2x+y=-9 -3x+11=y whathe is the answer

1 answer:

8 0

[ 2x + y = -9 ] x -3
[- 3x +11= y ] x 2

-6x -3y = 27
-6x + 22 = 2y

-6x +27 = 3y
-6x +22 = 2y

5 = y

-6x -3x5 = 27
-6x -15 = 27
-6x = 42
-x = 42/6
-x = 7
x = -7

answer

x = -7
y = 5

You might be interested in

Please Help! This is a trigonometry question.

liraira [26]

$\large\begin{array}{l} \textsf{From the picture, we get}\\\\ \mathsf{tan\,\theta=\dfrac{2}{3}}\\\\ \mathsf{\dfrac{sin\,\theta}{cos\,\theta}=\dfrac{2}{3}}\\\\ \mathsf{3\,sin\,\theta=2\,cos\,\theta}\qquad\mathsf{(i)} \end{array}$

$\large\begin{array}{l} \textsf{Square both sides of \mathsf{(i)} above:}\\\\ \mathsf{(3\,sin\,\theta)^2=(2\,cos\,\theta)^2}\\\\ \mathsf{9\,sin^2\,\theta=4\,cos^2\,\theta}\qquad\quad\textsf{(but }\mathsf{cos^2\theta=1-sin^2\,\theta}\textsf{)}\\\\ \mathsf{9\,sin^2\,\theta=4\cdot (1-sin^2\,\theta)}\\\\ \mathsf{9\,sin^2\,\theta=4-4\,sin^2\,\theta}\\\\ \mathsf{9\,sin^2\,\theta+4\,sin^2\,\theta=4} \end{array}$

$\large\begin{array}{l} \mathsf{13\,sin^2\,\theta=4}\\\\ \mathsf{sin^2\,\theta=\dfrac{4}{13}}\\\\ \mathsf{sin\,\theta=\sqrt{\dfrac{4}{13}}}\\\\ \textsf{(we must take the positive square root, because }\theta \textsf{ is an}\\\textsf{acute angle, so its sine is positive)}\\\\ \mathsf{sin\,\theta=\dfrac{2}{\sqrt{13}}} \end{array}$

________

$\large\begin{array}{l} \textsf{From (i), we find the value of }\mathsf{cos\,\theta:}\\\\ \mathsf{3\,sin\,\theta=2\,cos\,\theta}\\\\ \mathsf{cos\,\theta=\dfrac{3}{2}\,sin\,\theta}\\\\ \mathsf{cos\,\theta=\dfrac{3}{\diagup\!\!\!\! 2}\cdot \dfrac{\diagup\!\!\!\! 2}{\sqrt{13}}}\\\\ \mathsf{cos\,\theta=\dfrac{3}{\sqrt{13}}}\\\\ \end{array}$

________

$\large\begin{array}{l} \textsf{Since sine and cosecant functions are reciprocal, we have}\\\\ \mathsf{sin\,2\theta\cdot csc\,2\theta=1}\\\\ \mathsf{csc\,2\theta=\dfrac{1}{sin\,2\theta}\qquad\quad\textsf{(but }}\mathsf{sin\,2\theta=2\,sin\,\theta\,cos\,\theta}\textsf{)}\\\\ \mathsf{csc\,2\theta=\dfrac{1}{2\,sin\,\theta\,cos\,\theta}}\\\\ \mathsf{csc\,2\theta=\dfrac{1}{2\cdot \frac{2}{\sqrt{13}}\cdot \frac{3}{\sqrt{13}}}} \end{array}$

$\large\begin{array}{l} \mathsf{csc\,2\theta=\dfrac{~~~~1~~~~}{\frac{2\cdot 2\cdot 3}{(\sqrt{13})^2}}}\\\\ \mathsf{csc\,2\theta=\dfrac{~~1~~}{\frac{12}{13}}}\\\\ \boxed{\begin{array}{c}\mathsf{csc\,2\theta=\dfrac{13}{12}} \end{array}}\qquad\checkmark \end{array}$

<span>If you're having problems understanding this answer, try seeing it through your browser: brainly.com/question/2150237

$\large\textsf{I hope it helps.}$

Tags: <em>trigonometry trig function cosecant csc double angle identity geometry</em>

</span>

8 0

3 years ago

Lady_Fox [76]

They pay $10.74 each

3 0

3 years ago

The mass of a certain window is 18kg. What is the mass of 5 of those windows?

Daniel [21]

18 kg times 5 kg would be 90 kg because each window was 18 kg, hence 90 kg

3 0

3 years ago

What free resources would you recommend for assistance in maths?

Andru [333]

Usually bbc bitesize is what I use for basically everything. Here’s a link:
https://www.bbc.com/bitesize

3 0

3 years ago

Read 2 more answers

A flagpole has a height of 10 yards. It will be supported by 3 cables, each of which is attached to the flagpole at a point 4 ya

Contact [7]

Pytagorean theorem

height10
attached at 10-4=6 yards from bottom
8 yards from base to away

a^2+b^2=c^2
a and b are side legnths and c=hypotonuse
so therfore we are given a and b and we must solve for c

so
6=height
8=base

6^2+8^2=c^2
36+64=c^2
100=c^2
square root
10=c

so therfor the legnth of each cable is 10 yards
3 cables so 3 times 10=30

30 yards needed

7 0

3 years ago