Question

Darpana solved the equation s = a+b+c/3 for a. Her steps are shown below:

1. Multiply by 3: s=a+b+c/3
3s=a+b+c

2. Subtract b: 3s-b=a+b+c-b
3s-b=a+c
3. Divide by c: 3s-b/c=a
Which statement about Darpana’s work is true?
In step 1 she needed to divide by 3 rather than multiply.
In step 2 she needed to add b rather than subtract.
In step 3 she needed to subtract c rather than divide.
Darpana solved the equation correctly.

Answer 1

Answer: Third option (In step 3 she needed to subtract c rather than divide)

Explanation:

When she subtract b from both sides of the equation, she obtained:

`3s-b=a+c`

Therefore, to leave the a alone at one member of the equation, she needs to subtract c from both sides of the equation.

Then, she would obtain the following:

`3s-b-c=a+c-c\\3s-b-c=a\\a=3s-b-c`

Therefore the answer is the third option: In step 3 she needed to subtract c rather than divide.

Answer 2

The true statement is: In step 3 she needed to subtract c rather than divide.

Explanation:

Lets solve our equation
`s=(a+b+c)/(3)` step by step.

Step 1. Since 3 is the denominator of the right hand side, we need to multiply both sides of the equation by 3:

`3s=(3(a+b+c))/(3)`

Now we can cancel the 3 in the numerator and the 3 in the denominator to get

`3s=a+b+c`

As you can see, the first statement is false

Step 2. Since we want to isolate the variable
`a`, we need to subtract b from both sides of the equation:

`3s=a+b+c`

`3s-b=a+b+c-b`

`3s-b=a+c`

The second statement is also false

Step 3. The last thing we to do to isolate
`a` (and solve for it) is subtract c from both sides of the equation:

`3s-b-c=a+c-c`

`3s-b-c=a`

`a=3s-b-c`

Therefore, the third statement is true: In step 3 she needed to subtract c rather than divide.