- For the first system \(\begin{cases} y = 3 \ y = \frac{1}{4}x + 5 \end{cases}\), solving yields \(x = -8\), so the solution is \((-8, 3)\).
- For the second system \(\begin{cases} y = x + 6 \ y = 9 \end{cases}\), solving yields \(x = 3\), so the solution is \((3, 9)\).
- Verification confirms both solutions satisfy their respective systems.
- The final answers are \(\boxed{(-8, 3)}\) and \(\boxed{(3, 9)}\).
Explanation
- Problem Analysis
We are given two separate systems of linear equations to solve and verify.
System 1:
\begin{cases} y = 3 \ y = \frac{1}{4}x + 5 \end{cases}
System 2:
\begin{cases} y = x + 6 \ y = 9 \end{cases}
Our goal is to find the values of \(x\) and \(y\) for each system and confirm the results by substituting them back into the original equations. We will use the method of substitution to solve for the intersection point in each case.
- Solving System 1
To solve the first system, we substitute the value \(y = 3\) into the second equation:
\[3 = \frac{1}{4}x + 5\]
Subtracting 5 from both sides gives:
\[3 - 5 = \frac{1}{4}x\]
\[-2 = \frac{1}{4}x\]
Multiplying both sides by 4, we find the value of \(x\):
\[x = -2 \times 4 = -8\]
Thus, the solution is \((-8, 3)\).
To verify, substitute \(x = -8\) into the second equation: \(y = \frac{1}{4}(-8) + 5 = -2 + 5 = 3\). Since this matches the first equation \(y = 3\), the solution is correct.
- Solving System 2
To solve the second system, we substitute the value \(y = 9\) into the first equation:
\[9 = x + 6\]
Subtracting 6 from both sides gives:
\[x = 9 - 6 = 3\]
Thus, the solution is \((3, 9)\).
To verify, substitute \(x = 3\) into the first equation: \(y = 3 + 6 = 9\). Since this matches the second equation \(y = 9\), the solution is correct.
Examples
Systems of linear equations are used in various real-life scenarios, such as finding the break-even point in business. For example, if a company has a fixed cost plus a variable cost per unit (represented by \(y = mx + b\)), they can compare this to their revenue function to determine how many units must be sold to cover their costs. By setting the cost equation and the revenue equation equal to each other, the intersection point tells the business exactly when their total revenue equals their total costs.