Answer:
[tex]x=18[/tex]
[tex]y=60[/tex]
[tex]z=40[/tex]
Step-by-step explanation:
From the relations stablished in the problem we have the following equation system:
[tex]y-z=20[/tex] (equation 1)
[tex]\frac{1}{2} xy+420=\frac{1}{2}z(x+30)[/tex] (equation 2)
[tex]\frac{y}{y+z} =\frac{x}{30}[/tex] (equation 3)
From equation 1 we can find an expression of [tex]y[/tex] in terms of [tex]z[/tex] which we're going to call equation 4
[tex]y-z=20[/tex]
[tex]y=z+20[/tex] (equation 4)
We can then replace the equation 4 in the equation 2 in order to find an expression of [tex]x[/tex] in terms of [tex]z[/tex]
[tex]\frac{1}{2} xy+420=\frac{1}{2}z(x+30)[/tex]
[tex]\frac{1}{2} (xy+840)=\frac{1}{2}z(x+30)[/tex]
[tex]xy+840=z(x+30)[/tex]
[tex]x(z+20)+840=z(x+30)[/tex] (here we replaced the eq.4)
[tex]xz+20x+840=xz+30x[/tex]
[tex]xz+20x-xz=30z-840[/tex]
[tex]20x=30z-840[/tex]
[tex]10(2x)=10(3z-84)[/tex]
[tex]x=\frac{1}{2} (3z-84)[/tex] (equation 5)
Now, we can replace equations 4 & 5 inside the equation 3 so we can find the value of [tex]z[/tex]
[tex]\frac{y}{y+z} =\frac{x}{30}[/tex]
[tex]\frac{z+20}{z+20+z} =\frac{1}{30}*\frac{1}{2} (3z-84)[/tex]
[tex]\frac{z+20}{2z+20} =\frac{1}{30}*\frac{1}{2} (3z-84)[/tex]
[tex]\frac{z+20}{2(z+10)} =\frac{1}{2}*\frac{1}{30} (3z-84)[/tex]
[tex]\frac{1}{2}*\frac{z+20}{z+10} =\frac{1}{2}*\frac{1}{30} (3z-84)[/tex]
[tex]\frac{z+20}{z+10} =\frac{1}{10} (\frac{3z}{3}-\frac{84}{3})[/tex]
[tex]\frac{z+20}{z+10} =\frac{1}{10} (z-28)[/tex]
[tex]z+20 =\frac{1}{10} (z-28)*(z+10)[/tex]
[tex]10(z+20) =z^{2}+10z-28z-280[/tex]
[tex]10z+200 =z^{2}-18z-280[/tex]
[tex]z^{2}-28z-480=0[/tex]
This is a quadratic equation which has the form [tex]a*z^{2} +b*z+c=0[/tex]
where
[tex]a=1[/tex]
[tex]b=-28[/tex]
[tex]c=-480[/tex]
Then, we can find the solutions to this quadratic equation using the well-know quadatric formula which says that
[tex]z=\frac{-b}{2a}[/tex]±[tex]\frac{\sqrt{b^{2}-4ac} }{2a}[/tex]
then, replacing the values of a, b and c we find the values of z
[tex]z_{1}=\frac{-(-28)+\sqrt{(-28)^{2}-4(1)(-480)} }{2(1)}[/tex]
[tex]z_{1}=40[/tex]
[tex]z_{2}=\frac{-(-28)-\sqrt{(-28)^{2}-4(1)(-480)} }{2(1)}[/tex]
[tex]z_{2}=-12[/tex]
We have two possible values of z, but because we're trying to find the measure of trapezoid's height the result shouldn't be negative, so we keep only the positive value of z, then
[tex]z=40[/tex]
Now we may replace this value of z in the equations 4 & 5 in order to find the values of x & y.
[tex]y=z+20[/tex] (equation 4)
[tex]y=40+20[/tex]
[tex]y=60[/tex]
[tex]x=\frac{1}{2} (3z-84)[/tex] (equation 5)
[tex]x=\frac{1}{2} (3(40)-84)[/tex]
[tex]x=18[/tex]
So we've found the values of x, y, and z.
[tex]x=18[/tex]
[tex]y=60[/tex]
[tex]z=40[/tex]
