heureka

help please

\(\text{Given that $13^{-1} \equiv 29 \pmod{47}$, find $34^{-1} \pmod{47}$, as a residue modulo 47. $\\$(Give a number between 0 and 46, inclusive.)}\)

\(\begin{array}{|rcll|} \hline &&\mathbf{ 34^{-1} \pmod{47} } \\ &\equiv& (34-47)^{-1} \pmod{47} \\ &\equiv& (-13)^{-1} \pmod{47} \\ &\equiv& -(13)^{-1} \pmod{47} \quad | \quad 13^{-1}\pmod{47} = 29 \\ &\equiv& -29 \pmod{47} \\ &\equiv& -29+47 \pmod{47} \\ &\equiv& \mathbf{18 \pmod{47} } \\ \hline \end{array}\)

If x - y = 4 and xy = 2, then find |x + y|.

\(\begin{array}{|rcll|} \hline \mathbf{\dfrac{(x+y)^2-(x-y)^2}{2}} &=& \dfrac{x^2+2xy+y^2-x^2+2xy-y^2}{2} \\\\ \dfrac{(x+y)^2-(x-y)^2}{2} &=& \dfrac{4xy}{2} \\\\ (x+y)^2-(x-y)^2 &=& 4xy \quad | \quad x - y = 4,\ xy = 2 \\\\ (x+y)^2-4^2 &=& 4* 2 \\ (x+y)^2-16 &=& 8 \\ (x+y)^2 &=& 8+16 \\ (x+y)^2 &=& 24 \\ x+y &=& \sqrt{24} \\ x+y &=& \sqrt{4*6} \\ \mathbf{ x+y } &=& \mathbf{2\sqrt{6}} \\ \hline \end{array}\)

Find the solutions to \(m^2 + 105 = n^2\) in positive integers.

\(\begin{array}{|rcll|} \hline m^2 + 105 &=& n^2 \\ n^2-m^2 &=& 105 \\ \mathbf{(n-m)(n+m)} &=& \mathbf{105} \\ \mathbf{x*y} &=& \mathbf{105} \\ && \text{find all integer products }\ x*y = 105, \\ && \text{where}\ x=n-m,\ \text{ and }\ y=n+m \\ \hline \end{array} \)

All integer products

\(\begin{array}{|ccl|r|r|r|r|r|} \hline && x*y & x & y & n=\frac{y+x}{2} & m=\frac{y-x}{2} \\ \hline 105 &=& 1\times 105 & 1 & 105 & \frac{105+1}{2}=53 & \frac{105-1}{2}=52 & \mathbf{52^2+105 =53^2} \\ 105 &=& 3\times 35 & 3 & 35 & \frac{ 35+3}{2}=19 & \frac{ 35-3}{2}=16 & \mathbf{16^2+105 =19^2} \\ 105 &=& 5\times 21 & 5 & 21 & \frac{ 21+5}{2}=13 & \frac{ 21-5}{2}= 8 & \mathbf{ 8^2+105 =13^2} \\ 105 &=& 7\times 15 & 7 & 15 & \frac{ 15+7}{2}=11 & \frac{ 15-7}{2}= 4 & \mathbf{ 4^2+105 =11^2} \\ \hline \end{array}\)

Find all integers x and y such that \(x + y = xy\).

\(\begin{array}{|rcll|} \hline \mathbf{ x + y} &=& \mathbf{ = xy} \\ xy-x-y &=& 0 \\ x(y-1)-y &=& 0 \quad | \quad +1 \\ x(y-1)-y+1 &=& 1 \\ x(y-1)-(y-1) &=& 1 \\ (x-1)(y-1) &=& 1 \\ \mathbf{(x-1)(y-1)} &=& \mathbf{1} \\ \hline \end{array}\)

Solutions:

\(\begin{array}{|rcll|} \hline 1) & 1 &=& 1*1 \\ 2) & 1 &=& (-1)(-1) \\ \hline \end{array}\)

\(\begin{array}{|r|r|r|l|} \hline (x-1) & (y-1) & x & y & \text{$x$ is integer and $y$ is integer}\\ \hline 1 & 1 & x-1=1 & y-1 = 1 \\ & & x =2 & y =2 & \checkmark \\ \hline -1 & -1 & x-1=-1 & y-1 = -1 \\ & & x =0 & y =0 \\ \hline \end{array} \)

In this multi-part problem, we will consider this system of simultaneous equations:
\(\begin{array}{r@{~}c@{~}l l} 3x+5y-6z &=&2, & \textrm{(i)} \\ 5xy-10yz-6xz &=& -41, & \textrm{(ii)} \\ xyz&=&6. & \textrm{(iii)} \end{array}\)
Let \(a=3x\), \(b=5y\), and \(c=-6z\).

Given that (x, y, z) is a solution to the original system of equations, determine all distinct possible values of x+y.
(Suggestion: Using the substitutions in part (a), first determine all possible values of the ordered triple (a, b, c), then determine the possible solutions (x,y, z).)

