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ivanzaharov [21]
2 years ago
14

What is the solution to the system of equations? (You can either do the substitution or elimination method) x - 2y = 19 x + 3y =

14 (1.-17) (-1, 17) (-17, 1) (17,-1)​
Mathematics
1 answer:
masha68 [24]2 years ago
4 0

Answer:

D. (17,-1)

Step-by-step explanation:

We'll start by canceling out x. To do so, multiple the first equation by -1

-1( x - 2y)=( 19 )(-1)

This gives us:

-x +2y = -19

x + 3y = 14

Add the equations together:

→ 5y = -15

→ y = -1

Plug in y = -1 into an equation:

→ x + 3(-1) = 14

→ x - 3 = 14

→ x = 17

(17,-1)

Hope this helps!

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ruslelena [56]

Step-by-step explanation:

\underline{\textsf{Given:}}

Given:

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\underline{\textsf{To find:}}

To find:

\mathsf{Factors\;of\;x^3+7x^2+7x-15}Factorsofx

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\underline{\textsf{Solution:}}

Solution:

\textsf{Factor theorem:}Factor theorem:

\boxed{\mathsf{(x-a)\;is\;a\;factor\;P(x)\;\iff\;P(a)=0}}

(x−a)isafactorP(x)⟺P(a)=0

\mathsf{Let\;P(x)=x^3+7x^2+7x-15}LetP(x)=x

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\mathsf{Sum\;of\;the\;coefficients=1+7+7-15=0}Sumofthecoefficients=1+7+7−15=0

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\mathsf{When\;x=-3}Whenx=−3

\mathsf{P(-3)=(-3)^3+7(-3)^2+7(-3)-15}P(−3)=(−3)

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\mathsf{P(-3)=0}P(−3)=0

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\mathsf{When\;x=-5}Whenx=−5

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\mathsf{P(-5)=175-175}P(−5)=175−175

\mathsf{P(-5)=0}P(−5)=0

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\underline{\textsf{Answer:}}

Answer:

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\underline{\textsf{Find more:}}

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6 0
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[∵ No. of ways to choose r things out of n =^nP_r=\dfrac{n!}{(n-r)!} ]

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