Question 773373
I believe you meant
{{{system(10x+3y=8,2x+y=2)}}}
 
When solving a system by substitution, you start by solving for one of the variables in one of the equations.
It is easier if you look for a variable with a coefficient of 1 or -1, meaning that you look for a variable that does not have a number in front, at most it has a + or - sign.
For example,
in {{{2x-y=2}}}, it is easiest to solve for {{{y}}},
while in {{{x+4y=5}}}, it is easiest to solve for {{{x}}}.
Solving for the variable gives you an expression for that variable as a function of the other variable(s).
Then you substitute that expression into the other equation(s).
 
{{{system(10x+3y=8,2x+y=2)}}}
Solving for {{{y}}} in {{{2x+y=2}}} is the easiest way to start:
{{{2x+y=2}}} --> {{{y=2-2x}}} (subtracting {{{2x}}} from both sides of the equal sign.
Next we substitute {{{2-2x}}} for {{{y}}} in {{{10x+3y=8}}}:
{{{10x+3(2-2x)=8}}}-->{{{10x+3*2-3*2x=8}}} (applying the distributive property)
{{{10x+3*2-3*2x=8}}}-->{{{10x+6-6x=8}}} (doing the indicated operations)
{{{10x+6-6x=8}}}-->{{{10x-6x=8-6}}}-->{{{10x-6x=2}}} (subtracting {{{6}}} from both sides)
{{{10x-6x=2}}}-->{{{(10-6)x=2}}}-->{{{4x=2}}} (taking out {{{x}}} as a common factor, sometimes called "collecting like terms")
{{{4x=2}}}-->{{{x=2/4}}}-->{{{highlight(x=1/2)}}} (dividing both sides by 4)
Now we substitute {{{1/2}}} for {{{x}}} in the solved-for-y equation {{{y=2-2x}}} above:
{{{y=2-2*(1/2)}}}-->{{{y=2-1}}}-->{{{highlight(y=1)}}}
 
Checking the answer {{{highlight(system(x=1/2,y=1))}}}:
{{{10x+3y=10*(1/2)+3*1=5+3=8}}} verifies {{{10x+3y=8}}}
{{{2x+y=2*(1/2)+1=1+1=2}}} verifies {{{2x+y=2}}}