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3 years ago

7

How do you solve this?

1 answer:

Law Incorporation [45]3 years ago

6 0

Find the soluiton

2x+2y=16
3x-y=4

x+y=8
<u>3x-y=4+</u>
4x=12
x=3
3(3)-y=4
9-y=4
y=5
(3,5)
test them to see if get true statment
obviously the first equatons of 1,2,4 work
1 doesn't work
2, works
4. doesn't work

2 is the same as the first except both euations are just doubled

answer is 2

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Thanks so much for the help!!

saveliy_v [14]

Answer: Domain: x=-2. Range: -2<=y<1

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Tems11 [23]

Divide through everything by <em>b</em> :

$\dfrac{a+c}{b+d} = \dfrac{\dfrac ab + \dfrac cb}{1 + \dfrac db}$

Since <em>a/b</em> < <em>c/d</em>, it follows that

$\dfrac{a+c}{b+d} < \dfrac{\dfrac cd+\dfrac cb}{1 + \dfrac db}$

Multiply through everything on the right side by <em>b/d</em> to get

$\dfrac{a+c}{b+d} < \dfrac{\dfrac{bc}{d^2}+\dfrac cd}{\dfrac bd+1} = \dfrac{\dfrac cd\left(\dfrac bd+1\right)}{\dfrac bd+1} = \dfrac cd$

and so (<em>a</em> + <em>c</em>)/(<em>b</em> + <em>d</em>) < <em>c/d</em>.

For the other side, you can do something similar and divide through everything by <em>d</em> :

$\dfrac{a+c}{b+d} = \dfrac{\dfrac ad + \dfrac cd}{\dfrac bd + 1}$

and <em>a/b</em> < <em>c/d</em> tells us that

$\dfrac{a+c}{b+d} > \dfrac{\dfrac ad + \dfrac ab}{\dfrac bd + 1}$

Then

$\dfrac{a+c}{b+d} > \dfrac{\dfrac ab + \dfrac{ad}{b^2}}{1 + \dfrac db} = \dfrac{\dfrac ab\left(1+\dfrac db\right)}{1 + \dfrac db} = \dfrac ab$

and so (<em>a</em> + <em>c</em>)/(<em>b</em> + <em>d</em>) > <em>a/b</em>.

Then together we get the desired inequality.

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Bad White [126]

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