Find two numbers whose sum is 28 and the difference is 4

asked by guest
on Nov 13, 2024 at 1:30 am



MathBot Answer:

The numbers are \(16\) and \(12\).


The sum of \(n_{1}\) and \(n_{2}\) is \(28\), and their difference is \(4\). This translates to the two equations: \[\begin{aligned}n_{1} + n_{2} &= 28\\n_{1} - n_{2} &= 4\\\end{aligned}\] Adding the two equations together yields:\[\begin{gather*} 2 n_{1} = 32\\ n_{1} = 16\end{gather*}\] Substituting \(16\) for \(n_{1}\) into \(n_{1} + n_{2} = 28\) results in \(16 + n_{2} = 28\), so \(n_{2} = 12\).