Using the substitutions in part (a):

\(\begin{array}{|rcll|} \hline && \begin{array}{c} a=3x,\ b=5y,\ \text{and} \ c=-6z \\ \end{array} \\ \hline 3x+5y-6z &=&2 \\ \mathbf{a+b+c} &=& \mathbf{2} \\\\ \hline abc &=& 3x5y(-6z) \\ abc &=& -90xyz \quad | \quad xyz = 6 \\ abc &=& -90*6 \\ \mathbf{abc} &=& \mathbf{-540} \\ \hline \end{array} \begin{array}{|rcl|} \hline \boxed{ ab=15xy,\ bc=-30yz,\ \text{and} \ ac=-18xz \\ xy = \frac{ab}{15},\ yz = -\frac{bc}{30},\ xz = -\frac{ac}{18} } \\ \hline 5xy-10yz-6xz &=& -41 \\ 5*\frac{ab}{15}+10*\frac{bc}{30}+6*\frac{ac}{18} &=& -41 \\ \frac{ab}{3}+\frac{bc}{3}+\frac{ac}{3} &=& -41 \\ ab+bc+ac &=& -3*41 \\ \mathbf{ab+bc+ac} &=& \mathbf{-123} \\ \hline \end{array} \)

\(\begin{array}{|rcl|} \hline \begin{array}{r@{~}c@{~}l l} \mathbf{a+b+c} &=& \mathbf{2} \\ \mathbf{ab+bc+ac} &=& \mathbf{-123} \\ \mathbf{abc} &=& \mathbf{-540} \\ \end{array} \\ \hline \text{If $a$, $b$, $c$ are the roots of $(x-a)(x-b)(x-c)=0 \\$expandig we get $x^3-(a+b+c)x^2+(ab+bc+ac)x^2-abc = 0$} \\ \mathbf{x^3-2x^2-123x^2+540} &=& \mathbf{0} \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline x^3-2x^2-123x^2+540 &=& 0 \quad | \quad \text{re-write as} \\ x^3 - 2x^2 - 63x - 60x + 540 &=& 0 \\ x(x^2 - 2x - 63) - 60x + 540 &=& 0 \\ x(x -9)(x + 7) - 60 (x - 9) &=& 0 \\ (x - 9) [ x(x + 7) - 60 ] &=& 0 \\ (x - 9) [ x^2 + 7x - 60] &=& 0 \\ (x - 9) ( x + 12)(x - 5) &=& 0 \\\\ \text{The roots: $5$, $9$, and $-12$ } \\ \hline \end{array}\)

In the solution for 

https://web2.0calc.com/questions/question-about-conics-pls-help#r3, what do you do to get to \(-\dfrac{b^2}{a^2}\dfrac{x_t}{y_t}=1\)

The sum of the ages of Sam and Tomas is 27 years.

Seven years ago, Tomas was three years more than one- fourth as old as Sam.

Find there present age.

x = sam age
y = thomas age

\(\begin{array}{|rclcrcl|} \hline x+y &=& 27 \quad \text{or} \quad \mathbf{y = 27-x} \\\\ \mathbf{y-7} &=& \mathbf{\dfrac{1}{4}(x-7) + 3} \\ y &=& \dfrac{1}{4}(x-7) + 3 +7 \\\\ y &=& \dfrac{1}{4}(x-7) + 10\quad | \quad y = 27-x \\\\ 27-x &=& \dfrac{1}{4}(x-7) + 10 \\\\ -x &=& \dfrac{1}{4}(x-7) + 10-27 \\\\ -x &=& \dfrac{1}{4}(x-7) -17 \\\\ x + \dfrac{1}{4}(x-7) &=& 17 \quad | \quad *4 \\\\ 4x + x-7 &=& 4*17 \\\\ 5x &=& 4*17 +7 \\\\ 5x &=& 75 \\\\ \mathbf{x} &=& \mathbf{15} \\ \hline \\ \mathbf{y} &=& \mathbf{27-x} \\ y &=& 27-15 \\ \mathbf{y} &=& \mathbf{12} \\ \hline \end{array}\)

sam is 15 and thomas is 12

Heureka, what do you do to get to \(-\dfrac{b^2}{a^2}\dfrac{x_t}{y_t}=1\)

\(\begin{array}{|l|} \hline \mathbf{y} = \mathbf{\dfrac{b^2}{y_t}-\dfrac{b^2}{a^2}\dfrac{x_t}{y_t}x}\quad \boxed{\text{compare with tangent line }\\ y=k+1x} \\ \hline \end{array} \\ \begin{array}{rclcrl} y&=&\dfrac{b^2}{y_t}-\dfrac{b^2}{a^2}\dfrac{x_t}{y_t}x &\color{darkgrey}\text{compare}& y&=&k+1*x \\\\ y&=& {\color{red}\dfrac{b^2}{y_t}}{\color{green}-\dfrac{b^2}{a^2}\dfrac{x_t}{y_t}}x &\color{darkgrey}\text{compare}& y&=&{\color{red}k}{\color{green}+1}*x \\ &&&\begin{array}{|c|} \hline {\color{red}\dfrac{b^2}{y_t}} = {\color{red}k} \\\\ {\color{green}-\dfrac{b^2}{a^2}\dfrac{x_t}{y_t}} = {\color{green}+1}\\ \hline \end{array} \end{array}\\\)

BC is tangent to both a circle with center at A and a circle with center at D.
The area of the circle with center at A is 225p and the area of the circle with center at D is 36p.
If BC=16, find the distance between the centers of the two circles.

Formula area of the circle: \(A=\pi r^2\)

\(\begin{array}{|rcll|} \hline A=\pi r^2 \\ \hline 225\pi &=& \pi R^2 \\ 225 &=& R^2 \\ \mathbf{15} &=& \mathbf{R} \\\\ 36\pi &=& \pi r^2 \\ 36 &=& r^2 \\ \mathbf{6} &=& \mathbf{r} \\ \hline \end{array} \)

Pythagoras' theorem:

\(\begin{array}{|rcll|} \hline AD^2 &=& 16^2+(R+r)^2 \\ AD^2 &=& 16^2+(15+6)^2 \\ AD^2 &=& 16^2+21^2 \\ AD^2 &=& 256+441 \\ AD^2 &=& 697 \\ AD &=& 26.4007575649 \\ \mathbf{AD} &\approx& \mathbf{26.4} \\ \hline \end{array}\)

The distance between the centers of the two circles is \(\approx \mathbf{26.4}\)