Answer (#1) if ever typo ung given (check picture): It is correct, but I suggest using the GCF to get the answer right away instead.
[tex]24x^2y+36xy=12xy(2x+3)[/tex]
Answer (#1):
It is wrong because you cannot factor by suddenly inserting a coefficient into a term.
[tex]x^2y+36xy=xy(x+36)[/tex]
Answer (#2):
[tex]a^2+4a=21\\\\a^2+4a-21=0\\\\(a-3)(a+7)=0\\\\a-3=0;a+7=0\\\\a=3;a=-7[/tex]
Answer (#3):
[tex]A_{rectangle}=LW\\\\9=(2W-3)(W)\\\\9=2W^2-3W\\\\2W^2-3W-9=0\\\\2W^2+(-6W+3W)-9=0\\\\2W^2-6W+3W-9=0\\\\(2W^2-6W)+(3W-9)=0\\\\2W(W-3)+3(W-3)=0\\\\(2W+3)(W-3)=0\\\\2W+3=0;W-3=0\\\\2W=-3;W=3\\\\W=-\frac{3}{2};W=3[/tex]
note that: [tex]L=2W-3[/tex]
Case 1: [tex]W=-\frac{3}{2}\\\\[/tex]
[tex]L=2W-3=2(-\frac{3}{2})-3=3-3=0[/tex]
Case 2: [tex]W=3[/tex]
[tex]L=2W-3=2(3)-3=6-3=3[/tex]
Therefore the length is 3 sq. ft. and the width is 3 sq. ft.
Answer (#4):
note: [tex]A_{rectangle}=LW[/tex]
Step 1: Form equation to solve for shaded area
[tex]A_{shaded}=A_{big,rectangle}-A_{unshaded}\\\\\\[/tex]
Step 2: Get area equation for big rectangle
[tex]A_{big,rectangle}=LW\\\\A_{big,rectangle}=(3+2x+3)(3+x+3)\\\\A_{big,rectangle}=(2x+6)(x+6)\\\\A_{big,rectangle}=2x^2+12x+6x+36\\\\A_{big,rectangle}=2x^2+18x+36[/tex]
Step 3: Get area equation for unshaded rectangle
[tex]A_{unshaded}=LW\\\\A_{unshaded}=(2x)(x)\\\\A_{unshaded}=2x^2[/tex]
Step 4: Substitute steps 3 & 4 to equation in step 1
[tex]A_{shaded}=A_{big,rectangle}-A_{unshaded}\\\\A_{shaded}=(2x^2+18x+36)-(2x^2)\\\\A_{shaded}=2x^2+18x+36-2x^2\\\\A_{shaded}=18x+36[/tex]
Step 5: Recall the statement, "...if the area of the big rectangle is 6 times the area of the unshaded rectangle"
[tex]A_{big,rectangle}=6A_{unshaded}[/tex]
Step 6: Substitute step 5 to equation in step 4
[tex]A_{shaded}=A_{big,rectangle}-A_{unshaded}\\\\A_{shaded}=6A_{unshaded}-A_{unshaded}\\\\A_{shaded}=5A_{unshaded}\\\\A_{shaded}=5(2x^2)\\\\A_{shaded}=10x^2[/tex]
Step 7: Equate step 6 and step 4, the solve for x
[tex]10x^2=18x+36\\\\10x^2-18x-36=0\\\\10x^2+(-30x+12x)-36=0\\\\(10x^2-30x)+(12x-36)=0\\\\10x(x-3)+12(x-3)=0\\\\(10x+12)(x-3)=0\\\\10x+12=0;x-3=0\\\\x=-\frac{6}{5};x=3[/tex]
Final Step: Substitute step 7 with equation of area of shaded region (step 4 or 6) using positive value of x (since measurement/dimension of shapes cannot be negative)
using step 4 equation:
[tex]18x+36=18(3)+36=54+36=90 sq. units[/tex]
using step 6 equation:
[tex]10x^2=10(3)^2=10(9)=90 sq. units[/tex]
