}{4}\) cups for a specific topping, and you have exactly 3 cups of flour total, solving this equation tells you exactly what fraction of the batch (\(m\)) you can prepare. Mastering these algebraic steps allows you to adjust resources and manage quantities efficiently in daily tasks.