The length of a rectangle is 2 units more than the width. The area of the rectangle is 63 units. What is the length, in units, of the rectangle?

Length = 9units

Width = 7units

Step-by-step explanation:

It is said that the length is 2units more than the width

Assume that the width is x, then the length will be 2 + x

ie

Width = x

Length = 2 + x

Area of the rectangle = 63units

Area of rectangle = l * b

l - length of the rectangle

b - width of the rectangle

A = l * b

63 = (2 + x) * x

63 = ( 2 + x) x

63 = 2x + x^2

Let's rearrange it

x^2 + 2x - 63 = 0

Let's find the factor of 63

A factor that can be multiplied to give -63 and that can be added to give +2

Let's use -7 and +9

x^2 - 7x + 9x - 63 = 0

Separate with brackets

( x^2 - 7x) + ( 9x - 63) = 0

x( x - 7) + 9(x - 7) = 0

( x + 9)(x - 7) = 0

( x + 9) = 0

( x - 7) = 0

x + 9 = 0

x = -9

x - 7 = 0

x = 7

Note: the length of a rectangle can not be negative

So therefore,

x = 7

Length = 2 + x

= 2 + 7

= 9units

Width = x

= 7units

(0,-3) see attached graph

step-by-step explanation:

a. to solve the system, graph the two equations.

x = y+3 becomes y=x-3 has m=1/1 and b=-3. start at (0,-3) on the y-axis and plot a point. then move up 1 unit and 1 unit to the right. plot a point and connect.

y=-4x+3 has m =-4/1 and b=-3. start at (0,-3) on the y-axis and plot a point. then move down four units and to the right 1 unit. plot a point and connect.

the solution is the point (x,y) where the two lines intersect. by the graph below it is (0,-3).

b. to solve by substitution, substitute y = -4x-3 into x= y+3

x= (-4x-3) +3

x=-4x

5x=0

x=0

now substitute x = 0 back in to find y.

y= -4(0)-3

y=0-3

y=-3

solution is (0,-3)

c. use elimination by adding the equations together.

x-y=3

4x+y=-3

5x=0

x=0

then substitute into one equation to find y.

y=-4(0)-3

y=-3