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Marysya12 [62]
2 years ago
12

The length of the sides of a square measure 2x-5. The length of a rectangle measures 2x, and the width measures x + 2. For what

value of x is the perimeter of the square the same as the perimeter of the rectangle?
Mathematics
1 answer:
vladimir1956 [14]2 years ago
3 0

Answer:

x = 12

Step-by-step explanation:

Perimeter of a square:

The perimeter of a square of side x is given by:

P_s = 4x

Perimeter of a rectangle:

The perimeter of a rectangle of length l and width w is given by:

P_r = 2(l + w)

The length of the sides of a square measure 2x-5.

This means that the perimeter of the square is:

P_s = 4(2x - 5) = 8x - 20

The length of a rectangle measures 2x, and the width measures x + 2.

This means that the perimeter of the rectangle is:

P_r = 2(2x + x + 2) = 2(3x + 2) = 6x + 4

For what value of x is the perimeter of the square the same as the perimeter of the rectangle?

This is x for which:

P_s = P_r

So

8x - 20 = 6x + 4

8x - 6x = 20 + 4

2x = 24

x = \frac{24}{2}

x = 12

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Step-by-step explanation:

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VikaD [51]

Let's work to solve this system of equations:

y = 2x ~~~~~~~~\gray{\text{Equation 1}}y=2x        Equation 1

x + y = 24 ~~~~~~~~\gray{\text{Equation 2}}x+y=24        Equation 2

The tricky thing is that there are two variables, xx and yy. If only we could get rid of one of the variables...

Here's an idea! Equation 11 tells us that \goldD{2x}2x and \goldD yy are equal. So let's plug in \goldD{2x}2x for \goldD yy in Equation 22 to get rid of the yy variable in that equation:

\begin{aligned} x + \goldD y &= 24 &\gray{\text{Equation 2}} \\\\ x + \goldD{2x} &= 24 &\gray{\text{Substitute 2x for y}}\end{aligned}  

x+y

x+2x

​    

=24

=24

​    

Equation 2

Substitute 2x for y

​  

Brilliant! Now we have an equation with just the xx variable that we know how to solve:

x+2x3x 3x3x=24=24=243=8Divide each side by 3

Nice! So we know that xx equals 88. But remember that we are looking for an ordered pair. We need a yy value as well. Let's use the first equation to find yy when xx equals 88:

\begin{aligned} y &= 2\blueD x &\gray{\text{Equation 1}} \\\\ y &= 2(\blueD8) &\gray{\text{Substitute 8 for x}}\\\\ \greenD y &\greenD= \greenD{16}\end{aligned}  

y

y

y

​    

=2x

=2(8)

=16

​    

Equation 1

Substitute 8 for x

​  

Sweet! So the solution to the system of equations is (\blueD8, \greenD{16})(8,16). It's always a good idea to check the solution back in the original equations just to be sure.

Let's check the first equation:

\begin{aligned} y &= 2x \\\\ \greenD{16} &\stackrel?= 2(\blueD{8}) &\gray{\text{Plug in x = 8 and y = 16}}\\\\ 16 &= 16 &\gray{\text{Yes!}}\end{aligned}  

y

16

16

​    

=2x

=

?

2(8)

=16

​    

Plug in x = 8 and y = 16

Yes!

​  

Let's check the second equation:

\begin{aligned} x +y &= 24 \\\\ \blueD{8} + \greenD{16} &\stackrel?= 24 &\gray{\text{Plug in x = 8 and y = 16}}\\\\ 24 &= 24 &\gray{\text{Yes!}}\end{aligned}  

x+y

8+16

24

​    

=24

=

?

24

=24

​    

Plug in x = 8 and y = 16

Yes!

​  

Great! (\blueD8, \greenD{16})(8,16) is indeed a solution. We must not have made any mistakes.

Your turn to solve a system of equations using substitution.

Use substitution to solve the following system of equations.

4x + y = 284x+y=28

y = 3xy=3x

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ik it doesnt make any sinse but hope this helps

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