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A number, x, is 11 less than 3 times a smaller number, y,. The sum of the numbers is 24. solve for x and y

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the numbers: x, y

[tex]x=3y-11 \\ x+y=24 \\ \\ \hbox{substitute 3y-11 for x in the second equation:} \\ 3y-11+y=24 \\ 3y+y=24+11 \\ 4y=35 \\ y=\frac{35}{4}=8 \frac{3}{4} \\ \\ x=3 \cdot \frac{35}{4}-11 =\frac{105}{4}-\frac{44}{4}=\frac{61}{4}=15 \frac{1}{4}[/tex]
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