What is the solution for $p$ in the equation
$\frac{1}{3} p+\frac{1}{2} p+7=\frac{7}{6} p+5+4$
A. $p=1$
B. $p=6$
C. $p=-1$
D. $p=-6$
Answer (1)
Combine like terms on both sides of the equation: 6 5 p + 7 = 6 7 p + 9 .
Isolate the variable p by subtracting 6 5 p from both sides: 7 = 6 2 p + 9 .
Subtract 9 from both sides: − 2 = 6 2 p .
Solve for p : p = − 6 .
Explanation
Problem Analysis We are given the equation 3 1 p + 2 1 p + 7 = 6 7 p + 5 + 4 and we want to find the value of p that satisfies this equation.
Simplifying the Equation First, let's simplify both sides of the equation. On the left side, we have 3 1 p + 2 1 p + 7 . To combine the terms with p , we need a common denominator, which is 6. So we rewrite the left side as 6 2 p + 6 3 p + 7 = 6 5 p + 7 . On the right side, we have 6 7 p + 5 + 4 = 6 7 p + 9 . Now our equation is 6 5 p + 7 = 6 7 p + 9 .
Isolating the Variable Next, we want to isolate the terms with p on one side and the constant terms on the other side. Subtract 6 5 p from both sides to get 7 = 6 7 p − 6 5 p + 9 , which simplifies to 7 = 6 2 p + 9 . Now subtract 9 from both sides to get 7 − 9 = 6 2 p , which simplifies to − 2 = 6 2 p .
Solving for p Now, we solve for p . We have − 2 = 6 2 p . Multiply both sides by 6 to get − 12 = 2 p . Divide both sides by 2 to get p = − 6 .
Verification Finally, we can check our solution by substituting p = − 6 back into the original equation: 3 1 ( − 6 ) + 2 1 ( − 6 ) + 7 = 6 7 ( − 6 ) + 5 + 4 . This simplifies to − 2 − 3 + 7 = − 7 + 5 + 4 , which gives 2 = 2 . Since the equation holds true, our solution is correct.
Final Answer Therefore, the solution for p is − 6 .
Examples
When solving mixture problems, such as determining the amount of a solution to add to another to achieve a desired concentration, you often need to set up and solve linear equations. For example, if you have two solutions of different concentrations of acid, say 1/3 and 1/2, and you want to combine them to get a certain amount of a solution with a concentration of 7/6, you would use a similar equation to the one we solved to find the required amounts. This type of problem is common in chemistry and other fields where mixtures are used.
