Question 165268
"There are two numbers whose sum is 50" ---> {{{x+y=50}}}


"Three times the first is five more than twice the second" ---> {{{3x=2y+5}}}



{{{x+y=50}}} Start with the first equation.



{{{y=50-x}}} Subtract x from both sides.



{{{3x=2y+5}}} Move onto the second equation



{{{3x=2(50-x)+5}}} Plug in {{{y=50-x}}}



{{{3x=100-2x+5}}} Distribute.



{{{3x=-2x+105}}} Combine like terms on the right side.



{{{3x+2x=105}}} Add {{{2x}}} to both sides.



{{{5x=105}}} Combine like terms on the left side.



{{{x=(105)/(5)}}} Divide both sides by {{{5}}} to isolate {{{x}}}.



{{{x=21}}} Reduce.



{{{y=50-x}}} Go back to the previously isolated equation



{{{y=50-21}}} Plug in {{{x=21}}}



{{{y=29}}} Subtract



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Answer:


So the solutions are {{{x=21}}} and {{{y=29}}} which form the ordered pair (21,29)