Question 798014
{{{system(x+y=0.5,x-y=1)}}}
 
To solve a system of equations by substitution, you start by solving one of the equations for one of the variables.
In this case, we could solve {{{x-y=1}}} for {{{x}}}:
{{{x-y=1}}}-->{{{x=y+1}}}
 
Next, you substitute the expression found for that variable into the other equation.
In this case, we replace {{{y+1}}} for {{{x}}} in {{{x+y=0.5}}} to get
{{{y+1+y=0.5}}}
 
Then, you simplify and solve the resulting equation:
{{{y+1+y=0.5}}}-->{{{2y+1=0.5}}}-->{{{2y=0.5-1}}}-->{{{2y=-0.5}}}-->{{{y=-0.5/2}}}-->{{{highlight(y=-0.25)}}}
 
Finally, you go back to the equation initially "solved" and plug in the value found:
Plugging {{{y=-0.25}}} into {{{x=y+1}}} we find
{{{x=-0.25+1}}}-->{{{highlight(x=0.75)}}}
 
The solution to the system of equations is
{{{highlight(system(x=0.75,y=-0.25))}}}
 
It is good practice to verify that your solution is correct by plugging the numbers found into the original equations.
Sometimes the problem instructions tell you to check that way.
Plugging {{{x=0.75}}} and {{{y=-0.25}}} into {{{x+y=0.5}}} we find
{{{0.75+(-0.25)=0.5}}}<-->{{{0.75-0.25=0.5}}}, which is true.
Plugging the solution found into {{{x-y=1}}} we find
{{{0.75-(-0.25)=1}}}<-->{{{0.75+0.25=1}}}, which is also true.
Now we know that the solution found is correct. We did not make a mistake.
 
NOTE:
It is not always that easy. To start, you should choose wisely what variable to solve for. Look for a variable that has no visible coefficient in front.
Otherwise, you may end up working with fractions, and it is more likely to make a mistake.
If the system is {{{system(3x+y=4,7x-2y=5)}}},
the easiest way to start is to solve {{{3x+y=4}}} for {{{y}}}:
{{{3x+y=4}}}-->{{{y=4-3x}}}.