Answer:
The numbers 'x' and 'y' are:
[tex]x=56,\:y=36[/tex]
Step-by-step explanation:
Let 'x' and 'y' be the two numbers
As the sum of the two numbers is 92.
so
[tex]x+y = 92[/tex]
Given that One-eighth of the larger number plus one-third of the smaller number is 19.
so
[tex]\frac{1}{8}x\:+\:\frac{1}{3}y=19[/tex]
now solving both equations to determine the numbers 'x' and 'y'.
[tex]\begin{bmatrix}\frac{1}{8}x+\frac{1}{3}y=19\\ x+y=92\end{bmatrix}[/tex]
[tex]\mathrm{Multiply\:}\frac{1}{8}x+\frac{1}{3}y=19\mathrm{\:by\:}8\:\mathrm{:}\:\quad \:x+\frac{8}{3}y=152[/tex]
[tex]\begin{bmatrix}x+\frac{8}{3}y=152\\ x+y=92\end{bmatrix}[/tex]
[tex]x+y=92[/tex]
[tex]-[/tex]
[tex]\underline{x+\frac{8}{3}y=152}[/tex]
[tex]-\frac{5}{3}y=-60[/tex]
[tex]\begin{bmatrix}x+\frac{8}{3}y=152\\ -\frac{5}{3}y=-60\end{bmatrix}[/tex]
solve for y
[tex]-\frac{5}{3}y=-60[/tex]
[tex]-5y=-180[/tex]
[tex]\mathrm{Divide\:both\:sides\:by\:}-5[/tex]
[tex]\frac{-5y}{-5}=\frac{-180}{-5}[/tex]
[tex]y=36[/tex]
[tex]\mathrm{For\:}x+\frac{8}{3}y=152\mathrm{\:plug\:in\:}y=36[/tex]
[tex]x+\frac{8}{3}\cdot \:36=152[/tex]
[tex]x=56[/tex]
Thus, the numbers 'x' and 'y' are:
[tex]x=56,\:y=36[/tex]
