The rectangle has a perimeter of 126cm and an area of 612cm squared. Its length is 3 more than 4 times the width.Write and solve a system of equations.

Question

Need help with this. What are the dimensions of the rectangle, and what is the equation to solve it?

Answer 1

The perimeter of a rectangle is the sum of twice the length and twice the width.  The area of the same shape is length times width.

 

Let width = x

Let length = 4x + 3

 

 

Using these variables, write the equations that represents the perimeter and area.

 

2(x) + 2(4x + 3) = 126

2x + 8x + 6 = 126

10x = 120              ------>  perimeter

 

 

x(4x + 3) = 612

4x² + 3x = 612

4x² + 3x - 612 = 0          ----->   area

 

 

Since the bolded equations have only one variable, we can rearrange the perimeter equation so that the right side equals zero.  Subtract 120 on both sides of that equation to get the right side of the equation equal to zero and then equate the two expressions.  So now you will have

 

10x - 120 = 4x² + 3x - 612

 

 

Now, we make the left side of this new equation equal to zero by subtracting 10x and adding 120.

 

0 = 4x² - 7x - 492

 

 

Use the quadratic equation to solve for x:

 

x = (-b ± √(b² - 4ac)) / 2a

 

 

where:

a = 4

b = -7

c = -492

 

Plug in these values into the formula to solve for x.  Once you have solved for x, select the positive value of x.