Solver Linear Systems by Addition

Source code of 'Linear Systems by Addition'

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==section input
Variable names: *[input X=x] *[input Y=y]

*[input x1=1]X + *[input y1=-1]Y = *[input b1=0]

*[input x2=1]X + *[input y2=1]Y =  *[input b2=2]

==section solution
perl

print "We'll solve the system:
{{{$x1*$X + $y1*$Y = $b1}}}
{{{$x2*$X + $y2*$Y = $b2}}}
by elimination by addition.";

my $neg=-1*$x1;
my $diff=$neg/$x2;

print "To eliminate by addition, we need to set both coefficients of x to numbers with changed signs, i.e a and -a. Since in the second equation we have $x2 as our coefficient for x, to get  $neg we have to multiply all terms of the second equation by {{{$neg/$x2}}} which is equal to $diff.

Multiplying, we get on our second equation:";
my $x3=$x2*$diff; 
my $y3=$y2*$diff; 
my $b3=$b2*$diff;

print "{{{($x2*$diff)$X + ($y2*$diff)$Y = $b2*$diff}}}
{{{$x3*$X + $y3*$Y = $b3}}}

Adding both equations we get:

{{{($x1+$x3)$X + ($y1+$y3)$Y = ($b1+$b3)}}}

Since $x1 and $x3 cancel out, we have a linear equation:";

my $y4=$y1+$y3;
my $b4=$b1+$b3;
my $y=$b4/$y4;
my $y5=$y1*$y;
my $x=($b1 - $y5)/$x1;
my $temp=$b1 - $y5;

my $y = solve( "linear_equation", $Y, $y4, 0, $b4);

print "Therefore, we know that $Y = $y.

Plugging that in into the first equation gives us:

{{{$x1*$X + $y1*$Y = $b1}}}
{{{$x1*$X + $y1*$y = $b1}}} 
{{{$x1*$X + $y5 = $b1}}}
{{{$x1*$X = $b1 - $y5}}}
{{{$X = ($b1 - $y5)/$x1}}} 
{{{$X = $temp/$x1}}}
{{{$X = $x}}}

Therefore, our answer is:

{{{system( $X=$x, $Y=$y )}}}";

==section output
x
y
==section check
 X=x Y=y x1=1 y1=-1 b1=0 x2=1 y2=1 b2=2 x= y=