Question Video: Addition and Subtraction of Algebraic Expressions | Nagwa Question Video: Addition and Subtraction of Algebraic Expressions | Nagwa

# Question Video: Addition and Subtraction of Algebraic Expressions Mathematics • First Year of Preparatory School

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Consider three rectangles whose areas are represented by the expressions. All three rectangles are put together such that none of them overlap. Write an expression to represent the area of the new combined shape. If shape A is removed, and instead it sticks to a fourth rectangle, D, such that the shapes A and D do not overlap, the combined area of A and D is 16π¦π§ β 5. Write an expression to represent the area of rectangle D.

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### Video Transcript

Consider three rectangles whose areas are represented by the expressions below. All three rectangles are put together such that none of them overlap. Write an expression to represent the area of the new combined shape. If shape A is removed and instead it sticks to a fourth rectangle, D, such that the shapes A and D do not overlap, the combined area of A and D is 16π¦π§ minus five. Write an expression to represent the area of rectangle D.

This question has two parts, and the first part asks us to find the combined area of the three given rectangles. Since the rectangles do not overlap, the area of the combined shape will be the sum of their areas. Letβs clear some space so we can calculate this.

The area of rectangle A is 13π₯ minus seven π¦π§. The area of rectangle B is two π₯ plus four π¦π§. And the area of rectangle C is negative four π₯ plus nine π¦π§. The combined area is therefore as shown. We can simplify this expression by combining like terms. 13π₯ plus two π₯ plus negative four π₯ is equal to 11π₯. And negative seven π¦π§ plus four π¦π§ plus nine π¦π§ is equal to six π¦π§.

We can therefore conclude that the combined area is equal to 11π₯ plus six π¦π§. This is the answer to the first part of this question.

In the second part of the question, rectangle A is combined with a fourth rectangle, D. We are told that the combined area of rectangles A and D is equal to 16π¦π§ minus five. This means that 13π₯ minus seven π¦π§ plus the area of rectangle D must be equal to 16π¦π§ minus five. Subtracting 13π₯ and adding seven π¦π§ to both sides of this equation, we have the area of rectangle D is equal to 16π¦π§ minus five minus 13π₯ plus seven π¦π§.

Once again, we can combine like terms. The area of rectangle D is equal to negative 13π₯ plus 23π¦π§ minus five. This is the answer to the second part of the question.

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