Answer:
The wide of the mat strip showing around the picture is 2 in.
Step-by-step explanation:
Please refer to the diagram attached at the bottom of these answer.
Since the mat's perimeter is 86 in,
[tex]2L+2W=86 in[/tex] (eq.1)
Now, the paper is laid son there is a uniformly wide strip of the map showing all around the picture, so all four distances from the perimeter of the mat to the paper are the same; we will call this distance "a".
As you can see in the diagram:
W - 15 in = 2a
L - 20 in = 2a
So:
W - 15 in = L - 20 in (eq.2)
So with eq.1 and eq.2 we have a system of two equations and two unknowns.
We will solve for L in eq.1:
[tex]L=\frac{86in-2W}{2}=\frac{86in}{2}-\frac{2W}{2}=43in-W[/tex]
Solving for W in eq.2:
[tex]W=L-20in+15in=L-5in[\tex]
Replacing L in this solution:
[tex]W=(43in-W)-5in[\tex]
[tex]W=38in-W[\tex]
[tex]W+W=38in[\tex]
[tex]2W=38in[\tex]
[tex]W=\frac{38in}{2}=19in[\tex]
As we said before, W - 15 in = 2a, then:
[tex]19in-15in=2a[\tex]
[tex]4in=2a[\tex]
[tex]a=\frac{4in}{2}=2in[\tex]
Which is the answer to the question.
We can also solve eq.1 to find L:
L = 43 in - W = 43 in - 19 in = 24 in