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3 years ago

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Is (4,-3) a solution to this system? 3x-3y=21 5x+4y=24

2 answers:

Elena-2011 [213]3 years ago

8 0

Welcome to Brainly!

We have the following system of linear equations:

3x-3y=21

5x+4y=24

We could solve this system through elimination: Multiply the first equation by 5 and the second equation by -3. When you add together the results, x will drop out. This is a great approach.

But you could take a short cut by subst. the coordinates (4,-3) into both of the given linear equations. IF both equations are true after this substitution, then you know that (4,-3) is a solution. Let's try it:

3(4) - 3(-3) = 21 => 12 + 9 = 21. This is true. But we must ALSO check whether (4,-3) satisfies 5x+4y=24: 5(4) + 4(-3) = 24 => 20 - 12 = 12. This is NOT equal to 24, so we must conclude that (4,-3) is NOT a solution of this system of linear equations.

Arturiano [62]3 years ago

7 0

No it is not. although it fits the first equation, it doesn’t fit the second.

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Answer:

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Step-by-step explanation:

Given : An urn contains 3 red and 7 black balls. Players A and B take turns (A goes first) withdrawing balls from the urn consecutively.

To find : What is the probability that A selects the first red ball?

Solution :

A wins if the first red ball is drawn 1st,3rd,5th or 7th.

A red ball drawn first, there are $E(1)= ^9C_2$ places in which the other 2 red balls can be placed.

A red ball drawn third, there are $E(3)= ^7C_2$ places in which the other 2 red balls can be placed.

A red ball drawn fifth, there are $E(5)= ^5C_2$ places in which the other 2 red balls can be placed.

A red ball drawn seventh, there are $E(7)= ^3C_2$ places in which the other 2 red balls can be placed.

The total number of total event is $S= ^{10}C_3$

The probability that A selects the first red ball is

$P(A \text{wins})=\frac{(^9C_2)+(^7C_2)+(^5C_2)+(^3C_2)}{^{10}C_3}$

$P(A \text{wins})=\frac{36+21+10+3}{120}$

$P(A \text{wins})=\frac{70}{120}$

$P(A \text{wins})=0.5833$

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3 years ago

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igor_vitrenko [27]

Answer:

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Step-by-step explanation:

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