heureka

convert -1.7 rad to degrees

$$-1.7 \mathrm{~rad~} \cdot \frac{360\ensurement{^{\circ}}}{2\pi \mathrm{~rad~}}\\\\
=-1.7 \cdot \frac{180\ensurement{^{\circ}}}{\pi} \\\\
= -97.4028251722\ensurement{^{\circ}}\\
\mathrm{ }\\
\hline
\\$$

Die Fläche A eines  Dreiecks mit Determinante berechnen:

Gegeben sind drei Punkte :

$$\small{\text{$
S_1(x_1|y_1) = (4|-3) \qquad S_2 (x_2|y_2)=(-2|0) \qquad S_3(x_3|y_3)=(-1|2)
$}}$$

$$\small{\text{$
\begin{array}{|l|l|l|} \hline & x & y \\
\hline
S_1 & x_1=4 & y_1=-3 \\
S_2 & x_2=-2 & y_2= 0 \\
S_3 & x_3=-1 & y_3=2 \\
\hline
\end{array} $}}\\\\$$

$$\small{\text{$
A = \dfrac{1}{2}\cdot
\begin{vmatrix}
x_1 & y_1 & 1 \\
x_2 & y_2 & 1 \\
x_3 & y_3 & 1 \\
\end{vmatrix}
$ }}\\\\$$

Auflösung nach der 1. Spalte:

$$\small{\text{$
\begin{array}{lll}
A &=& \dfrac{1}{2}\cdot
\begin{vmatrix}
\textcolor[rgb]{1,0,0}{x_1} & y_1 & 1 \\
\textcolor[rgb]{1,0,0}{x_2} & y_2 & 1 \\
\textcolor[rgb]{1,0,0}{x_3} & y_3 & 1 \\
\end{vmatrix}
=\frac12\cdot \left(
\textcolor[rgb]{1,0,0}{x_1}\cdot
\begin{vmatrix}
y_2 & 1 \\
y_3 & 1 \\
\end{vmatrix}
-\textcolor[rgb]{1,0,0}{x_2}\cdot
\begin{vmatrix}
y_1 & 1 \\
y_3 & 1 \\
\end{vmatrix}
+\textcolor[rgb]{1,0,0}{x_3}\cdot
\begin{vmatrix}
y_1 & 1 \\
y_2 & 1 \\
\end{vmatrix}
\right)\\\\
&=& \frac12\cdot \left(~~ [~x_1\cdot
(y_2-y_3)~]-[~x_2\cdot
(y_1-y_3)~]+[~x_3\cdot
(y_1-y_2)~] ~~\right)\\\\
&=&
\mathbf{ \frac12\cdot \left(~~ [~x_1\cdot
(y_2-y_3)~]+[~x_2\cdot
(y_3-y_1)~]+[~x_3\cdot
(y_1-y_2)~] ~~\right)
} \\ \\
&=& \frac12\cdot \left(~~ [~4\cdot
(0-2)~]-[~2\cdot
(2-(-3))~]-[~1\cdot
(-3-0)~] ~~\right)\\\\
&=& \frac12\cdot \left(~~ [~4\cdot
(-2)~]-[~2\cdot
(2+3))~]-[~1\cdot
(-3)~] ~~\right)\\\\
&=& \frac12\cdot \left(~~ -8-10+3 ~~\right)\\\\
&=& -7,5\\\\
A&=& 7,5
\end{array}
$ }}\\\\$$

