Solution:
note: [tex]A_{rectangle}=LW[/tex]
Step 1: "A rectangular is 5 meters longer than its width"
[tex]A=LW=(W+5)(W)[/tex]
Step 2: "If the length is shortened by 2 meters and the width is increased by 1 meter"
[tex]A=LW=(L-2)(W+1)[/tex]
Step 3: Recall from step 1 that "rectangular is 5 meters longer than its width"
[tex]so: L=W+5[/tex]
Step 4: Equate step 1 & step 2 and substitute "L" with the equation from step 3 and solve
[tex](W+5)(W)=(L-2)(W+1)\\\\(W+5)(W)=[(W+5)-2](W+1)\\\\W^2+5W=(W+5-2)(W+1)\\\\W^2+5W=(W+3)(W+1)\\\\W^2+5W=W^2+W+3W+3\\\\W^2+5W=W^2+4W+3\\\\W^2-W^2+5W-4W=3\\\\W=3[/tex]
Step 5: Solve for L using equation in step 3
[tex]L=W+5\\\\L=3+5\\\\L=8[/tex]
Step 6: Solve for Area
[tex]A=LW=(8)(3)=24m^2[/tex]