Problem 8 - Algebra A Combined Function

Question

Determine whether each ordered pair is a solution of the system of linear equations. See Examples 1 and \(2 .\) \(\left\\{\begin{array}{l}4 x=1-y \\ x-3 y=-8\end{array}\right.\) a. (0,1) b. \(\left(\frac{1}{6}, \frac{1}{3}\right)\)

Step-by-Step Solution

Answer

Neither (0, 1) nor \(\left( \frac{1}{6}, \frac{1}{3} \right)\) is a solution to the system.

Step 1: Substitute the ordered pair (0, 1) into first equation

The first equation is \(4x = 1 - y\). Substitute \(x = 0\) and \(y = 1\) into the equation: \(4(0) = 1 - 1\). Simplifying, we get \(0 = 0\), which is true. So (0,1) satisfies the first equation.

Step 2: Substitute the ordered pair (0, 1) into second equation

The second equation is \(x - 3y = -8\). Substitute \(x = 0\) and \(y = 1\) into the equation: \(0 - 3(1) = -8\). Simplifying, we get \(-3 = -8\), which is false. So (0,1) does not satisfy the second equation.

Step 3: Conclusion for ordered pair (0, 1)

Since (0,1) satisfies the first equation but not the second, (0,1) is not a solution to the system of equations.

Step 4: Substitute the ordered pair \(\left( \frac{1}{6}, \frac{1}{3} \right)\) into first equation

Substitute \(x = \frac{1}{6}\) and \(y = \frac{1}{3}\) into the first equation: \(4 \left( \frac{1}{6} \right) = 1 - \frac{1}{3}\). Simplifying: \(\frac{4}{6} = \frac{3}{3} - \frac{1}{3}\) gives \(\frac{2}{3} = \frac{2}{3}\), which is true.

Step 5: Substitute the ordered pair \(\left( \frac{1}{6}, \frac{1}{3} \right)\) into second equation

Substitute \(x = \frac{1}{6}\) and \(y = \frac{1}{3}\) into the second equation: \(\frac{1}{6} - 3 \left( \frac{1}{3} \right) = -8\). Simplifying: \(\frac{1}{6} - 1 = -8\) gives \(-\frac{5}{6} = -8\), which is false.

Step 6: Conclusion for ordered pair \(\left( \frac{1}{6}, \frac{1}{3} \right)\)

Since \(\left( \frac{1}{6}, \frac{1}{3} \right)\) satisfies the first equation but not the second, \(\left( \frac{1}{6}, \frac{1}{3} \right)\) is not a solution to the system of equations.

Key Concepts

Ordered Pairs

Ordered pairs are a way to represent a specific point on a coordinate system, denoted by \((x, y)\). They tell you two things; where to go horizontally (left or right, determined by \(x\)) and where to go vertically (up or down, determined by \(y\)). In the context of solving systems of linear equations, you are often tasked with testing if a particular ordered pair is a solution, meaning it satisfies all conditions or equations in the system.

• For example, the ordered pair \((0, 1)\) means you start at the origin \((0, 0)\), move 0 units right, and then move 1 unit up on the coordinate plane.
• Another example is \(\left(\frac{1}{6}, \frac{1}{3}\right)\), where you move slightly right (by \(\frac{1}{6}\)) and slightly up (by \(\frac{1}{3}\)) from the origin.

Using ordered pairs to determine solutions involves substituting \(x\) and \(y\) into each equation of the system to verify if they hold true. This will reveal if the ordered pair actually belongs to the solution set of the given system of equations.

Solution Verification

To verify if an ordered pair is a solution to a system of equations, you need to check each equation individually. This involves substituting the pair into both equations and checking for truthfulness.
Here's a simple rundown on how to do that:

• Take the first equation, substitute the \(x\) and \(y\) values from your ordered pair, and simplify to see if the equation is true. For instance, if using \((0, 1)\), you substitute into each equation.
• Do the exact same process with the second equation. If either results in a false statement, the ordered pair is not a solution to the system.

In our example with \((0, 1)\), the first equation becomes \(0 = 0\), true. But the second equation simplifies to \(-3 = -8\), false. So \((0, 1)\) isn't a valid solution. Similarly, when substituting \(\left( \frac{1}{6}, \frac{1}{3} \right)\), the first equation holds true, but the second doesn't, confirming it's not a solution either. Thus, full verification requires checking both equations.

Substitution Method

The substitution method is a powerful tool to solve systems of equations and check if an ordered pair fits that system. This technique involves substituting one of the variables with its expression from another equation in the system.Here’s how you can apply the substitution method:

• Start by isolating one variable in one of the equations. For example, consider \(x\) as expressed in terms of \(y\): \(x = \frac{1 - y}{4}\).
• Substitute this expression into the other equation. By doing this, you're essentially "plugging in" one equation into another, allowing you to solve for "\(y\)" or "\(x\)" independently.
• Once you solve for one variable, use its value to find the other variable by replacing it back into the first expression.

This method is especially useful when verifying solutions because it can streamline checking ordered pairs against complex equations. If the substitution leads to a true equation, the ordered pair is a solution. Looking back at our given problem, using substitution helps determine if a specific ordered pair results in valid equations or not.