So what we have to do to solve this problem is to write down those values in 2 equations (one that represents what you sold and the other what your friend sold) compare them and find how much each ticket is worth.
First equation : 11x + 8y = 158
Where x = how much each adult ticket is
and y = how much each student ticket is
The second equation is : 5x + 17y = 152
Using the method of substitution , we can compare each equation side by side:
11x + 8y = 158
5x + 17y = 152
Now we need to set one of the variables of both equations so they are equal:
11(5)x + 8(5) = 158(5)
5(11)x + 17(11)x = 152(11)
55x + 40y = 790
55x + 187y = 1672
Then we subtract the second equation by the first one
55x-55x + 187y - 40y = 1672 - 790
147y = 882
y = 6
The we apply y to one of the equations to discover x :
11x + 8y = 158
11x + 8(6) = 158
11x + 48 = 158
11x = 110
x = 10
So the awnser is :
Each adult ticket (x) is $10
And each student ticket (y) is $8
I hope you understood my explanation,
Divide it and get 0.6 then multiply by 100 so 60%
Slope of line = tan(120) = -tan(60) = - √3
Distance from origin = 8
Let equation be Ax+By+C=0
then -A/B=-√3, or
B=A/√3.
Equation becomes
Ax+(A/√3)y+C=0
Knowing that line is 8 units from origin, apply distance formula
8=abs((Ax+(A/√3)y+C)/sqrt(A^2+(A/√3)^2))
Substitute coordinates of origin (x,y)=(0,0) =>
8=abs(C/sqrt(A^2+A^2/3))
Let A=1 (or any other arbitrary finite value)
solve for positive solution of C
8=C/√(4/3) => C=8*2/√3 = (16/3)√3
Therefore one solution is
x+(1/√3)+(16/3)√3=0
or equivalently
√3 x + y + 16 = 0
Check:
slope = -1/√3 .....ok
distance from origin
= (√3 * 0 + 0 + 16)/(sqrt(√3)^2+1^2)
=16/2
=8 ok.
Similarly C=-16 will satisfy the given conditions.
Answer The required equations are
√3 x + y = ± 16
in standard form.
You can conveniently convert to point-slope form if you wish.
4 times 1/3 is 1.333.. or 1 1/3 feet.