Auflösung nach der 2. Spalte:

$$\small{\text{$
\begin{array}{lll}
A &=& \dfrac{1}{2}\cdot
\begin{vmatrix}
x_1 & \textcolor[rgb]{1,0,0}{y_1} & 1 \\
x_2 & \textcolor[rgb]{1,0,0}{y_2} & 1 \\
x_3 & \textcolor[rgb]{1,0,0}{y_3} & 1 \\
\end{vmatrix}
=\frac12\cdot \left(
-\textcolor[rgb]{1,0,0}{y_1}\cdot
\begin{vmatrix}
x_2 & 1 \\
x_3 & 1 \\
\end{vmatrix}
+\textcolor[rgb]{1,0,0}{y_2}\cdot
\begin{vmatrix}
x_1 & 1 \\
x_3 & 1 \\
\end{vmatrix}
-\textcolor[rgb]{1,0,0}{y_3}\cdot
\begin{vmatrix}
x_1 & 1 \\
x_2 & 1 \\
\end{vmatrix}
\right)\\\\
&=& \frac12\cdot
\left(~~
-[~y_1\cdot (x_2-x_3)~]
+[~y_2\cdot (x_1-x_3)~]
-[~y_3\cdot (x_1-x_2)~]
~~\right)\\\\
&=& \mathbf{
\frac12\cdot
\left(~~
[~y_1\cdot (x_3-x_2)~]
+[~y_2\cdot (x_1-x_3)~]
+[~y_3\cdot (x_2-x_1)~]
~~\right)
}\\\\
&=& \frac12\cdot
\left(~~
[~-3\cdot (-1-(-2))~]
+[~0\cdot (4-(-1))~]
+[~2\cdot (-2-4)~]
~~\right)\\\\
&=& \frac12\cdot
\left(~~
[~-3\cdot 1~]
+[~2\cdot (-6)~]
~~\right)\\\\
&=& \frac12\cdot
\left(~~
-3 - 12
~~\right)\\\\
&=& \frac12\cdot
\left(~~
-15~~\right)\\\\
&=& -7,5\\\\
A&=& 7,5
\end{array}
$ }}\\\\$$

Vielen Dank radix

Heureka

Beispiel:  Wie löse ich das nach x auf?

$$\mathbf{
50= -10 \cdot \log_{10}{ \left( 10^{(-5,7)} + 10^{(\frac{x}{10})} \right) }
}\\\\\\
\begin{array}{rcl}
50 &=& -10 \cdot \log_{10}{ \left( 10^{(-5,7)} + 10^{(\frac{x}{10})} \right) } \qquad | \qquad : -10 \\\\
-5 &=& \log_{10}{ \left[10^{(-5,7)} + 10^{(\frac{x}{10})} \right] } \\\\
-5 &=& \log_{10}{ \left[10^{(-5,7)} + 10^{(\frac{x}{10})} \right] }
\qquad | \qquad : 10^x \\\\
10^{-5} &=& 10^{(-5,7)} + 10^{(\frac{x}{10})} \\\\
10^{(\frac{x}{10})} &=&10^{(-5)} - 10^{(-5,7)} \qquad | \qquad \log_{10} \\\\
\log_{10}{ (10^{(\frac{x}{10})}) } &=&\log_{10}{ (10^{(-5)} - 10^{(-5,7)}) }\\\\
\frac{x}{10}\cdot \log_{10}{ (10) } &=&\log_{10}{ (10^{(-5)} - 10^{(-5,7)}) } \qquad | \qquad \log_{10} =1 \\\\
\frac{x}{10} &=&\log_{10}{ (10^{(-5)} - 10^{(-5,7)}) } \\\\
\end{array}$$

$$\begin{array}{rcl}
x &=& 10 \cdot \log_{10}{ (10^{(-5)} - 10^{(-5,7)}) } \\\\
x &=& 10 \cdot \log_{10}{ ( 0,00001 - 0,00000199526 ) } \\\\
x &=& 10 \cdot \log_{10}{ ( 0,00000800474) } \\\\
x &=& 10 \cdot -5,09665289533 \\\\
\mathbf{x} &\mathbf{=}&\mathbf{-50,9665289533}
\end{array}$$

