Question #e4fda

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Question #e4fda

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Answer 1 · 2017-10-17T01:35:21

The width is $16 cm$ $"16 cm"$ and the length is $23 cm$ $"23 cm"$ .

Explanation:

1) Choose a variable to represent the length and another for the width. Here I chose

$x = length$ $x = "length"$

$y = width$ $y = "width"$

2) Write an equation to represent what you know. The length $(x)$ $(x)$ is $7 cm$ $"7 cm"$ more than the width $(y)$ $(y)$ , or

$x = y + 7$ $x=y+7$

3) Write an equation for your perimeter. REMEMBER, the perimeter is the total of all four sides, so $2 \cdot length \cdot 2 \cdot width$ $2 * "length" * 2 * "width"$ . or

$y + y + x + x = 78$ $y+y+x+x=78$

which can be written as

$2 y + 2 x = 78 cm$ $2y+2x="78 cm"$

4) You cannot solve an equation with two unknowns, so replace $x$ $x$ with $y + 7$ $y+7$

5) Now your equation is

$2 y + 14 + 2 y = 78 cm$ $2y+14+2y= "78 cm"$

6) Simply and combine like terms

$4 y + 14 = 78 cm$ $4y+14 = "78 cm"$

7) Subtract $14$ $14$ from both sides to get the variable by itself

$4 y = 64 cm$ $4y = "64 cm"$

8) Divide both sides by $4$ $4$ to solve to for one $y$ $y$

$y = 16 cm$ $y = "16 cm"$

9) Now you know the width is $16 cm$ $"16 cm"$ , $y = 16$ $y=16$

10) Plug it back into the equation $x = y + 7$ $x=y+7$ to solve for the length

$x = 16 + 7$ $x=16+7$

$x = 23$ $x=23$

The length is $23 cm$ $"23 cm"$ .

EXTRA: You can plug these variables into the perimeter equation $(2 y + 2 x = 78)$ $(2y+2x=78)$ to see if it matches to check your answer.

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