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3 years ago

Which ordered pair (x, y) is the solution to the given system of linear equations?

Mathematics

2 answers:

QveST [7]3 years ago

4 0

Answer:(-1,0)

Step-by-step explanation:

3x + 3y=-3. Solve by the elimination method.

-2x+y=2

2(3x + 3y)=-3(2) New equation 6x + 6y=-6

3(-2 + y)=2(3). New equation -6x +3y=6

9y= 0

y=0

6x +6(0)=-6

6x =-6

X=-1

Stella [2.4K]3 years ago

3 0

(-1,0)

y=2x+2 *new equation for second*
3x+3(2x+2)=-3 *plug y into first equation*
3x+(6x+6)=-3 *combine like terms*
9x+6=-3 *bring 6 to the other side by subtracting 6 on both sides*
9x = -9 *divide by 9 to get x by itself*
x = -1

to find y (now that we have x):
-2(-1) +y=2
2+y=2
y=0

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Step-by-step explanation:

Considering the expression

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Lets determine the expansion of the expression

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Expanding summation

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$i=4\quad :\quad \frac{4!}{4!\left(4-4\right)!}3^0x^4$

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$\frac{4!}{1!\left(4-1\right)!}\cdot \:3^3x^1:\quad 108x$

$\frac{4!}{2!\left(4-2\right)!}\cdot \:3^2x^2:\quad 54x^2$

$\frac{4!}{3!\left(4-3\right)!}\cdot \:3^1x^3:\quad 12x^3$

$\frac{4!}{4!\left(4-4\right)!}\cdot \:3^0x^4:\quad x^4$

so equation becomes

$=81+108x+54x^2+12x^3+x^4$

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$\left(3+x\right)^4:\quad x^4+12x^3+54x^2+108x+81$

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