Probe: 

$$50= -10 \cdot \log_{10}{ \left( 10^{(-5,7)} + 10^{(\frac{x}{10})} \right) } \\\\
50= -10 \cdot \log_{10}{ \left( 10^{(-5,7)} + 10^{(\frac{-50,9665289533}{10})} \right) } \\\\
50= -10 \cdot \log_{10}{ \left( 10^{(-5,7)} + 10^{( -5,09665289533 )} \right) } \\\\
50= -10 \cdot \log_{10}{ \left( 0,00000199526 + 0,00000800474 \right) } \\\\
50= -10 \cdot \log_{10}{ \left( 0,00001\right) } \\\\
50= -10 \cdot (-5) \\\\
50= 50 \qquad \mathrm{okay} \\\\$$

Integrate (cos^-1x)^2 /(1-x^2)^1/2

$$\small{\text{$
\begin{array}{lrcl}
& \int{
\dfrac{[ \cos^{-1}{(x)} ]^2 }
{ (1-x^2)^{\frac{1}{2}} } \ dx }
= \mathbf{ \int{
[\arccos{(x)}]^2 \cdot
\dfrac{ 1 }{ \sqrt{ (1-x^2) } } \ dx } }\\\\
\end{array}
$}}\\\\$$

$$\small{\text{$
\boxed{
\begin{array}{lrcl}
\mathrm{Formula:~}
& y &=& \dfrac{ f(x)^{n+1} } {n+1} \\\\
& y' &=& \left(\dfrac{n+1}{n+1}\right) \cdot f(x)^{n+1-1}\cdot f'(x) \\\\
& y' &=& f(x)^{n}\cdot f'(x) \\\\
&\mathbf{
\int{ f(x)^{n}\cdot f'(x) \ dx }
} & \mathbf{=} & \mathbf{ \dfrac{ f(x)^{n+1} } {n+1} }
\end{array}}
$}}$$

$$\small{\text{$
\begin{array}{lrcl}
&\mathbf{
\int{ f(x)^{n}\cdot f'(x) \ dx }
} & \mathbf{=} & \mathbf{ \dfrac{ f(x)^{n+1} } {n+1} }
\qquad f(x) = \arccos{(x)}
\qquad f'(x)= -\dfrac{ 1 }{ \sqrt{ (1-x^2) } }\\\\
& \int{ \left[ \arccos{(x)} \right]^2 \cdot
\left( -\dfrac{ 1 }{ \sqrt{ (1-x^2) } }
\right) \ dx }
& = &
\dfrac{ \left[ \arccos{(x)} \right]^{3} } {3} \\\\
& -\int{ \left[ \arccos{(x)} \right]^2 \cdot
\left( \dfrac{ 1 }{ \sqrt{ (1-x^2) } }
\right) \ dx }
& = &
\dfrac{ \left[ \arccos{(x)} \right]^{3} } {3} \\\\
& \int{ \left[ \arccos{(x)} \right]^2 \cdot
\left( \dfrac{ 1 }{ \sqrt{ (1-x^2) } }
\right) \ dx }
& = &
-\dfrac{ \left[ \arccos{(x)} \right]^{3} } {3} \\\\
& \int{ \left[ \arccos{(x)} \right] ^2 \cdot
\left( \dfrac{ 1 }{ \sqrt{ (1-x^2) } }
\right) \ dx }
& = &
-\dfrac{1}{3} \cdot \left[ \arccos{(x)} \right]^{3} \\\\
& \int{ \left[ \arccos{(x)} \right]^2 \cdot
\left( \dfrac{ 1 }{ \sqrt{ (1-x^2) } }
\right) \ dx }
& = &
-\dfrac{1}{3} \cdot \left[\cos^{-1}{(x)}\right]^3 \\\\
\end{array}
$}}$$

