Solve the simultaneous equations8x – 3y = 303x + y = 7

1 answer:

6 0

Solve for y in 3x+7=y

y=7−3x

Substitute y=7−3x

y=7−3x into 8x−3y=30

8x−3y=30.

17x−21=30

Solve for x

x in 17x−21=30

17x−21=30.

X=3

Substitute x=3 into y=7−3x

y=-2

Therefore,

x=3

y=−2

You might be interested in

A) Compute the sum

avanturin [10]

A)

To calculate this sum, we could use trigonometric identity:

$\arcsin(x)-\arcsin(y)=\arcsin\left(x\sqrt{1-y^2}-y\sqrt{1-x^2}\right)$

We have:

$\sum\limits_{k=1}^n\arcsin\left[\dfrac{\sqrt{k^2+2k}-\sqrt{k^2-1}}{k(k+1)}\right]=\\\\\\=
\sum\limits_{k=1}^n\arcsin\left[\dfrac{\sqrt{k^2+2k}}{k(k+1)}-\dfrac{\sqrt{k^2-1}}{k(k+1)}\right]=\\\\\\=
\sum\limits_{k=1}^n\arcsin\left[\dfrac{\sqrt{k^2+2k+1-1}}{k(k+1)}-\dfrac{\sqrt{k^2-1}}{k(k+1)}\right]=\\\\\\=
\sum\limits_{k=1}^n\arcsin\left[\dfrac{\sqrt{(k+1)^2-1}}{k(k+1)}-\dfrac{\sqrt{k^2-1}}{k(k+1)}\right]=\\\\\\
$

$=\sum\limits_{k=1}^n\arcsin\left[\dfrac{1}{k}\cdot\dfrac{\sqrt{(k+1)^2-1}}{\sqrt{(k+1)^2}}-\dfrac{1}{k+1}\cdot\dfrac{\sqrt{k^2-1}}{\sqrt{k^2}}\right]=\\\\\\=
\sum\limits_{k=1}^n\arcsin\left[\dfrac{1}{k}\cdot\sqrt{\dfrac{(k+1)&^2-1}{(k+1)^2}}-\dfrac{1}{k+1}\cdot\sqrt{\dfrac{k^2-1}{k^2}}\right]=\\\\\\=
\sum\limits_{k=1}^n\arcsin\left[\dfrac{1}{k}\cdot\sqrt{1-\dfrac{1}{(k+1)^2}}-\dfrac{1}{k+1}\cdot\sqrt{1-\dfrac{1}{k^2}}\right]=\\\\\\
$

$=\sum\limits_{k=1}^n\arcsin\left[\dfrac{1}{k}\cdot\sqrt{1-\left(\dfrac{1}{k+1}\right)^2}-\dfrac{1}{k+1}\cdot\sqrt{1-\left(\dfrac{1}{k}\right)^2}\right]=\\\\\\=
\sum\limits_{k=1}^n\left[\arcsin\left(\dfrac{1}{k}\right)-\arcsin\left(\dfrac{1}{k+1}\right)\right]=\\\\\\$

$=\bigg[\arcsin(1)-\arcsin\left(\frac{1}{2}\right)\bigg]+\bigg[\arcsin\left(\frac{1}{2}\right)-\arcsin\left(\frac{1}{3}\right)\bigg]+\\\\\\+
\bigg[\arcsin\left(\frac{1}{3}\right)-\arcsin\left(\frac{1}{4}\right)\bigg]+\dots+
\bigg[\arcsin\left(\frac{1}{n}\right)-\arcsin\left(\frac{1}{n+1}\right)\bigg]=\\\\\\$

$=\arcsin(1)-\arcsin\left(\frac{1}{2}\right)+\arcsin\left(\frac{1}{2}\right)-\arcsin\left(\frac{1}{3}\right)+\arcsin\left(\frac{1}{3}\right)-\\\\\\-\arcsin\left(\frac{1}{4}\right)+\dots+\arcsin\left(\frac{1}{n}\right)-\arcsin\left(\frac{1}{n+1}\right)=\\\\\\=
\arcsin(1)-\arcsin\left(\frac{1}{n+1}\right)=\dfrac{\pi}{2}-\arcsin\left(\frac{1}{n+1}\right)$

So the answer is:

$\sum\limits_{k=1}^n\arcsin\left[\dfrac{\sqrt{k^2+2k}-\sqrt{k^2-1}}{k(k+1)}\right]=\dfrac{\pi}{2}-\arcsin\left(\dfrac{1}{n+1}\right)}$

B)

$\sum\limits_{k=1}^\infty\arcsin\left[\dfrac{\sqrt{k^2+2k}-\sqrt{k^2-1}}{k(k+1)}\right]=\\\\\\=
\lim\limits_{n\to\infty}\sum\limits_{k=1}^n\arcsin\left[\dfrac{\sqrt{k^2+2k}-\sqrt{k^2-1}}{k(k+1)}\right]=\\\\\\=
\lim\limits_{n\to\infty}\Bigg(\dfrac{\pi}{2}-\arcsin\left(\dfrac{1}{n+1}\right)\Bigg)=\dfrac{\pi}{2}-\lim\limits_{n\to\infty}\arcsin\left(\dfrac{1}{n+1}\right)=\\\\\\=
\Bigg\{\dfrac{1}{n+1}\xrightarrow{n\to\infty}0\Bigg\}=\dfrac{\pi}{2}-\arcsin(0)=\dfrac{\pi}{2}-0=\dfrac{\pi}{2}$

So we prove that:

$\sum\limits_{k=1}^\infty\arcsin\left[\dfrac{\sqrt{k^2+2k}-\sqrt{k^2-1}}{k(k+1)}\right]=\dfrac{\pi}{2}$

7 0

3 years ago

Max and Sasha exercise a total of 20 hours each week. Max exercises 15 hours less than 4 times the

navik [9.2K]

Answer:

Max exercises 13 hours a week, and Sasha 7.

Step-by-step explanation:

To find the number of hours each of them exercises during the week, we solve the system of equations.

In the second equation:

$x = 4y - 15$