Question 748931
1.	If Joseph can paint a house in 10 days, Joseph can paint {{{1/10}}} of the house per day (that's his rate/speed of work).
If George can paint the house in 12 days, George can paint {{{1/12}}} of the house per day.
(Joseph paints a little faster than George).
During the 5 days that both work together,
Joseph will paint {{{5*(1/10)=1/2}}} of the house, and
George will paint {{{5*(1/12)=5/12}}} of the house.
Between the two of them, they almost have it finished in 5 days.
The fraction of the house painted during those 5 days is
{{{1/2+5/12=6/12+5/12=11/12}}}
The fraction of the house left to be painted is
{{{1-11/12=12/12-11/12=1/12}}}
That is the amount of work that George could do in 1 day.
Joseph, who is a little faster and could paint {{{1/10}}} finishes the work in a little less than one day and gets to go home a little earlier after finishing the job the 6th day. (or maybe he will use the time to clean up his brushes and put away all materials and equipment).
 
2. Let's name variables.
{{{G}}} = Garry's age (in years)
{{{B}}} = Badong's age (in years)
{{{N}}} = Narva’s age (in years)
Each sentence of the information given will translate into an equation and we will end up with a system of equations to solve, finding values for {{{G}}}, {{{B}}} and {{{N}}}.
If Garry is one year more than twice as old as Badong, Garry's age is
{{{G=2B+1}}} (1 year more than twice B)
If the Garry and Badong together are ten years older than Narva,
{{{G+B=N+10}}}
Narva is 3 years younger than Garry translates as
{{{N=G-3}}}
The 3 equations together form the system of equations
{{{system(G=2B+1,G+B=N+10,N=G-3)}}}
Systems of equations are solved by changing one equation at a time.
You can combine 2 equations to make a more convenient equation that can replace one of the 2 equations you combined.
You do not need to write as much as I did, but you have to keep track of what direction you are going so you do not go around in circles.
We start by substituting {{{2B+1}}} for {{{G}}}
{{{system(G=2B+1,G+B=N+10,N=G-3)}}} --> {{{system(G=2B+1,G+B=N+10,N=2B+1-3)}}} --> {{{system(G=2B+1,G+B=N+10,N=2B-2)}}} --> {{{system(G=2B+1,2B+1+B=2B-2+10,N=2B-2)}}} --> {{{system(G=2B+1,3B+1=2B+8,N=2B-2)}}} --> {{{system(G=2B+1,B=7,N=2B-2)}}} --> {{{system(G=2*7+1,B=7,N=2*7-2)}}} --> {{{system(G=15,B=-3,highlight(N=12))}}}
 
3. Let's name variables.
{{{x}}} = the first number
{{{y}}} = the second number
{{{z}}} = the third number
The average of three numbers is 10 translates as
{{{(x+y+z)/3=10}}} --> {{{x+y+z=30}}}
The second (number) is one more than twice the first translates as
{{{y=2x+1}}}
The third (number) is 5 more than three times the first translates as
{{{z=3x+5}}}
That gives you an easy system {{{system(x+y+z=30,y=2x+1,z=3x+5)}}}
Substituting into the first equation the expressions for {{{y}}} and {{{z}}} given in the second and third equation, we get
{{{x+(2x+1)+(3x+5)=30}}} --> {{{6x+6=30}}} --> {{{6x=24}}} --> {{{highlight(x=4)}}}
{{{system(x+y+z=30,y=2x+1,z=3x+5)}}} --> {{{system(highlight(x=4),y=2x+1,z=3x+5)}}} --> {{{system(highlight(x=4),y=2*4+1,z=3*4+5)}}} --> {{{system(highlight(x=4),highlight(y=9),highlight(z=17))}}}
 
4. Let's name variables.
{{{x}}} = liters of solution A needed
{{{y}}} = liters of solution B needed
To get 10 liters of mixture we need to make {{{x+y=10}}} <--> {{{y=10-x}}}
That could be one equation in a system of equations, or we could decide to use
{{{10-x}}} = liters of solution B needed
and have just one variable all along.
Since brine solution A is 5% salt, {{{x}}} liters of it will contain an amount of salt equal to
{{{0.05x}}}
Since brine solution B is 15% salt, {{{10-x}}} liters of it will contain an amount of salt equal to
{{{0.15*(10-x)=1.5-0.15x}}}
The total amount of salt in the mix will be
{{{0.05x+1.5-0.15x=1.5-0.10x}}}
which must equal 12% of 10 liters or {{{0.12*10=1.2}}}
So {{{1.5-0.10x=1.2}}} --> {{{1.5-1.2=0.10x}}} --> {{{0.3=0.1x}}} --> {{{x=0.3/0.1}}} --> {{{highlight(x=3)}}}
So you need {{{highlight(3)}}} liters of solution A and {{{10-3=highlight(7)}}} liters of solution B.

NOTE:
Notice that I did not give you units for the amount of salt.
I did not because the problem does not deserve it.
If it made a little more sense I would have said the amounts were in kilograms, but I would have to explain it as a chemical engineering mass balance problem.
To a chemist/chemical engineer (like me) this problem is total nonsense for at least 2 reasons:
1) mixing 3 liters of one solution with 7 liters of another do not necessarily give you 10 liters, and
2) concentrations should be clearly specified as in 5% w/v, where w/v means weight (w) in volume (v), meaning 5 kilograms of salt per 100 liters of brine solution (or 5 grams in 100 milliliters). If brine solution A was specified as 5% w/w (5 kilogram salt per