Equation  X^6-X^3=2

$$\small{\text{$
\begin{array}{rcl}
x^6-x^3&=&2 \qquad \mathrm{we ~substitute~and ~set~} z = x^3 \\
\mathrm{then~we~have~} \quad z^2-z &=& 2 \\
z^2-z -2 &=& 0\\
z_{1,2} &=& \dfrac{1\pm \sqrt{1-4\cdot(-2)}}{2} \\
z_{1,2} &=& \dfrac{1\pm \sqrt{9}}{2} \\\\
z_{1,2} &=& \dfrac{1\pm 3 }{2} \\\\
z_1 &=& \frac{1 + 3 }{2} = 2 \\\\
z_2 &=& \frac{1 - 3 }{2} = - 1\\\\\\
x_1 & =& \sqrt[3]{z_1}= \sqrt[3]{2}= 1.25992104989\\\\
\mathbf{x_1} & \mathbf{=}& \mathbf{1.25992104989}\\\\\\
x_2 &=& \sqrt[3]{z_2}= \sqrt[3]{-1}= -1\\\\
\mathbf{x_2} &\mathbf{=}& \mathbf{-1}\\\\
\end{array}
$}}$$

Flächeninhalt und Umfang der eingeschlossenen Fläche berechnen?

y1 = -x+1

y2 =-1/2*x-1

y3 = 2*x+4

Gegebene Geradengleichungen der Form y = mx + b  oder y - mx = b

$$\small{\text{$
\begin{array}{llll}
(1) & y = -x+1 & m_1 = -1 & b_1 = 1 \\
(2) & y = -\frac{1}{2}x-1 & m_2 = -\frac{1}{2} & b_2 = -1 \\
(3) & y = 2x+4 & m_3 = 2 & b_3 = 4 \\
\end{array}
$}}$$

I. Berechnung der Schnittpunkte:

1. Schnitt $$\small{\text{$S_1$}}$$ Gerade (1) mit Gerade (2 ):

$$\small{\text{$
\begin{array}{rcl}
1\cdot y_s - m_1 \cdot x_s &=& b_1 \\
1\cdot y_s - m_2 \cdot x_s &=& b_2 \\
\\
\hline
\\
y_s = \dfrac
{\begin{vmatrix} b_1 & -m_1\\ b_2 &-m_2 \end{vmatrix}
}
{\begin{vmatrix} 1 & -m_1\\ 1 &-m_2 \end{vmatrix}
}
=\dfrac
{\begin{vmatrix} 1 & -(-1)\\ -1 & -(-\frac{1}{2}) \end{vmatrix}
}
{\begin{vmatrix} 1 & -(-1)\\ 1 & -(-\frac{1}{2}) \end{vmatrix}
}
=\dfrac
{\begin{vmatrix} 1 & 1\\ -1 & \frac{1}{2} \end{vmatrix}
}
{\begin{vmatrix} 1 & 1\\ 1 & \frac{1}{2} \end{vmatrix}
}
=\dfrac
{ 1\cdot \frac{1}{2} - (-1)\cdot 1 }
{ 1\cdot \frac{1}{2} - 1\cdot 1 }
=\dfrac
{ \frac{1}{2} + 1 }
{ \frac{1}{2} - 1 }
=\dfrac
{ \frac{3}{2} }
{ -\frac{1}{2} }
= -3\\\\
x_s = \dfrac
{\begin{vmatrix} 1 & b_1\\ 1 & b_2 \end{vmatrix}
}
{\begin{vmatrix} 1 & -m_1\\ 1 &-m_2 \end{vmatrix}
}
=\dfrac
{\begin{vmatrix} 1 & 1\\ 1 & -1\end{vmatrix}
}
{\begin{vmatrix} 1 & -(-1)\\ 1 & -(-\frac{1}{2}) \end{vmatrix}
}
=\dfrac
{\begin{vmatrix} 1 & 1\\ 1 & -1\end{vmatrix}
}
{\begin{vmatrix} 1 & 1\\ 1 & \frac{1}{2} \end{vmatrix}
}
=\dfrac
{ 1\cdot (-1) - 1\cdot 1 }
{ 1\cdot \frac{1}{2} - 1\cdot 1 }
=\dfrac
{ -1-1 }
{ \frac{1}{2} - 1 }
=\dfrac
{ -2 }
{ -\frac{1}{2} }
= 4
\end{array}
$}}$$

