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Anettt [7]
2 years ago
5

How does the method for solving equations with fractional or decimal coefficients and constants compare with the method for solv

ing equations with integer coefficients and constants?
The methods are essentially ____. The only extra step is that you ____ solving by eliminating the fractions or the decimals from the equation.

In the first missing spot it should be different or the same
In the sec missing spot it should be finish or begin
Mathematics
1 answer:
I am Lyosha [343]2 years ago
6 0

Answer:

Step-by-step explanation:

When solving equations with fractional or decimal coefficients, the equations needs to be multiplied by the multiple of denominator such that the equations have integer coefficients and constants

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Let f(x)=x+a/x+b such that f(f(1)=0 & f(2)=-3 then a&b resctivily are.
Yuliya22 [10]

Answer:

The value of a = - 1  ,   And b = \frac{ - 7 }{3}    

Step-by-step explanation:

Given as :

Function f(x) = \frac{(x + a)}{(x + b)}

And f(f(1)) = 0     And f(2) = - 3

Now For , x = 2 , y = - 3

I.e  f(2) = \frac{(2 + a)}{(2 + b)}

or,  - 3 = \frac{(2 + a)}{(2 + b)}

I.e 2 + a = - 6 - 3b

Or, a + 3b = - 8               ....... 1

Again  f(f(1)) = 0

So,  \frac{(1 + a)}{(1 + b)} = 0

Or,  1 + a = 0

∴    a = - 1

So , put htis value of a i n eq 1 , we get value of  b

So , - 1 + 3b = - 8

Or,   3b = - 7

∴       b = \frac{ - 7 }{3}

Hence The value of a = - 1  ,   And b = \frac{ - 7 }{3}    Answer

3 0
3 years ago
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