Given:
Width of the rectangle =3 and length of the rectangle = 4+x.
Consider the area of the rectangle.
[tex]A=lw[/tex]
Substitute l=4+x and w=3 in the formula.
[tex]A=3(4+x)[/tex][tex]A=12+3x[/tex]
[tex]A=3x+12[/tex]
The area of the rectangle is at least 27 centimeters squared.
[tex]A\le27[/tex]
Substitute A=12+3x in the inequality, we get
[tex]3x+12\le27[/tex]
The inequality is
[tex]3x+12\le27[/tex]
Subtracting 12 from both sides of the inequalities, we get
[tex]3x+12-12\le27-12[/tex]
[tex]3x\le15[/tex]
Dividing both sides by 5, we get
[tex]\frac{3x}{3}\le\frac{15}{3}[/tex][tex]x\le5[/tex]
Hence the solution is
[tex]x\le5[/tex]