2. Schnitt $$\small{\text{$S_2$}}$$ Gerade (2) mit Gerade (3):

$$\small{\text{$
\begin{array}{rcl}
1\cdot y_s - m_2 \cdot x_s &=& b_2 \\
1\cdot y_s - m_3 \cdot x_s &=& b_3 \\
\\
\hline
\\
y_s = \dfrac
{\begin{vmatrix} b_2 & -m_2\\ b_3 &-m_3 \end{vmatrix}
}
{\begin{vmatrix} 1 & -m_2\\ 1 &-m_3 \end{vmatrix}
}
=\dfrac
{\begin{vmatrix} -1 & -\frac{1}{2}\\ 4 & 2\end{vmatrix}
}
{\begin{vmatrix} 1 & -(-\frac{1}{2})\\ 1 & -2\end{vmatrix}
}
=\dfrac
{\begin{vmatrix} -1 & -\frac{1}{2}\\ 4 & 2\end{vmatrix}
}
{\begin{vmatrix} 1 & \frac{1}{2} \\ 1 & -2\end{vmatrix}
}
=\dfrac
{ -1\cdot 2 - 4 \cdot(- \frac{1}{2}) }
{ 1\cdot (- 2) - 1 \cdot \frac{1}{2} }
=\dfrac
{ -2+2}
{ -2 - \frac{1}{2} }
=\dfrac
{ 0 }
{ -\frac{5}{2} }
= 0\\\\
x_s = \dfrac
{\begin{vmatrix} 1 & b_2\\ 1 & b_3 \end{vmatrix}
}
{\begin{vmatrix} 1 & -m_2\\ 1 &-m_3 \end{vmatrix}
}
=\dfrac
{\begin{vmatrix} 1 & -1\\ 1 & 4\end{vmatrix}
}
{\begin{vmatrix} 1 & -(-\frac{1}{2})\\ 1 & -2\end{vmatrix}
}
=\dfrac
{\begin{vmatrix} 1 & -1\\ 1 & 4\end{vmatrix}
}
{\begin{vmatrix} 1 & \frac{1}{2} \\ 1 & -2\end{vmatrix}
}
=\dfrac
{ 1\cdot 4 - 1\cdot (- 1) }
{ 1\cdot (- 2) - 1 \cdot \frac{1}{2} }
=\dfrac
{ 4+1 }
{ -2 - \frac{1}{2} }
=\dfrac
{ 5 }
{ -\frac{5}{2} }
= -2
\end{array}
$}}$$

