a+b=33 b=2a

(a-5)+(b+2)

asked by guest
on Oct 24, 2024 at 5:18 am



You asked:

Solve the system of equations:\[\begin{aligned}a + b &= 33\\b &= 2 \cdot a\left( a - 5 \right) + b + 2\end{aligned}\]

MathBot Answer:

The system of equations has \(2\) solutions.

\[a = \frac{5}{2} - \frac{\sqrt{21}}{2}, b = \frac{\sqrt{21}}{2} + \frac{61}{2}\]\[a = \frac{\sqrt{21}}{2} + \frac{5}{2}, b = \frac{61}{2} - \frac{\sqrt{21}}{2}\]


Solve \(a + b = 33\) for \(a\). \[a = 33 - b\]Substitute \(33 - b\) for \(a\) in \(b = 2 a \left(a - 5\right) + b + 2\) and simplify. $$\begin{aligned}b &amp= 2 a \left(a - 5\right) + b + 2 \\ b &= 2 \left(33 - b\right) \left(\left(33 - b\right) - 5\right) + b + 2 \\ b^{2} - 61 b &= -925 \\b^{2} - 61 b + 925 &= 0 \\ b &= \frac{-(-61) \pm \sqrt{(-61)^{2} - 4(1)(925)}}{2(-61)} \\ b = \frac{61}{2} - \frac{\sqrt{21}}{2}&, b = \frac{\sqrt{21}}{2} + \frac{61}{2}\end{aligned}$$Substitute \(\frac{61}{2} - \frac{\sqrt{21}}{2}\) into \(a + b = 33\) to solve for \(a\). \[\begin{aligned}a - \frac{\sqrt{21}}{2} + \frac{61}{2} &= 33\\a + \left(\frac{61}{2} - \frac{\sqrt{21}}{2}\right) &= 33\\a &= \frac{\sqrt{21}}{2} + \frac{5}{2}\end{aligned}\]This yields the following solution. $$\begin{aligned}a = \frac{\sqrt{21}}{2} + \frac{5}{2},\,b = \frac{61}{2} - \frac{\sqrt{21}}{2}\end{aligned}$$Substitute \(\frac{\sqrt{21}}{2} + \frac{61}{2}\) into \(a + b = 33\) to solve for \(a\). \[\begin{aligned}a + \frac{\sqrt{21}}{2} + \frac{61}{2} &= 33\\a + \left(\frac{\sqrt{21}}{2} + \frac{61}{2}\right) &= 33\\a &= \frac{5}{2} - \frac{\sqrt{21}}{2}\end{aligned}\]This yields the following solution. $$\begin{aligned}a = \frac{5}{2} - \frac{\sqrt{21}}{2},\,b = \frac{\sqrt{21}}{2} + \frac{61}{2}\end{aligned}$$