3x+2y=23 1/2x-4=y What should I do for this one...? I dont really understand

Mathematics

1 answer:

Hitman42 [59]3 years ago

8 0

Since you have y alone on one side of the equation......

Use substitution and plug in 1/2x-4 for y in the other equation.

$3x+2(1/2x-4)=23\\$

Distribute

$3x+x-8=23$

Combine like terms

$4x-8=23$

Add 8 to both sides

$4x=31$

Divide by 4

and $x=31/4$

Plug this value in for x

$31/4*1/2-4=y$

Simplify

$31/8-4=y$

Multiply everything by 8 and simplify

$31-32=8y\\-1=8y$

Divide.....

$y=-1/8$

Your answers are:

$y=-1/8\\x=31/4$

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Find all solutions to the equation in the interval [0,2n) cos x = sin 2x

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Answer:

$\huge\boxed{x\in\left\{\dfrac{\pi}{6};\ \dfrac{\pi}{2}\ \dfrac{3\pi}{2};\ \dfrac{5\pi}{6}\right\}}$

Step-by-step explanation:

$\cos x=\sin2x\qquad|\text{use}\ \sin\theta=2\sin\theta\cos\theta\\\\\cos x=2\sin x\cos x\qquad|\text{subtract}\ \cos x\ \text{from both sides}\\\\0=2\sin x\cos x-\cos x\\\\2\sin x\cos x-\cos x=0\qquad|\text{distribute}\\\\\cos x(2\sin x-1)=0\iff\underbrace{\cos x=0}_{(1)}\ \vee\ \underbrace{2\sin x-1=0}_{(2)}$

$(1)\\\cos x=0\Rightarrow x=\dfrac{\pi}{2}+k\pi;\ k\in\mathbb{Z}$

$(2)\\2\sin x-1=0\qquad|\text{add 1 to both sides}\\\\2\sin x=1\qquad|\text{divide sides by 2}\\\\\sin x=\dfrac{1}{2}\Rightarrow x=\dfrac{\pi}{6}+2k\pi\ \vee\ x=\dfrac{5\pi}{6}+2k\pi$

$\text{From}\ (1)\ \text{and}\ (2)\ \text{we have}\\\\x=\dfrac{\pi}{2}+k\pi\ \vee\ x=\dfrac{\pi}{6}+2k\pi\ \vee\ x=\dfrac{5\pi}{6}+2k\pi\\\\\text{We have the interval}\ x\in[0;2\pi).\ \text{Therefore the solution is:}\\\\x\in\left\{\dfrac{\pi}{6};\ \dfrac{\pi}{2}\ \dfrac{3\pi}{2};\ \dfrac{5\pi}{6}\right\}$