Processing Math: Done

From Förberedande kurs i matematik 1

First, let's decide to determine all distance in dm (decimeters), so that we have all the distances as integers.

Call the length of the washing line from the trees to the hanger y and z, as in the figure below, and introduce two auxiliary triangles which have y and z as their hypotenuses. (As an approximation, we suppose that the taut washing line consists of two straight parts.)

Because the line is 54 dm long, we have

y+z=54.

(1)

Then, the Pythagorean theorem gives the relations

y2=x2+122

(2)

z2=(x+6)2+362.

(3)

The idea now is to solve the system of equations (1)-(3) by first eliminating z, so that we get two equations which only contain x and y. Then, eliminate y from one of these equations, so that we get an equation which determines x.

From (1), we have z=54−y, and substituting this into (3) gives us the equation

(54−y)2=(x+6)2+362.

(3')

Equations (2) and (3') together give a smaller system for x and y,

y2=x2+122(54−y)2=(x+6)2+362.

(2)(3)

Expand the quadratic terms on both sides of (3'),

542−254y+y2=x2+26x+62+362

and simplify

2916−108y+y2=x2+12x+1332.

Use (2) and replace y2 with x2+12 in this equation,

2916−108y+x2+144=x2+12x+1332

which gets rid of the x²-term,

2916−108y+144=12x+1332

and further simplification gives the equation

12x+108y=1728

(3")

If we pause for a moment and summarize the situation, we see that we have succeeded in simplifying the equation system (2) and (3') to a system (2) and (3"), where one of the equations is linear

y2=x2+12212x+108y=1728.

(2)(3")

In this system, we can make y the subject in (3"),

y=1081728−12x=16−x9

and substitute into (2),

16−x92=x2+144.

This is an equation which only contains x, and if we solve it, we will get our answer.

Expand the quadratic on the left-hand side,

162−216x9+x92=x2+144

and collect together all terms on one side,

x2−x281+932x+144−162=0

which gives the equation

8180x2+932x−112=0.

Multiply both sides by 8180 so that we get the equation in standard form,

x2+518x−5567=0.

Completing the square on the left-hand side gives

x+592−592−5567=0

and then

x+592=2581+5567=252916

i.e.

x=−59252916=−59554.

This means that the equation has the solutions

x=−59−554=−563andx=−59+554=9.

The answer is thus x=9 dm (the negative root must be discarded).

To be sure that we have calculated correctly, we also look at the values of y and z, and check that the original equations (1) to (3) are satisfied.

Equation (3") gives

y=16−x9=16−1=15

and equation (1) gives

z=54−y=54−15=39.

Now, we check that x=9, y=15 and z=39 satisfy the equations (1), (2) and (3),

LHS of (1)RHS of (1)LHS of (2)RHS of (2)LHS of (3)RHS of (3)=15+39=54=54=152=225=92+122=81+144=225=392=1521=(9+6)2+362=152+362=225+1296=1521.