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3 years ago

Slove algebraically3x - 4y = -24x +4y=8

Mathematics

1 answer:

Vanyuwa [196]3 years ago

5 0

Solve algebrically 3x - 4y = -24 and x + 4y = 8 is x = -4 and y = 3

<u>Solution:</u>

We have been given two equations which are as follows:

3x - 4y = -24 ----- eqn 1

x + 4y = 8 -------- eqn 2

We have been asked to solve the equations which means we have to find the value of ‘x’ and ‘y’.

We rearrange eqn 2 as follows:

x + 4y = 8

x = 8 - 4y ------eqn 3

Now we substitute eqn 3 in eqn 1 as follows:

3(8 - 4y) -4y = -24

24 - 12y - 4y = -24

-16y = -48

y = 3

Substitute "y" value in eqn 3. Therefore the value of ‘x’ becomes:

x = 8 - 4(3)

x = 8 - 12 = -4

Hence on solving both the given equations we get the value of x and y as -4 and 3 respectively.

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Suppose you have two urns with poker chips in them. Urn I contains two red chips and four white chips. Urn II contains three red

Neporo4naja [7]

Answer:

Multiple answers

Step-by-step explanation:

The original urns have:

Urn 1 = 2 red + 4 white = 6 chips
Urn 2 = 3 red + 1 white = 4 chips

We take one chip from the first urn, so we have:

The probability of take a red one is : $\frac{1}{3}$ (2 red from 6 chips(2/6=1/2))

For a white one is: $\frac{2}{3}$ (4 white from 6 chips(4/6=(2/3))

Then we put this chip into the second urn:

We have two possible cases:

First if the chip we got from the first urn was white. The urn 2 now has 3 red + 2 whites = 5 chips
Second if the chip we got from the first urn was red. The urn two now has 4 red + 1 white = 5 chips

If we select a chip from the urn two:

In the first case the probability of taking a white one is of: $\frac{2}{5}$ = 40% ( 2 whites of 5 chips)
In the second case the probability of taking a white one is of: $\frac{1}{5}$ = 20% ( 1 whites of 5 chips)

This problem is a dependent event because the final result depends of the first chip we got from the urn 1.

For the fist case we multiply :

$\frac{4}{6}$ x $\frac{2}{5}$ = $\frac{4}{15}$ = 26.66% ( $\frac{4}{6}$ the probability of taking a white chip from the urn 1, $\frac{2}{5}$ the probability of taking a white chip from urn two)

For the second case we multiply:

$\frac{1}{3}$ x $\frac{1}{5}$ = $\frac{1}{30}$ = .06% ( $\frac{1}{3}$ the probability of taking a red chip from the urn 1, $\frac{1}{5}$ the probability of taking a white chip from the urn two)