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2 years ago

Question 1)Solve 8x + 3 = 3x + 18

Mathematics

2 answers:

Dafna1 [17]2 years ago

8 0

Step-by-step explanation:

8x-3x = 18-3(like terms)

or,5x = 15

or,x=15/5

or,x=3

Naddik [55]2 years ago

6 0

Answer:

x = 3

Step-by-step explanation:

in order to solve for the value of x in the expression

8x + 3 = 3x + 18

we are going to simply combine the like terms then later evaluate for the value of x

8x + 3 = 3x + 18

8x + 3 - 3x = 18

8x -3x = 18 -3

5x = 15

divide both sides by the coefficient of x

5x/5 = 15/5

x = 3

therefore the value of x in the expression

8x + 3 = 3x + 18 is equals to 3

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2 years ago

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2 years ago

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Simplify: cos2x-cos4 all over sin2x + sin 4x

GrogVix [38]

Answer:

$\frac{\cos\left(2x\right)-\cos\left(4x\right)}{\sin\left(2x\right)+\sin\left(4x\right)}=\tan\left(x\right)$

Step-by-step explanation:

$\frac{\cos\left(2x\right)-\cos\left(4x\right)}{\sin\left(2x\right)+\sin\left(4x\right)}$

Apply formula:

$\cos\left(A\right)-\cos\left(B\right)=-2\cdot\sin\left(\frac{A+B}{2}\right)\cdot\sin\left(\frac{A-B}{2}\right)$ and

$\sin\left(A\right)+\sin\left(B\right)=2\cdot\sin\left(\frac{A+B}{2}\right)\cdot\sin\left(\frac{A-B}{2}\right)$

We get:

$=\frac{-2\cdot\sin\left(\frac{2x+4x}{2}\right)\cdot\sin\left(\frac{2x-4x}{2}\right)}{2\cdot\sin\left(\frac{2x+4x}{2}\right)\cdot\cos\left(\frac{2x-4x}{2}\right)}$

$=\frac{-\sin\left(\frac{2x-4x}{2}\right)}{\cos\left(\frac{2x-4x}{2}\right)}$

$=\frac{-\sin\left(\frac{-2x}{2}\right)}{\cos\left(\frac{-2x}{2}\right)}$

$=\frac{-\sin\left(-x\right)}{\cos\left(-x\right)}$

$=\frac{-\cdot-\sin\left(x\right)}{\cos\left(x\right)}$

$=\frac{\sin\left(x\right)}{\cos\left(x\right)}$

$=\tan\left(x\right)$

Hence final answer is

$\frac{\cos\left(2x\right)-\cos\left(4x\right)}{\sin\left(2x\right)+\sin\left(4x\right)}=\tan\left(x\right)$

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masha68 [24]

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Question 1)Solve 8x + 3 = 3x + 18​

Question 1)Solve 8x + 3 = 3x + 18