-5x + 9y = 14 -3x + 9y = -6

1 answer:

4 0

You can solve it by several ways, but easiest and shortest would be "By Elimination"

So, subtract 2nd equation from 1st,
-2x = 20
x = -10

Substitute it in 2nd equation,
-3(-10) + 9y = -6
30 + 9y = -6
9y = -6 - 30
y = -36/9
y = -4

In short, Your Answer would be: (-10, -4)

Hope this helps!

You might be interested in

Determine between which consecutive integers the real zeros of y(x) = x2 - 4x – 2 are located.

Sholpan [36]

Answer:

The zeros are x1=4.45 and x2=-0.45.

x1 is between 4 and 5.

x2 is between -1 and 0.

Step-by-step explanation:

We have the function:

$y(x) = x2 - 4x-2$

As this is a quadratic function, we can calculate the zeros of the function with the quadratic equation:

$x_{1,2}=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\\\x=\dfrac{-(-4)\pm\sqrt{(-4)^2-4\cdot 1\cdot(-2)}}{2\cdot 1}\\\\\\x=\dfrac{4\pm\sqrt{16+8}}{2}\\\\\\x=\dfrac{4\pm\sqrt{24}}{2}\\\\\\x=\dfrac{4\pm4.9}{2}=2\pm2.45\\\\\\x_1=2+2.45=4.45\\\\x_2=2-2.45=-0.45$

The zeros are x1=4.45 and x2=-0.45.

x1 is between 4 and 5.

x2 is between -1 and 0.

7 0

3 years ago

Read 2 more answers

Consider a discrete random variable x with pmf px (1) = c 3 ; px (2) = c 6 ; px (5) = c 3 and 0 otherwise, where c is a positive

PtichkaEL [24]

Looks like the PMF is supposed to be

$\mathbb P(X=x)=\begin{cases}\dfrac c3&\text{for }x\in\{1,5\}\\\\\dfrac c6&\text{for }x=2\\\\0&\text{otherwise}\end{cases}$

which is kinda weird, but it's not entirely clear what you meant...

Anyway, assuming the PMF above, for this to be a valid PMF, we need the probabilities of all events to sum to 1:

$\displaystyle\sum_{x\in\{1,2,5\}}\mathbb P(X=x)=\dfrac c3+\dfrac c6+\dfrac c3=\dfrac{5c}6=1\implies c=\dfrac65$

Next,

$\mathbb P(X>2)=\mathbb P(X=5)=\dfrac c3=\dfrac25$

$\mathbb E(X)=\displaystyle\sum_{x\in\{1,2,5\}}x\,\mathbb P(X=x)=\dfrac c3+\dfrac{2c}6+\dfrac{5c}3=\dfrac{7c}3=\dfrac{14}5$

$\mathbb V(X)=\mathbb E\bigg((X-\mathbb E(X))^2\bigg)=\mathbb E(X^2)-\mathbb E(X)^2$
$\mathbb E(X^2)=\displaystyle\sum_{x\in\{1,2,5\}}x^2\,\mathbb P(X=x)=\dfrac c3+\dfrac{4c}6+\dfrac{25c}3=\dfrac{28c}3=\dfrac{56}5$
$\implies\mathbb V(X)=\dfrac{56}5-\left(\dfrac{14}5\right)^2=\dfrac{84}{25}$

If $Y=X^2+1$ , then $X^2=Y-1\implies X=\sqrt{Y-1}$ , where we take the positive root because we know $X$ can only take on positive values, namely 1, 2, and 5. Correspondingly, we know that $Y$ can take on the values $1^2+1=2$ , $2^2+1=5$ , and $5^2+1=26$ . At these values of $Y$ , we would have the same probability as we did for the respective value of $X$ . That is,

$\mathbb P(Y=y)=\begin{cases}\dfrac c3&\text{for }y=2\\\\\dfrac c6&\text{for }y=5\\\\\dfrac c3&\text{for }y=26\\\\0&\text{otherwise}\end{cases}$

Part (5) is incomplete, so I'll stop here.

8 0

3 years ago

What is a mathematical statement that two expressions are equal

natulia [17]

It is called an equation.

6 0

3 years ago

Write 7x10^-8 in standard form

attashe74 [19]

Answer:

The answer would be 0.00000007

6 0