3. Schnitt $$\small{\text{$S_3$}}$$ Gerade (3) mit Gerade (1):

$$\small{\text{$
\begin{array}{rcl}
1\cdot y_s - m_3 \cdot x_s &=& b_3 \\
1\cdot y_s - m_1 \cdot x_s &=& b_1 \\
\\
\hline
\\
y_s = \dfrac
{\begin{vmatrix} b_3 & -m_3\\ b_1 &-m_1 \end{vmatrix}
}
{\begin{vmatrix} 1 & -m_3\\ 1 &-m_1 \end{vmatrix}
}
=\dfrac
{\begin{vmatrix} 4 & -2\\ 1 & -(-1)\end{vmatrix}
}
{\begin{vmatrix} 1 & -2\\ 1 & -(-1)\end{vmatrix}
}
=\dfrac
{\begin{vmatrix} 4 & -2\\ 1 & 1\end{vmatrix}
}
{\begin{vmatrix} 1 & -2\\ 1 & 1\end{vmatrix}
}
=\dfrac
{ 4\cdot 1 - 1 \cdot (-2) }
{ 1\cdot 1 - 1 \cdot (-2) }
=\dfrac
{ 4+2}
{ 1+2 }
=\dfrac
{ 6 }
{ 3 }
= 2\\\\
x_s = \dfrac
{\begin{vmatrix} 1 & b_3\\ 1 & b_1 \end{vmatrix}
}
{\begin{vmatrix} 1 & -m_3\\ 1 &-m_1 \end{vmatrix}
}
=\dfrac
{\begin{vmatrix} 1 & 4\\ 1 & 1\end{vmatrix}
}
{\begin{vmatrix} 1 & -2\\ 1 & -(-1)\end{vmatrix}
}
=\dfrac
{\begin{vmatrix} 1 & 4\\ 1 & 1\end{vmatrix}
}
{\begin{vmatrix} 1 & -2\\ 1 & 1\end{vmatrix}
}
=\dfrac
{ 1\cdot 1 - 1\cdot 4 }
{ 1\cdot 1 - 1 \cdot (-2) }
=\dfrac
{ 1-4 }
{ 1+2 }
=\dfrac
{ -3 }
{ 3 }
= -1
\end{array}
$}}$$

Die 3 Schnittpunkte zusammengefasst:

$$\small{\text{$
\begin{array}{|l|r|c|}
\hline
& x & y \\
\hline
S_1 &4 & -3 \\
S_2 &-2 & 0 \\
S_3 &-1 & 2 \\
\hline
\end{array}
$}}$$

Die Strecke von $$\small{\text{$S_1 $ nach $ S_2$}}$$:

$$\small{\text{$ \begin{array}{lrc} \overline{S_1S_2} =\sqrt{ [(4)-(-2)]^2 + [0-(-3)]^2 } =\sqrt{ 6^2 + 3^2 } =\sqrt{ 36 + 9 } =\sqrt{ 45 } =6,70820393250 \end{array} $}}$$

Die Strecke von $$\small{\text{$S_2 $ nach $ S_3$}}$$:

$$\small{\text{$ \begin{array}{lrc} \overline{S_2S_3} =\sqrt{ [(-1)-(-2)]^2 + (2-0)^2 } =\sqrt{ (-1+2)^2 + 2^2 } =\sqrt{ 1^2 + 4 } =\sqrt{ 5 } =2.23606797750 \end{array} $}}$$

Die Strecke von $$\small{\text{$S_3 $ nach $ S_1$}}$$:

$$\small{\text{$ \begin{array}{lrc} \overline{S_2S_3} =\sqrt{ [4-(-1)]^2 + (-3-2)^2 } =\sqrt{ (4+1)^2 + (-5)^2 } =\sqrt{ 5^2 + 5^2 } =\sqrt{ 50 } =7.07106781187 \end{array} $}}$$

Der Umfang der eingeschlossenen Fläche ( Dreieck) beträgt: $$\small{\text{$\overline{S_1S_2}+\overline{S_2S_3}+\overline{S_2S_3} =6,70820393250+2,23606797750+7,07106781187 =16,0153397219 $}}$$

Die Fläche des Dreiecks nach Heron:

$$\small{\text{$
\boxed{~~ A =\sqrt{ s\cdot(s-a)\cdot(s-b)\cdot(s-c)} \qquad
s = (a + b + c)/2 ~~}
$}}\\\\
\small{\text{$
\begin{array}{lcl}
a= \overline{S_1S_2} = 6,70820393250\\
b= \overline{S_2S_3} = 2,23606797750\\
c= \overline{S_3S_1} = 7,07106781187 \\\\
s = \dfrac{16,0153397219}{2}= 8,00766986093\\
\end{array}
$}}\\\\\\
\small{\text{$
\begin{array}{lcl}
A =\sqrt{ 8,00766986093\cdot(8,00766986093-6,70820393250)\cdot(8,00766986093-2,23606797750)\cdot(8,00766986093-7,07106781187)}\\
A =\sqrt{ 8,00766986093\cdot 1,29946592843 \cdot 5,77160188343\cdot 0,93660204906}\\
A =\sqrt{ 56.2499999994 }\\
A=7,5
\end{array}
$}}\\\\$$

