Question 264711
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I'll just do the 3rd one, the hardest one:

{{{system(3x+8y = 81,5x+6y=-39)}}}

To eliminate the x's:

The coefficients of x are 3 and 5.
The least common multiple of 3 and 5 is 15. We want to
make one of the coefficients of x 15 and the other one
-15.  To do that we multiply the first equation through
by 5 and the second one through by -3:

{{{system(15x+40y=405,-15x-18y=117)}}}

Add the two equations term by term
 
{{{0x+22y=522}}}
{{{22y=522}}}
{{{y=522/22}}}
{{{y=261/11}}}

Go back to the original system:

{{{system(3x+8y = 81,5x+6y=-39)}}}

To eliminate the y's:
The coefficients of y are 8 and 6.
The least common multiple of 8 and 6 is 24. We want to
make one of the coefficients of y 24 and the other one
-24.  To do that we multiply the first equation through
by 3 and the second one through by -4:

{{{system(9x+24y=243,-20x-24y=156)}}}

Add the two equations term by term
 
{{{-11x+0y=399}}}

{{{-11x=399}}}

{{{x=-399/11}}}

Edwin</pre>