Question

Can someone see if my answer to this question is right?

Answer 1

The work of a student to find the dimensions of a rectangle of area 6 + 8x and width 2 is shown below:

Step 1: 6 + 8x
Step 2: 2(4) + 2(6x)
Step 3: 2(4 + 6x)
Step 4: Dimensions of the rectangle are 2 and 4 + 6x

In which step did the student first make an error and what is the correct step?

Step 3; 2 + (4 + 6x)
Step 3; 2 + (4 ⋅ 6x)
Step 2; 2(4) + 2(4x)
Step 2; 2(3) + 2(4x)

Answer 2

Is it D?

Answer 3

hmmm how about if you solve it? what do you get?

notice, you have the Area and the Width
only part missing is the Length

Answer 4

That's true, but how would you get the length?

Answer 5

Area of a rectangle = length * width

we know the width = 2
we know the area is (6+8x)

thus \(\bf 6+8x = length \times 2\) by solving for length

Answer 6

dividing both sides by 2
you should get

\(\bf 6+8x=length\cdot 2\implies \cfrac{6+8x}{2}=\cfrac{length\cdot \cancel{2}}{\cancel{2}}
\\ \quad \\
\cfrac{6+8x}{2}=length\implies \cfrac{\cancel{6}}{\cancel{2}}+\cfrac{\cancel{8}x}{\cancel{2}}=length\)

Answer 7

now.... none of those steps, show a division by 2

so.....hmmm none really correct any errors made