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3 years ago

X+6y=7 24y=-4x find x and y

Mathematics

2 answers:

antoniya [11.8K]3 years ago

7 0

Answer:

Solve for the first variable in one of the equations, then substitute the result into the other equation.

y = − X/ 6 + 7/ 6

x = X − 7

Step-by-step explanation:

Y_Kistochka [10]3 years ago

6 0

No solution to this problem

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1/6

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3 years ago

Please show me how you find x and determine what x is <br> Thank you I appreciate it :)

AfilCa [17]

well, let's notice something, we have two 45° angles, now each of those will give use an equal opposite side, namely if that one side is "x", the other is her twin, "x" as well.

is a right-triangle, so we can use the pythagorean theorem, and we know the sides are x,x,6.

$\bf \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies 6^2=x^2+x^2 \qquad \begin{cases} c=\stackrel{hypotenuse}{6}\\ a=\stackrel{adjacent}{x}\\ b=\stackrel{opposite}{x}\\ \end{cases} \\\\\\ 36=2x^2\implies \cfrac{36}{2}=x^2\implies 18=x^2\implies \sqrt{18}=x\implies \sqrt{9\cdot 2}=x \\\\\\ \sqrt{3^2\cdot 2}=x\implies \boxed{3\sqrt{2}=x}$

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3 years ago

Rewrite 2tan3x in terms of tanx

blsea [12.9K]

$2\tan3x=2\tan(2x+x)=\dfrac{2\tan2x+2\tan x}{1-\tan2x\tan x}$

Use the same identity to expand $\tan2x$ .

$\tan2x=\tan(x+x)=\dfrac{\tan x+\tan x}{1-\tan x\tan x}=\dfrac{2\tan x}{1-\tan^2x}$

$\implies2\tan3x=\dfrac{2\dfrac{2\tan x}{1-\tan^2x}+2\tan x}{1-\dfrac{2\tan x}{1-\tan^2x}\tan x}$
$2\tan3x=\dfrac{2\tan x\left(\dfrac2{1-\tan^2x}+1\right)}{1-\dfrac{2\tan^2x}{1-\tan^2x}}$
$2\tan3x=\dfrac{2\tan x\left(2+1-\tan^2x\right)}{1-\tan^2x-2\tan^2x}$
$2\tan3x=\dfrac{2\tan x(3-\tan^2x)}{1-3\tan^2x}$

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3 years ago