Question 845926
*[invoke linear_substitution "x", "y", 1, 1, 2, 1, -1, 0 ]I am not sure that your problems are displayed correctly, but I will make my best guess at what you meant.
Not knowing your age, your grade, or how much math you have already learned, I do not know how much explanation you need.
Just in case, I will explain a lot, so it may get slow and boring.
 
{{{highlight(6&5/9-9&3/4+7&1/3-1&5/6=59/9-39/4+22/3-11/6=236/36-351/36+264/36-66/36=(236-351+264-66)/36=83/36=2&11/36)}}}
To calculate {{{6&5/9-9&3/4+7&1/3-1&5/6}}} ,
the first step would be to convert those mixed numbers into improper fractions.
{{{6&5/9=6+5/9=54/9+5/9=59/9}}} so the mixed number {{{6&5/9}}} can be written as the improper fraction {{{59/9}}} .
Doing the same with the other numbers,we get
{{{6&5/9-9&3/4+7&1/3-1&5/6=59/9-39/4+22/3-11/6}}}
Now the problem is that all the fractions to be added have different denominators.
We need a common denominator, one that is a multiple of all the denominators in those fractions.
We need a number that is a multiple of {{{9}}} , {{{4}}} , {{{3}}} and {{{6}}} .
The smallest one we can find is {{{36=4*9}}}.
The multiples of 9 are 9, 18, 27, 36, 45, 54, and so on.
Of the multiples of {{{9}}}  ,
{{{9}}} is a multiple of {{{3}}} , but not a multiple of the other numbers;
{{{18}}} is a multiple of {{{3}}} and {{{6}}} , but not a multiple of {{{4}}} ,
and {{{27}}} is not a multiple of {{{4}}} or {{{6}}} either.
The next smallest multiple of {{{9}}} is {{{36}}} , and it is a multiple of {{{9}}} , {{{4}}} , {{{3}}} and {{{6}}} .
That is the least common multiple of {{{9}}} , {{{4}}} , {{{3}}} and {{{6}}} .
So we want to convert all those fractions to fractions with {{{36}}} for a denominator.
To convert a fraction into an equivalent fraction, we have to multiply the numerator and denominator times the same number.
Multiplying the numerator and denominator of {{{59/9}}} times {{{4}}} we get
{{{59/9=59*4/(9*4)=236/36}}} .
Multiplying the numerator and denominator of {{{39/4}}} times {{{9}}} we get
{{{39/4=39*9/(4*9)=351/36}}} .
Multiplying the numerator and denominator of {{{22/3}}} times {{{12}}} we get
{{{22/3=22*12/(3*12)=264/36}}} .
Multiplying the numerator and denominator of {{{11/6}}} times 6 we get
{{{11/6=11*6/(6*6)=66/36}}} .
So far we have
{{{6&5/9-9&3/4+7&1/3-1&5/6=59/9-39/4+22/3-11/6=236/36-351/36+264/36-66/36}}}
Now it is just a question of adding/subtracting the numerators to get
{{{(236-351+264-66)/36=83/36}}}
The way I calculate {{{236-351+264-66}}} has 3 steps:
1) I add together all the numbers with a + sign in front, and add together all the numbers with a - sign in front,
2) I figure out which total wins, to use the sign of the winning total, and
3) I calculate the difference of the totals.
I could write that as
{{{236-351+264-66=(236+264)-(351+66)=500-417=83}}}
Putting all of the above together, I would just write
{{{6&5/9-9&3/4+7&1/3-1&5/6=59/9-39/4+22/3-11/6=236/36-351/36+264/36-66/36=(236-351+264-66)/36=83/36}}}
(The rest of my calculations would be in my head, or on the margin of the paper, or on separate scrap paper).
{{{83/26}}} cannot be simplified into an equivalent fraction with a denominator smaller than {{{36}}},
so the last step I could do is converting that {{{83/36}}} improper fraction into the mixed number {{{2&11/36}}} .
I would do that in my head (or pretend I did), but my calculation would be
that {{{83}}} divided by {{{36}}} is {{{2}}} with a remainder of {{{11}}} ,
because {{{36}}} times {{{2}}} is {{{72}}} ,
and {{{83}}} is {{{83-72=11}}} more than {{{72}}} ,
so {{{83/36=(72+11)/36=72/36+11/36=2+11/36=2&11/36}}} .
On my paper, I would only write
{{{highlight(6&5/9-9&3/4+7&1/3-1&5/6=59/9-39/4+22/3-11/6=236/36-351/36+264/36-66/36=(236-351+264-66)/36=83/36=2&11/36)}}} ,
although a lot of calculating went into that.
 
I believe your second problem is {{{3&3/5}}} x {{{1&5/9}}} divided by {{{2/10}}} .
I would write that as
{{{highlight((3&3/5)*(1&5/9)/(2/10)=(18/5)*(14/9)/(1/5)=(18/5)*(14/9)*(5/1)=18*14*5/(5*9)=28)}}}
To begin with, {{{2/10=1/5}}} . It is strange that it was given as {{{2/10}}} ,
because {{{2/10}}} is the same as {{{1/5}}} and {{{1/5}}} is the simplest way to write that number.
You must also know that dividing by a fraction is the same as multiplying times its reciprocal (the fraction turned upside down), so dividing by {{{1/5}}} is the same as multiplying by  the fraction {{{5/1}}} , which is a fancy way of writing {{{5}}}.
The final {{{18*14*5/(5*9)=28}}} calculation can be done with a calculator,
but it is easy to do in my head if I simplify (cancel out) factors that appear in numerator and denominator.
In my head I see it like this:
{{{18*14*5/(5*9)=18*14*cross(5)/(cross(5)*9)=18*14/9=9*2*14/9=2*14=28}}} .
Some teachers do not like to see cross-out cancelling,
because sometimes students go cross-out crazy,
and they cross out numbers here and there without rhyme or reason,
ending with a result that is not equal to what they started with.
They think that it should be written as
{{{18*14*5/(5*9)=18*14*5/(9*1*5)=(18/9)*(14/1)*(5/5)=2*14*1=28}}} ,
That shows that the two {{{5}}}s cancel out because they can be rearranged into a factor that equals {{{1}}} , and multiplying times {{{1}}} does not change anything.