1.

2 answers:

6 0

Xy is the number

a.
unit digit is y
tens is x
units is twice tens digit
y=2x
if it is doubled, it will be 12 more than reversed
20x+2y=12+10y+x
20x+4x=12+20x+x
3x=12
x=4
y=2x
y=2(4)
y=8
48 is the number

b. 8 times sum of digis is bigger than that number by 19
8(x+y)=19+10x+y
x=y-2

8(y-2+y)=19+10(y-2)+y
16y-16=19+10-20+y
y=3
x=y-2
x=3-2
x=1
the number is 13

c.
y=2x
2(10x+y)=9+10y+x
2(10x+2x)=9+20x+x
x=3
y=2x
y=3(3)
y=6

the number is 36

a. 48
b. 13
c. 36

xeze [42]3 years ago

4 0

X: the unit digit
y: the tens digit

The unit digit is twice the tens digit so:
x=2y

The two digit number:
10y+x

The two digit number doubled:
2(10y+x)

The reversed number:
10x+y

So:
2(10y+x)=10x+y+12

substitute x for 2y

2(10y+2y)=10(2y)+y+12

24y=21y + 12

3y=12
y=4

x=2y=2*4=8

So the number is 48.

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(x^2y+e^x)dx-x^2dy=0

klio [65]

It looks like the differential equation is

$\left(x^2y + e^x\right) \,\mathrm dx - x^2\,\mathrm dy = 0$

Check for exactness:

$\dfrac{\partial\left(x^2y+e^x\right)}{\partial y} = x^2 \\\\ \dfrac{\partial\left(-x^2\right)}{\partial x} = -2x$

As is, the DE is not exact, so let's try to find an integrating factor µ(x, y) such that

$\mu\left(x^2y + e^x\right) \,\mathrm dx - \mu x^2\,\mathrm dy = 0$

*is* exact. If this modified DE is exact, then

$\dfrac{\partial\left(\mu\left(x^2y+e^x\right)\right)}{\partial y} = \dfrac{\partial\left(-\mu x^2\right)}{\partial x}$

We have

$\dfrac{\partial\left(\mu\left(x^2y+e^x\right)\right)}{\partial y} = \left(x^2y+e^x\right)\dfrac{\partial\mu}{\partial y} + x^2\mu \\\\ \dfrac{\partial\left(-\mu x^2\right)}{\partial x} = -x^2\dfrac{\partial\mu}{\partial x} - 2x\mu \\\\ \implies \left(x^2y+e^x\right)\dfrac{\partial\mu}{\partial y} + x^2\mu = -x^2\dfrac{\partial\mu}{\partial x} - 2x\mu$

Notice that if we let µ(x, y) = µ(x) be independent of y, then ∂µ/∂y = 0 and we can solve for µ :

$x^2\mu = -x^2\dfrac{\mathrm d\mu}{\mathrm dx} - 2x\mu \\\\ (x^2+2x)\mu = -x^2\dfrac{\mathrm d\mu}{\mathrm dx} \\\\ \dfrac{\mathrm d\mu}{\mu} = -\dfrac{x^2+2x}{x^2}\,\mathrm dx \\\\ \dfrac{\mathrm d\mu}{\mu} = \left(-1-\dfrac2x\right)\,\mathrm dx \\\\ \implies \ln|\mu| = -x - 2\ln|x| \\\\ \implies \mu = e^{-x-2\ln|x|} = \dfrac{e^{-x}}{x^2}$