Answer:
[tex]Length =5[/tex]
[tex]Width = 2[/tex]
Step-by-step explanation:
Given
[tex]Length =x[/tex]
[tex]Width = x -3[/tex]
[tex]Area = 10[/tex]
[tex]x(x-3) =10[/tex]
See comment for missing part of the question
Required
Complete the expression to determine the dimension of a rectangle
We have:
[tex]x(x-3) =10[/tex]
Open bracket
[tex]x^2 -3x = 10[/tex]
Equate to 0
[tex]x^2 -3x - 10 =0[/tex]
Expand
[tex]x^2 + 2x - 5x - 10 = 0[/tex]
Factorize
[tex]x(x + 2) - 5(x + 2) = 0[/tex]
Factor out x + 2
[tex](x - 5)(x + 2) = 0[/tex]
Solve for x
[tex]x - 5 = 0[/tex] or [tex]x + 2 = 0[/tex]
[tex]x = 5[/tex] or [tex]x = -2[/tex]
The value of x cannot be negative
So:
[tex]x = 5[/tex]
Recall that:
[tex]Length = x[/tex]
[tex]Width = x - 3[/tex]
So:
[tex]Length =5[/tex]
[tex]Width = 2[/tex] ---- i.e. 5 - 3
