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trasher [3.6K]
3 years ago
15

Find k so that the three equations 3x-y=2, 2x+8=3y, and y=kx have a common solution

Mathematics
1 answer:
andriy [413]3 years ago
8 0
2x+8=3kx (3k-2)x+8=0
3x-kx=2. (3-k)x-2=0
3k-2)x+10=3-k)x
4k-5=10/x
///////
3x-y-2=2x+8-3y
X+2y-10=0
X+2kx-10=0
K=(10-x)/2x
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Step-by-step explanation:

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Can someone please explain to me how to get this answer?
SIZIF [17.4K]
Answers:
A = 120
b = 45.0
c = 33.2

Side Note: only one triangle is possible
See attached for a visual. I used GeoGebra to draw the triangle.

-------------------------------------------------------------
-------------------------------------------------------------

Explanation:

We are given the following information
B = 35
C = 25
a = 68

We need to find the following
A, b, c

where the lower case letters represent the side lengths; the upper case letters are the angles. The angles are opposite their corresponding sides. For instance, side lowercase b is opposite angle uppercase B. The other letters are positioned the same way.

-----------------------

First use the idea that for any triangle, the three angles (A,B,C) must add to 180 degrees
A+B+C = 180
Replace B and C with 35 and 25. Solve for angle A
A+35+25 = 180
A+60 = 180
A+60-60 = 180-60
A = 120

Now that we know the three angles A = 120, B = 35, C = 25, we can find the missing sides 'b' and 'c'

-----------------------

We will use the law of sines to find side b
sin(A)/a = sin(B)/b
sin(120)/68 = sin(35)/b
b*sin(120) = 68*sin(35) <<--- cross multiply
b = 68*sin(35)/sin(120)
b = 45.037013350222 <<--- use a calculator; this value is approximate
b = 45.0 <<--- round to the nearest tenth

-----------------------

Do the same for side c
sin(A)/a = sin(C)/c
sin(120)/68 = sin(25)/c
c*sin(120) = 68*sin(25) <<--- cross multiply
c = 68*sin(25)/sin(120)
c = 33.1838323365404 <<--- use a calculator; this value is approximate
c = 33.2 <<--- round to the nearest tenth

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Thursday: 6.1

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