Which shows the correct solution of the equation [tex]$\frac{1}{2} a+\frac{2}{3} b=50$[/tex], when [tex]$b=30$[/tex]?[tex]$\begin{array}{r}\frac{1}{2} a+\frac{2}{3}(30)=50 \\ \frac{1}{2} a+20=50\end{array}$[/tex][tex]$\begin{array}{r}\frac{1}{2} a+20-20=50-20 \\ \frac{1}{2} a=30 \\ 2\left(\frac{1}{2} a\right)=\frac{1}{2}(30) \\ a=15\end{array}$[/tex][tex]$\begin{aligned}\frac{1}{2} a+\frac{2}{3}(30) & =50 \\ \frac{1}{2} a+20 & =50\end{aligned}$[/tex][tex]$\frac{1}{2} a+20+20=50+20$\frac{1}{a} a=70$[/tex]
Answer (2)
Substitute b = 30 into the equation 2 1 a + 3 2 b = 50 .
Simplify the equation to 2 1 a + 20 = 50 .
Isolate the term with a : 2 1 a = 30 .
Solve for a : a = 60 .
The correct solution is 60 .
Explanation
Problem Setup and Analysis We are given the equation 2 1 a + 3 2 b = 50 and the condition b = 30 . We need to find the correct solution for a . Let's analyze the given solutions step by step.
Substitution First, substitute b = 30 into the equation: 2 1 a + 3 2 ( 30 ) = 50
Simplification Simplify the equation: 2 1 a + 20 = 50
Isolating the Variable Now, isolate the term with a by subtracting 20 from both sides: 2 1 a = 50 − 20 2 1 a = 30
Solving for a Solve for a by multiplying both sides by 2: a = 2 ( 30 ) a = 60
Analyzing the First Solution Now let's examine the first provided solution: 2 1 a + 3 2 ( 30 ) = 50 2 1 a + 20 = 50
2 1 a + 20 − 20 = 50 − 20 2 1 a = 30 2 ( 2 1 a ) = 2 1 ( 30 ) a = 15 The error in this solution is in the last step: 2 ( 2 1 a ) = 2 1 ( 30 ) should be 2 ( 2 1 a ) = 2 ( 30 ) , which leads to a = 60 , not a = 15 .
Analyzing the Second Solution Now let's examine the second provided solution: 2 1 a + 3 2 ( 30 ) = 50 2 1 a + 20 = 50 2 1 a + 20 + 20 = 50 + 20 a 1 a = 70 This solution is incorrect because it adds 20 to both sides incorrectly and then makes a mistake in the last step.
Conclusion The correct solution is a = 60 . Neither of the provided solutions is correct. The first solution makes an arithmetic error in the last step, and the second solution makes multiple errors.
Examples
Imagine you're baking a cake and need to adjust the recipe. The original recipe calls for a certain amount of flour and sugar, but you want to use a different amount of sugar. By substituting the new sugar quantity into the recipe equation, you can solve for the adjusted amount of flour needed to maintain the cake's consistency. This is a practical application of solving equations with substitution in everyday cooking!
After substituting b = 30 into the equation and simplifying, we find that the correct value for a is 60 . This was determined by isolating a and solving the linear equation. Therefore, the solution is 60 .
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