Select the correct answer.
What is the solution for $t$ in the equation?
$\frac{2}{3} t-\frac{1}{5} t=2$
A. $t=6$
B. $t=\frac{30}{7}$
C. $t=\frac{7}{30}$
D. $t=\frac{2}{3}$
Answer (1)
Combine the fractions on the left side of the equation by finding a common denominator: 3 2 t − 5 1 t = 15 10 t − 15 3 t .
Simplify the left side of the equation: 15 7 t = 2 .
Multiply both sides by 7 15 to isolate t : t = 2 × 7 15 .
Simplify to find the value of t : t = 7 30 .
Explanation
Understanding the Problem We are given the equation 3 2 t − 5 1 t = 2 and asked to solve for t . This is a linear equation in one variable, and we will solve it by isolating t on one side of the equation.
Finding a Common Denominator First, we need to combine the terms on the left side of the equation. To do this, we need to find a common denominator for the fractions 3 2 and 5 1 . The least common denominator for 3 and 5 is 15. So we rewrite the fractions with the common denominator:
Rewriting the Equation 3 2 = 3 × 5 2 × 5 = 15 10 and 5 1 = 5 × 3 1 × 3 = 15 3 . Now we can rewrite the original equation as 15 10 t − 15 3 t = 2 .
Combining Like Terms Now, we can combine the terms on the left side: 15 10 t − 15 3 t = 15 10 − 3 t = 15 7 t So the equation becomes 15 7 t = 2 .
Isolating t To isolate t , we multiply both sides of the equation by the reciprocal of 15 7 , which is 7 15 :
15 7 t × 7 15 = 2 × 7 15 This simplifies to t = 7 30 .
Final Answer Therefore, the solution for t is 7 30 . Looking at the multiple choice options, we see that option B matches our solution.
Examples
When mixing ingredients for a recipe, you might need to solve an equation like this to determine the exact amount of a certain ingredient to use. For example, if you want to use a blend of two flours and need the total amount of flour to be 2 cups, and the blend is 3 2 of one flour and 5 1 of another flour, you would solve this equation to find out how much of the blend you need. This type of problem also appears in calculating resource allocation, determining rates of work, and solving problems related to proportions and ratios.
