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HACTEHA [7]
3 years ago
12

Solve the system. y= 3x2−2x+3​ y = x + 3 The solutions are (__,__) and (__,__).

Mathematics
1 answer:
photoshop1234 [79]3 years ago
8 0

Answer:

(0,3) and (1,4)

Step-by-step explanation:

Given that:

y= 3x2−2x+3​  ------------- eq1

y = x + 3  ------------------ eq2

Putting value from eq2 to eq1

x + 3 = 3x2−2x+3​

3x2−2x+3​  - x - 3 = 0

3x2−3x = 0

Taking 3 common

x2 - x = 0

x(x-1)= 0

So

x = 0      and   x = 1

Now put value in eq2

y = 0 + 3   and    y= 1 + 3

y = 3 and y = 4

So solution sets are:

(0,3) and (1,4)

i hope it will help you!

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Answer:
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Step-by-step explanation:
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How to answer h/4 ≤ 2/7
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Your answer would be  h ≤ 8/7

Step  1  :

           2

Simplify   —

           7

Equation at the end of step  1  :

 h    2

 — -  —  ≤ 0

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Step  2  :

           h

Simplify   —

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Equation at the end of step  2  :

 h    2

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Step  3  :

Calculating the Least Common Multiple :

3.1    Find the Least Common Multiple

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     The right denominator is :       7

After calculating the multipliers you get  h ≤ 8/7

Hope this helps!

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Answer:

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