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bezimeni [28]
3 years ago
15

What is the solution to the system of equations? X+y=5 X-2y=2 ( , )

Mathematics
2 answers:
kap26 [50]3 years ago
5 0
Y=-x+5
X-2(-x+5)=2
X+2x-10=2
3x=12
X=4

y+4=5
Y=1
The answer is (4,1)
Maurinko [17]3 years ago
3 0
X+y=5
x=5-y

x-2y=2
5-y-2y=2
5-3y=2
-3y=2-5
-3y=-3
-y=-1
y=1

x+1=5
x=4

x-2(1)=2
x-2=2
x=2+2
x=4

The solution of the system of equations is x=4, y=1, or (4,1)
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Hi there!

To find the Trigonometric Equation, we have to isolate sin, cos, tan, etc. We are also given the interval [0,2π).

<u>F</u><u>i</u><u>r</u><u>s</u><u>t</u><u> </u><u>Q</u><u>u</u><u>e</u><u>s</u><u>t</u><u>i</u><u>o</u><u>n</u>

What we have to do is to isolate cos first.

\displaystyle  \large{ cos \theta =  -  \frac{1}{2} }

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Since we know that cos is negative in Q2 and Q3. We will be using π + (ref. angle) for Q3. and π - (ref. angle) for Q2.

<u>F</u><u>i</u><u>n</u><u>d</u><u> </u><u>Q</u><u>2</u>

\displaystyle \large{ \pi -  \frac{ \pi}{3}  =  \frac{3 \pi}{3}  -  \frac{  \pi}{3} } \\  \displaystyle \large \boxed{ \frac{2 \pi}{3} }

<u>F</u><u>i</u><u>n</u><u>d</u><u> </u><u>Q</u><u>3</u>

<u>\displaystyle \large{ \pi  +   \frac{ \pi}{3}  =  \frac{3 \pi}{3}   +   \frac{  \pi}{3} } \\  \displaystyle \large \boxed{ \frac{4 \pi}{3} }</u>

Both values are apart of the interval. Hence,

\displaystyle \large \boxed{ \theta =  \frac{2 \pi}{3} , \frac{4 \pi}{3} }

<u>S</u><u>e</u><u>c</u><u>o</u><u>n</u><u>d</u><u> </u><u>Q</u><u>u</u><u>e</u><u>s</u><u>t</u><u>i</u><u>o</u><u>n</u>

Isolate sin(4 theta).

\displaystyle \large{sin 4 \theta =  -  \frac{1}{ \sqrt{2} } }

Rationalize the denominator.

\displaystyle \large{sin4 \theta =  -  \frac{ \sqrt{2} }{2} }

The problem here is 4 beside theta. What we are going to do is to expand the interval.

\displaystyle \large{0 \leqslant  \theta < 2 \pi}

Multiply whole by 4.

\displaystyle \large{0 \times 4 \leqslant  \theta \times 4 < 2 \pi \times 4} \\  \displaystyle \large \boxed{0 \leqslant 4 \theta < 8 \pi}

Then find the reference angle.

We know that sin(π/4) = √2/2. Hence π/4 is our reference angle.

sin is negative in Q3 and Q4. We use π + (ref. angle) for Q3 and 2π - (ref. angle for Q4.)

<u>F</u><u>i</u><u>n</u><u>d</u><u> </u><u>Q</u><u>3</u>

<u>\displaystyle \large{ \pi +  \frac{ \pi}{4}  =  \frac{ 4 \pi}{4}  +  \frac{ \pi}{4} } \\  \displaystyle \large \boxed{  \frac{5 \pi}{4} }</u>

<u>F</u><u>i</u><u>n</u><u>d</u><u> </u><u>Q</u><u>4</u>

\displaystyle \large{2 \pi -  \frac{ \pi}{4}  =  \frac{8 \pi}{4}  -  \frac{ \pi}{4} } \\  \displaystyle \large \boxed{ \frac{7 \pi}{4} }

Both values are in [0,2π). However, we exceed our interval to < 8π.

We will be using these following:-

\displaystyle \large{ \theta + 2 \pi k =  \theta \:  \:  \:  \:  \:  \sf{(k  \:  \: is \:  \: integer)}}

Hence:-

<u>F</u><u>o</u><u>r</u><u> </u><u>Q</u><u>3</u>

\displaystyle \large{ \frac{5 \pi}{4}  + 2 \pi =  \frac{13 \pi}{4} } \\  \displaystyle \large{ \frac{5 \pi}{4}  + 4\pi =  \frac{21 \pi}{4} } \\  \displaystyle \large{ \frac{5 \pi}{4}  + 6\pi =  \frac{29 \pi}{4} }

We cannot use any further k-values (or k cannot be 4 or higher) because it'd be +8π and not in the interval.

<u>F</u><u>o</u><u>r</u><u> </u><u>Q</u><u>4</u>

\displaystyle \large{ \frac{ 7 \pi}{4}  + 2 \pi =  \frac{15 \pi}{4} } \\  \displaystyle \large{ \frac{ 7 \pi}{4}  + 4 \pi =  \frac{23\pi}{4} } \\  \displaystyle \large{ \frac{ 7 \pi}{4}  + 6 \pi =  \frac{31 \pi}{4} }

Therefore:-

\displaystyle \large{4 \theta =  \frac{5 \pi}{4} , \frac{7 \pi}{4} , \frac{13\pi}{4} , \frac{21\pi}{4} , \frac{29\pi}{4}, \frac{15 \pi}{4} , \frac{23\pi}{4} , \frac{31\pi}{4}  }

Then we divide all these values by 4.

\displaystyle \large \boxed{\theta =  \frac{5 \pi}{16} , \frac{7 \pi}{16} , \frac{13\pi}{16} , \frac{21\pi}{16} , \frac{29\pi}{16}, \frac{15 \pi}{16} , \frac{23\pi}{16} , \frac{31\pi}{16}  }

Let me know if you have any questions!

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