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3 years ago

If 12y=75+5x+2x-1. What is the value of x

Mathematics

2 answers:

AleksandrR [38]3 years ago

6 0

Answer:

x = $\frac{12y - 74}{7}$

Step-by-step explanation:

12y = 75 + 5x + 2x - 1

combine like terms:

12y = 74 + 7x

subtract 74 from both sides:

7x = 12y - 74

divide both sides by 7:

x = $\frac{12y - 74}{7}$

natta225 [31]3 years ago

6 0

Answer:

x = 7/12

Step-by-step explanation:

So we are trying to put it in y = mx + b form

So first we just have to collect like terms

12y = 74 + 7x

Divide both sides by 12

y = 7/12x + 74/12

We have to reduce to its lowest terms

y = 7/12x + 37/6

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As a Bernoulli equation:

$x^2\dfrac{\mathrm dy}{\mathrm dx}=6y^2+6xy\iff x^2y^{-2}\dfrac{\mathrm dy}{\mathrm dx}-6xy^{-1}=6$

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$x^6\dfrac{\mathrm dz}{\mathrm dx}+6x^5z=-6x^4$
$\dfrac{\mathrm d}{\mathrm dx}[x^6z]=-6x^4$
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$z=-\dfrac6{5x}+\dfrac C{x^6}$
$y^{-1}=-\dfrac6{5x}+\dfrac C{x^6}$
$y=\dfrac1{\frac C{x^6}-\frac6{5x}}$
$y=\dfrac{5x^6}{C-6x^5}$

With $y(3)=6$ , we get

$6=\dfrac{5(3)^6}{C-6(3)^5}\implies C=\dfrac{4131}2$

so the solution is

$y=\dfrac{5x^6}{\frac{4131}2-6x^5}=\dfrac{10x^6}{4131-12x^5}$

which is valid as long as the denominator is not zero, which is the case for all $x\neq\sqrt[5]{\dfrac{4131}{12}}$ .

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3 years ago

Solve the exponential equation:<br> B.) 5e^x-7.1=22

brilliants [131]

5e^x=7.1+22
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ln(e^x)=ln(29.1/5)
x=ln(29.1/5)
Use a graphing calculator

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