Der Flächeninhalt der eingeschlossenen Fläche ( Dreieck) beträgt 7,5

Probe der Fläche:

$$\small{\text{$ \begin{array}{|l|r|c|} \hline & x & y \\ \hline S_1 &4 & -3 \\ S_2 &-2 & 0 \\ S_3 &-1 & 2 \\ \hline \end{array}
$}}\\\\
\small{\text{$
\boxed{
~~
A = \dfrac{1}{2}\cdot \begin{vmatrix}
1 & 1 & 1 \\
x_{s_1} & x_{s_2} & x_{s_3} \\
y_{s_1} & y_{s_2} & y_{s_3} \\
\end{vmatrix}
~~
}
$}}\\\\
\small{\text{$
\begin{array}{rcl}
A &=&\dfrac{1}{2}\cdot \begin{vmatrix}
1 & 1 & 1 \\
4 & -2 & -1 \\
-3 & 0 & 2 \\
\end{vmatrix} \\\\
&=&\frac{1}{2} \cdot
[ 1\cdot (-2) \cdot 2 + 4\cdot 0 \cdot 1 + (-3)\cdot 1 \cdot (-1)
- (-3)\cdot (-2) \cdot 1 - 0\cdot (-1) \cdot 1 - 4\cdot 1 \cdot 2
]\\\\
&=&\frac{1}{2} \cdot
[ 1\cdot (-2) \cdot 2 + (-3)\cdot 1 \cdot (-1)
- (-3)\cdot (-2) \cdot 1 - 4\cdot 1 \cdot 2
]\\\\
&=&\frac{1}{2} \cdot
[ -4 + 3 - 6 \cdot 1 - 8]\\\\
&=&\frac{1}{2} \cdot
[ -15 ]\\\\
&=& - 7,5 \\\\
|A| &=& 7,5 \qquad \mathrm{okay}
\end{array}
$}}$$

8(9s+15)=6(4s-4)

$$\small{\text{$
\begin{array}{rcl}
8(9s+15) &=& 6(4s-4) \\
8\cdot 9s+8\cdot 15 &=& 6\cdot 4s - 6\cdot 4 \\
72s + 120 &=& 24s - 24 \quad | \quad - 24s\\
72s-24s + 120 &=& -24 \\
48s +120 &=& - 24 \\
48s +120 &=& - 24 \quad | \quad - 120\\
48s &=& - 24- 120\\
48s &=& -144\\
48s &=& -144 \quad | \quad :48\\\\
s & =& -\dfrac{144}{48} \\\\
\mathbf{s} & \mathbf{=} &\mathbf{ - 3 }
\end{array}
$}}\\\\\\\\
\small{\text{$\mathbf{check:}
\begin{array}{rcl}
8\cdot [9(-3)+15] &=& 6\cdot [4(-3)-4]\\
8\cdot (-27+15) &=& 6\cdot (-12-4)\\
8\cdot (-12) &=& 6\cdot (-16) \\
-96 &=& -96 \qquad \mathrm{okay}
\end{array}
$}}$$

What is the slope of the line through the points (-2, -5) and (1, -7)

The formula of a line is y=mx+b

and m is the slop.

We have point 1 with x = -2 and y = -5 ... so we have   -5 = m*(-2) + b  or -5 = -2m+b

We have point 2 with x = 1 and y = -7... so we have      -7 = m*(1) + b   or -7= m+b

Now we subtract the equation 2 of the equation 1:        -7 -(-5) = m+b - (-2m+b)

so we have  -7+5 = m+b+2m-b

or  -2 = 3m

and the slope m  is $$\small{\text{$-\dfrac{2}{3}$}}$$