For solving system of equations, we can use either substitution where we plug one equation into the other, or elimination where we combine the equations.

- Using elimination, you would to eliminate one variable from both equations, so you automatically would get one equation with one variable!

- Using substitution means you are going to solve one equation for one variable and substitute with its value in the other equation in order to get also an equation with one variable.

Let's take an example ...

y+x=2 and y-2x = 1

Using elimination, we need to subtract these two equation; one from the other...

y+x=2
-
y-2x=1
-----------
0+3x=1

then x=1/3

and then substitute into any equation to get y-value
y+x=2
y+1/3 = 2 >>>>> y=5/3

NOW...Using substitution
y+x=2 and y-2x = 1 >>(y=1+2x)

Plug (y=1+2x) into y+x=2 and solve for x
y+x=2
(1+2x) + x =2
1+3x = 2
3x=1

again (and for sure) x = 1/3

plug in x=1/3 into any of the equations above to get y:
y+x=2
y+1/3=2
y=5/3

DOne !!!!!!

I hope you got the idea

If you still need help, just let me know.

7 0

3 years ago

Dose 8 has more factors or 9

Doss [256]

Answer:

Factors for 8: 1, 2, 4, 8

Factors for 9: 1, 3, 9

So 8 has more factors

7 0

2 years ago

A standard deck of cards contains 52 cards. One card is selected from the deck.(a)Compute the probability of randomly selecting

Aneli [31]

a. There are four 5s that can be drawn, and $\binom43=4$ ways of drawing any three of them. There are $\binom{52}3=22,100$ ways of drawing any three cards from the deck. So the probability of drawing three 5s is

$\dfrac{\binom43}{\binom{52}3}=\dfrac4{22,100}=\dfrac1{5525}\approx0.00018$

In case you're asked about the probability of drawing a 3 or a 5 (and NOT three 5s), then there are 8 possible cards (four each of 3 and 5) that interest you, with a probability of $\frac8{52}=\frac2{13}\approx0.15$ of getting drawn.

b. Similar to the second case considered in part (a), there are now 12 cards of interest with a probability $\frac{12}{52}=\frac3{13}\approx0.23$ of being drawn.

c. There are four 6s in the deck, and thirteen diamonds, one of which is a 6. That makes 4 + 13 - 1 = 16 cards of interest (subtract 1 because the 6 of diamonds is being double counted by the 4 and 13), hence a probability of $\frac{16}{52}=\frac4{13}\approx0.31$ .

- - -

Note: $\binom nk$ is the binomial coefficient,

$\dbinom nk=\dfrac{n!}{k!(n-k)!}={}_nC_k=C(n,k)=n\text{ choose }k$

6 0

3 years ago

15x ( 7 - 5y ) + 3 ( 5y - x ) - ( 4x + 3 ) ( 5y - 2 )

Sedbober [7]

15x ( 7 - 5y) + 3 ( 5y - x) - ( 4x + 3) (5y - 2) = -95xy + 110x + 6 is your solution hope it helps :)!

4 0

2 years ago

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