hectictar

\(6(y + 1) \,=\, 7(x - 2)\)

                                               Divide both sides of the equation by  7 .

\(\frac67(y + 1)\, =\, x - 2\)

                                               Add  2  to both sides of the equation.

\(\frac67(y + 1)+2\, =\, x\)

                                               We could stop here, but let's simplify it.

\(x\,=\,\frac67(y + 1)+2\)

                                               Distribute  \(\frac67\)  to the terms in parenthesees.

\(x\,=\,\frac67y + \frac67+2\)

                                               Add  \(\frac67+2\)

\(x\,=\,\frac67y + \frac67+\frac{14}{7}\\~\\ x\,=\,\frac67y + \frac{20}{7}\)

Here is the solution to the problem the way you interpreted it:

Let the number of km traveled on the first day  =  x

\(x+\frac12x+\frac13\Big(\frac12x\Big)+\frac14\Big[\frac13(\frac12x)\Big]\,=\,50\\~\\ x+\frac12x+\frac16x+\frac1{24}x\,=\,50\\~\\ x(1+\frac12+\frac16+\frac1{24})\,=\,50\\~\\ x(\frac{41}{24})\,=\,50\\~\\ x(\frac{41}{24})\cdot\frac{24}{41}\,=\,50\cdot\frac{24}{41}\\~\\ x\,=\,\frac{1200}{41}\\~\\ x\,\approx\,29.268\)

The number of km she traveled on the first day  ≈  29.268  

You solved a different problem. It is kind of unclear what "that distance" refers to.

Also, we can check whether 24 km is a solution (to the problem I solved) like this:

num of km walked on first day  +  num of km walked on second day  +  num of km walked on third day  +

num of km walked of fourth day  =  50  ?

24  +  half of 24  +  a third of 24  +  a quarter of 24  =  50   ?

24  +  12  +  8  +  6  =  50   ?

50  =  50  ?

True

Let the number of kilometers she traveled on the first day  =  x

When you say, "a third that distance on the third day and a quarter of that distance on the fourth day," I think "that distance" is  x .

\(x+\frac12x+\frac13x+\frac14x\,=\,50\\~\\ x(1+\frac12+\frac13+\frac14)\,=\,50\\~\\ x(\frac{25}{12})\,=\,50\\~\\ x(\frac{25}{12})\cdot\frac{12}{25}\,=\,50\cdot\frac{12}{25}\\~\\ x\,=\,24\)

She traveled 24 kilometers on the first day.

Note that   sin( 90° - a )  =  cos( a )      and      cos( 90° - a )  =  sin( a )      and      tan( 90° - a )  =  cot( a )

This might help you see why those are true....

sin( 90° - a )  =  cos( a )   is saying that line  r  is the same length as line  s .

cos( 90° - a )  =  sin( a )   is saying that line  b  is the same length as line  c .

tan( 90° - a )  =  cot( a )   is saying that   r / b  =  s / c

So.....

     sin( 38° )  -  cos( 52° )

=   sin( 90° - 52° )  -  cos( 52° )

=   cos( 52° )  -  cos( 52° )

=   0

     tan( 12° )  -  cot( 78° )

=   tan( 90° - 78° )  -  cot( 78° )

...Do you see where this one is going? Can you finish it?

\(\phantom{=}\qquad\frac{\cos(10°)}{\sin(80°)}\\~\\ =\qquad\frac{\cos(90° - 80°)}{\sin(80°)}\\~\\ =\qquad\frac{\sin(80°)}{\sin(80°)}\\~\\ =\qquad1\)

The last one is a really similar to previous one. Do you know how to do it now?

How about this website?

(x + 1)²   =   (x + 1)(x + 1)

Multiply each item in the first set of parenthesees by each item in the second set of parenthesees.

(x + 1)(x + 1)   =   (x)(x) + (x)(1) + (1)(x) + (1)(1)

Simplify.

(x)(x) + (x)(1) + (1)(x) + (1)(1)   =   x² + 2x + 1

So....

(x + 1)²   =   x² + 2x + 1

\(\ \phantom{=\qquad}3x^2+bx+10\\~\\ =\qquad3(x^2+\frac{b}{3}x)+10\\~\\ =\qquad3(x^2+\frac{b}{3}x+(\frac{b}{6})^2-(\frac{b}{6})^2)+10\\~\\ =\qquad3((x+\frac{b}{6})^2-(\frac{b}{6})^2)+10\\~\\ =\qquad3(x+\frac{b}{6})^2-3(\frac{b}{6})^2+10\)

Now it is in the form  a(x + m)² + n   where

\(a=3\qquad \text{and}\qquad m=\frac{b}{6}\qquad \text{and}\qquad n=-3(\frac{b}{6})^2+10\)

In order for  m  to be an integer,  b  must be a multiple of  6

In order for  n  to be an integer,  b  must be a multiple of  6

The largest integer that must be a divisor of  b  in order for both  m  and  n  to be integers is  6