Vinay brought a pizza and cut it into three pieces. When he weighed the first piece it was 7g lighter than the second piece and 4g heavier than the third piece. If the whole pizza weighed 300g. How much did each piece weigh?

Hint: Consider the weight of the first piece as x grams and of second piece as y grams and of third piece as z grams. Use the information given in the question to form three equations. Solve the equations to find the value of the variables, and hence the weights of the three pieces.

Complete step-by-step answer:
In the question, it is given that the weight of the pizza is 300g, and it is cut into three pieces. Let the weight of the first piece is x grams, the weight of the second piece is y grams and the weight of the third piece is z grams.
\[\Rightarrow x+y+z=300\]\[....(1)\]
It is also given that the first piece is 7g lighter than the second piece and 4g heavier than the third piece.
\[\Rightarrow \] x=y-7 and x=z+4
\[\Rightarrow \] y=x+7 \[....(2)\] and z=x-4 \[....(3)\]
To solve the equation, first let’s substitute equation (2) and equation (3) in equation (1). On substitution, we get:
\[x+y+z=300\]
 \[\Rightarrow \] \[x+(x+7)+(x-4)=300\]
\[\Rightarrow \] 3x+3=300
\[\Rightarrow \] x+1=100
\[\Rightarrow \]x=99\[...(4)\]
Therefore, the first piece weighs 99 grams. To find the value of y, we will substitute equation (4) in equation (2). On substitution, we get:
y = x + 7
\[\Rightarrow \] y = 99 + 7
\[\Rightarrow \] y = 106
Hence, the weight of the second piece is 106 grams.
Now, to find the value of z, we will substitute equation (4) in equation (3). On substitution, we get:
z = x – 4
\[\Rightarrow \] z = 99 – 4
\[\Rightarrow \] z = 95
Hence, the weight of the third piece is 95 grams.
Therefore, the weights of the first, second and third pieces are 99 grams, 106 grams and 95 grams respectively.

Note: While calculating, be careful about the signs. Sign mistakes are very common and can result in wrong answers. So, students should be very careful while calculating and while making substitutions.