- Solve system (a) by substituting the expression for \(y\), squaring, and solving the quadratic equation; verify for extraneous solutions to get \((-10, 6)\).
- Solve system (b) by rearranging to standard form and using elimination to find \((-2, 126)\).
- Final answers are \(\boxed{(-10, 6)}\) and \(\boxed{(-2, 126)}\).
Explanation
- Solving System (a)
We are given two systems of equations to solve algebraically and verify.
System (a):
\[ \begin{cases} 7x = y - 76 \ y = \sqrt{6 - 3x} \end{cases} \]
To solve this, we substitute the expression for \(y\) from the second equation into the first:
\[7x = \sqrt{6 - 3x} - 76\]
Rearranging terms, we have:
\[7x + 76 = \sqrt{6 - 3x}\]
Squaring both sides leads to:
\[(7x + 76)^2 = 6 - 3x\]
Expanding the left side:
\[49x^2 + 2(7x)(76) + 76^2 = 6 - 3x\]
\[49x^2 + 1064x + 5776 = 6 - 3x\]
Bringing all terms to one side gives the quadratic equation:
\[49x^2 + 1067x + 5770 = 0\]
Using the quadratic formula, the discriminant is \(\Delta = 1067^2 - 4(49)(5770) = 1138489 - 1130920 = 7569\).
The roots are \(x = \frac{-1067 \pm \sqrt{7569}}{2(49)} = \frac{-1067 \pm 87}{98}\).
This gives two potential values: \(x_1 = -10\) and \(x_2 = -\frac{1154}{98} \approx -11.78\).
Checking \(x_1 = -10\): \(7(-10) + 76 = 6\). Since \(\sqrt{6 - 3(-10)} = \sqrt{36} = 6\), this is a valid solution.
Checking \(x_2 \approx -11.78\): \(7(-11.78) + 76 \approx -82.46 + 76 = -6.46\). However, a square root must be non-negative, so this is an extraneous solution.
Thus, the solution is \((x, y) = (-10, 6)\).
- Solving System (b)
System (b):
\[ \begin{cases} 42x + y = 42 \ 92 - y = 17x \end{cases} \]
Rearranging the second equation into standard linear form \(Ax + By = C\):
\[17x + y = 92\]
Now we have the system:
\[ \begin{cases} 42x + y = 42 \ 17x + y = 92 \end{cases} \]
Subtracting the second equation from the first:
\[(42x - 17x) + (y - y) = 42 - 92\]
\[25x = -50\]
\[x = -2\]
Substituting \(x = -2\) into the first original equation:
\[42(-2) + y = 42\]
\[-84 + y = 42\]
\[y = 126\]
Verifying with the second equation: \(92 - 126 = 17(-2) \Rightarrow -34 = -34\). This is correct.
The solution is \((x, y) = (-2, 126)\).
- Final Results
In conclusion, we have solved both systems of equations. For system (a), the only solution is \((-10, 6)\). For system (b), the solution is \((-2, 126)\). The final answers are \(\boxed{(-10, 6)}\) and \(\boxed{(-2, 126)}\).
Examples
Systems of equations are fundamental in fields like engineering and economics. For example, if you are managing a logistics budget, one equation might represent the cost of transportation based on weight and distance, while another represents the capacity constraints of your vehicles. Solving these systems allows you to find the exact point where your budget and capacity intersect, ensuring your operations are both cost-effective and feasible within your limits. In this problem, we practiced finding those intersection points algebraically, a skill essential for any optimization task.