Answer:
The rectangle is 5 inches long by 19 inches wide.
Step-by-step explanation:
Let's start by listing what we know:
Let's represent the width with "w" and length with "l", and make a system of equations based on what we know.
[tex]\left \{ {{2w + 2l = 48} \atop {w = 3l + 4}} \right.[/tex]
Now, let's solve the system of equations using substitution:
[tex]2w + 2l = 48[/tex]
[tex]2 (3l + 4) + 2l = 48[/tex]
[tex]6l + 8 + 2l = 48[/tex]
[tex]8l + 8 = 48[/tex]
[tex]8l = 40[/tex]
[tex]l = 5[/tex]
The length of the rectangle is 5 inches.
Now, let's use this to solve for the width.
[tex]w = 3l + 4[/tex]
[tex]w = 3(5) + 4[/tex]
[tex]w = 15 + 4[/tex]
[tex]w = 19[/tex]
The width of the rectangle is 19 inches.
Let's check our work:
[tex]2w + 2l = 48[/tex]
[tex]2(19) + 2(5) = 48[/tex]
[tex]38 + 10 = 48[/tex]
[tex]48 = 48[/tex]
The results match, so our answer is correct.
The rectangle is 5 inches long by 19 inches wide.
