## Elementary Algebra

Let one number be $x$ and the other number be $y$. If 4 times the larger number minus the smaller number is equal to 64, then $4y-x=64$. This is the first of the two simultaneous equations. Also, twice the larger number plus the smaller number is 176. Therefore, $x+2y=176$. We write the equation $x+2y=176$ as $x=176-2y$ to make $x$ the subject of the equation. This is the second of the two simultaneous equations. Therefore, we now have a pair of simultaneous equations to solve. Substituting this second equation in the first equation, we obtain: $4y-x=64$ $4y-(176-2y)=64$ $4y-176+2y=64$ $4y+2y=64+176$ $6y=240$ $y=\frac{240}{6}=40$ We now substitute the value of $y$ in the first equation to find the value of $x$, $4y-x=64$ $4(40)-x=64$ $160-x=64$ $x=160-64$ $x=96$ Therefore, the smaller number is 40 while the larger number is 96.