which shows the correct solution of the equation $\frac{1}{2}a+\frac{2}{3}b = 50$, when $b =…

which shows the correct solution of the equation $\frac{1}{2}a+\frac{2}{3}b = 50$, when $b = 30$?\n$\frac{1}{2}a+\frac{2}{3}(30)=50$\n$\frac{1}{2}a + 20=50$\n$\frac{1}{2}a+20 - 20=50 - 20$\n$\frac{1}{2}a=30$\n$2(\frac{1}{2}a)=\frac{1}{2}(30)$\n$a = 15$\n$\frac{1}{2}a+\frac{2}{3}(30)=50$\n$\frac{1}{2}a + 20=50$\n$\frac{1}{2}a+20 + 20=50 + 20$\n$\frac{1}{2}a=70$

which shows the correct solution of the equation $\frac{1}{2}a+\frac{2}{3}b = 50$, when $b = 30$?\n$\frac{1}{2}a+\frac{2}{3}(30)=50$\n$\frac{1}{2}a + 20=50$\n$\frac{1}{2}a+20 - 20=50 - 20$\n$\frac{1}{2}a=30$\n$2(\frac{1}{2}a)=\frac{1}{2}(30)$\n$a = 15$\n$\frac{1}{2}a+\frac{2}{3}(30)=50$\n$\frac{1}{2}a + 20=50$\n$\frac{1}{2}a+20 + 20=50 + 20$\n$\frac{1}{2}a=70$

Answer

Explanation:

Step1: Substitute $b = 30$ into the equation

$\frac{1}{2}a+\frac{2}{3}(30)=50$ Since $\frac{2}{3}\times30 = 20$, the equation becomes $\frac{1}{2}a + 20=50$.

Step2: Isolate the term with $a$

Subtract 20 from both sides of the equation: $\frac{1}{2}a+20 - 20=50 - 20$, which simplifies to $\frac{1}{2}a=30$.

Step3: Solve for $a$

Multiply both sides of the equation by 2 to get $a$ alone. $2\times\frac{1}{2}a=2\times30$, so $a = 60$.

The correct steps should be: $\frac{1}{2}a+\frac{2}{3}(30)=50$ $\frac{1}{2}a + 20=50$ $\frac{1}{2}a+20 - 20=50 - 20$ $\frac{1}{2}a=30$ $2\times\frac{1}{2}a=2\times30$ $a = 60$

The given options have calculation - errors. The first option has an error in the last step where it multiplies the left - hand side by 2 and the right - hand side by $\frac{1}{2}$. The second option has an error in the step of subtracting 20, it adds 20 instead.

If we assume we are just checking the steps and not re - solving from scratch, the correct sequence of steps among the given ones (ignoring the final wrong calculation in the first option) is the first set of steps:

Answer:

The first set of steps (even though the final calculation of $a$ in it is wrong) shows the correct initial substitution and first few steps of isolating the variable $a$.