Solve the Inequality for x 30< 5x-5(x-6)/2< =60
The problem you've presented is an inequality that asks you to find the range of values for the variable x that satisfies two conditions at the same time. Specifically, you need to determine all the values of x where the value of 5x minus half of 5 times (x minus 6) is greater than 30 but less than or equal to 60. Thus, it's a two-part problem involving simplifying the algebraic expression and solving the resultant inequality to find the solution set for x.
$30 < 5 x - 5 \frac{x - 6}{2} \leq 60$
Combine $-5$ and $\frac{x - 6}{2}$ to get $30 < 5x + \frac{-5(x - 6)}{2} \leq 60$.
Distribute the negative sign across the fraction to obtain $30 < 5x - \frac{5(x - 6)}{2} \leq 60$.
Multiply $5x$ by $\frac{2}{2}$ to write it with a common denominator, resulting in $30 < \frac{5x \cdot 2}{2} - \frac{5(x - 6)}{2} \leq 60$.
Combine the numerators over the common denominator to get $30 < \frac{10x - 5(x - 6)}{2} \leq 60$.
Factor out the common term $5$ from the numerator to simplify the expression to $30 < \frac{5(2x - (x - 6))}{2} \leq 60$.
Apply the distributive property to get $30 < \frac{5(x + 6)}{2} \leq 60$.
Multiply each term by $2$ to eliminate the denominator, resulting in $60 < 5(x + 6) \leq 120$.
Distribute $5$ across $(x + 6)$ to obtain $60 < 5x + 30 \leq 120$.
Subtract $30$ from all parts of the inequality to isolate $x$, yielding $30 < 5x \leq 90$.
Divide each part by $5$ to solve for $x$, giving $6 < x \leq 18$.
The solution in inequality form is $6 < x \leq 18$, and in interval notation, it is $(6, 18]$.
To solve the given inequality $30 < 5x - \frac{5(x - 6)}{2} \leq 60$, we followed a systematic approach:
Simplification: We combined like terms and distributed negative signs where necessary.
Common Denominator: We expressed all terms as fractions over a common denominator to simplify the inequality.
Factoring: We factored out common terms to simplify the numerator of the fraction.
Distributive Property: We applied the distributive property to expand and simplify expressions.
Elimination of Denominator: We multiplied through by the denominator to eliminate fractions from the inequality.
Isolation of the Variable: We moved all terms not containing the variable to one side to isolate the variable.
Division: We divided by the coefficient of the variable to solve for the variable.
Solution Presentation: We presented the solution in both inequality form and interval notation.
This process involves algebraic manipulation, understanding of inequalities, and knowledge of interval notation. It is important to maintain the direction of the inequality signs throughout the process and to reverse them when multiplying or dividing by a negative number (which did not occur in this problem).