Question Video: Creating Quadratic Equations with One Variable Mathematics

A rectangular photograph measuring 6 cm by 4 cm is to be displayed in a card mount in a rectangular frame, as shown in the diagram. Which of the following equations can be used to find π‘₯ if the area of the mount is 64 cmΒ²? [A] (9 + 5π‘₯) (3 + 7π‘₯) βˆ’ 15 = 18 [B] (5 + 2π‘₯)(7 βˆ’ 2π‘₯) βˆ’ 24 = 28 [C] (3 + 2π‘₯)(6 βˆ’ 5π‘₯) βˆ’ 21 = 32 [D] (7 + 2π‘₯)(5 + 2π‘₯) βˆ’ 13 = 64 [E] (4 + 2π‘₯)(6 + 2π‘₯) βˆ’ 24 = 64

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Video Transcript

A rectangular photograph measuring six centimeters by four centimeters is to be displayed in a card mount in a rectangular frame, as shown in the diagram. Which of the following equations can be used to find π‘₯ if the area of the mount is 64 centimeters squared? (A) Nine plus five π‘₯ times three plus seven π‘₯ minus 15 equals 18. (B) Five plus two π‘₯ times seven minus two π‘₯ minus 24 equals 28. (C) Three plus two π‘₯ times six minus five π‘₯ minus 21 equals 32. (D) Seven plus two π‘₯ times five plus two π‘₯ minus 13 equals 64. Or (E) four plus two π‘₯ times six plus two π‘₯ minus 24 equals 64.

First, let’s think about what the area of the mount would be in the diagram. The area of the card mount would be this space. We can say that the card mount area will be equal to the area of the larger rectangle minus the area of the photograph. We know that for all rectangles, their area can be defined as the length times the width. The interior rectangle, the photograph, measures six centimeters by four centimeters. And therefore, its area can be found by multiplying six times four.

For the larger rectangle, we’ll need to think a bit more carefully. The distance lengthwise for the larger rectangle will be equal to π‘₯ plus six plus π‘₯. We can write that as six plus two π‘₯. Similarly, the width of the larger rectangle would then be equal to π‘₯ plus four plus π‘₯, which we can write as four plus two π‘₯. Our question says that the area of the mount is 64 centimeters squared. Our equation now looks like this: 64 equals six plus two π‘₯ times four plus two π‘₯ minus six times four. And six times four is 24. This exactly matches option (E).

If we wanted to eliminate the other options, option (A) has terms with five π‘₯ and seven π‘₯, which do not match the larger rectangle. Option (B) have numbers five and seven, which we don’t find for the larger rectangle. Option (C) again has five π‘₯ and is using subtraction. Option (D) has five and seven and additionally is subtracting 13 for the area of the photograph, which again shows that the only viable option is option (E). Four plus two π‘₯ times six plus two π‘₯ minus 24 equals 64.

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