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3 years ago

What is the solution of the linear-quadratic system of equations? y=x^2+5x-3 y-x=2

2 answers:

4 0

-x=-y+2
x=y-2
now substitute iin the first one
y=(y-2)^2+5 (y-2)-3
y=y^2-4y+4 +5y-10-3
y=y^2+y-9
y^2-9=0.
y^2=9
y=3

3-x=2
-x=2-3
-x=-1
x=1

the solution are x=1 and y=3
3=1^2+5 (1)-3
3=1+5-3
3=6-3
3=3

Rudiy273 years ago

3 0

Answer:

Step-by-step explanation:

-x=-y+2

x=y-2

now substitute iin the first one

y=(y-2)^2+5 (y-2)-3

y=y^2-4y+4 +5y-10-3

y=y^2+y-9

y^2-9=0.

y^2=9

y=3

3-x=2

-x=2-3

-x=-1

x=1

the solution are x=1 and y=3

3=1^2+5 (1)-3

3=1+5-3

3=6-3

3=3

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Answer:

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Step-by-step explanation:

Let the number of $3$ cent stamps be $x$ and $4$ cent stamps be $y$

We have

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The minimum number is obtained when more $4$ cent stamps are used

Here $y$ cannot be greater than $8$ since $\frac{33}{4}$

Substitute $y=8$

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Substitute $y=7$

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Not possible since $x$ is not a fraction

Substitute $y=6$

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Possible

Hence minimum number of stamps is

$=x+y\\\\=3+6\\\\=9$

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Step-by-step explanation:

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