Problem 40 - Precalculus: Mathematical for Calculus

Question

Find \(x\) and \(y\) in terms of \(a\) and \(b\). $$\left\\{\begin{array}{cc} a x+b y=0 & (a \neq b) \\ x+y=1 \end{array}\right.$$

Step-by-Step Solution

Answer

\( x = \frac{b}{b-a} \) and \( y = \frac{-a}{b-a} \).

Step 1: Express x in terms of y

Let's start by manipulating the second equation \( x+y=1 \) to express \( x \) in terms of \( y \). Subtract \( y \) from both sides: \[ x = 1 - y \]

Step 2: Substitute x in the first equation

Now, we substitute the expression for \( x \) from Step 1 into the first equation \( ax + by = 0 \). This gives us:\[ a(1-y) + by = 0 \]

Step 3: Simplify the equation

Expand the equation from Step 2:\[ a - ay + by = 0 \]Combine like terms:\[ a + (b-a)y = 0 \]

Step 4: Solve for y

Rearrange the equation to solve for \( y \). Subtract \( a \) from both sides:\[ (b-a)y = -a \]Divide both sides by \( b-a \) (note: \( a eq b \)):\[ y = \frac{-a}{b-a} \]

Step 5: Solve for x

With \( y \) solved, use the expression found in Step 1 to find \( x \):\[ x = 1 - y \]Substitute the expression for \( y \):\[ x = 1 - \left(\frac{-a}{b-a}\right) \]Simplify by finding a common denominator:\[ x = \frac{b-a + a}{b-a} = \frac{b}{b-a} \]

Key Concepts

Substitution Method

The substitution method is a popular technique used when solving systems of linear equations. The main idea is to solve one equation for one variable and then substitute that expression into the other equation. This allows you to reduce the number of variables and equations, making them easier to manage. In this exercise, we start by manipulating the simpler equation, \(x + y = 1\). We isolate one variable—in this case, \(x\)—by subtracting \(y\) from both sides, giving us \(x = 1 - y\).

• The substitution method is effective because it breaks complex problems into more manageable parts.
• It requires careful manipulation to ensure the substitution is correct.
• Always check the substituted equation for simplicity before proceeding, ensuring all terms of the chosen variable are combined.

By substituting the expression for \(x\) into the first equation \(a x + b y = 0\), we can focus on solving for \(y\), thus simplifying the system significantly.

Solving for Variables

Once we've substituted one variable, the next step is to solve for the remaining variable. This involves algebraically isolating the variable on one side of the equation. In the solution, we substitute \(x = 1 - y\) into the first equation \(a(1-y) + by = 0\), a critical step to solving for \(y\). First, expand the equation and combine like terms, giving us \(a + (b - a)y = 0\).

• Move constants or distinct terms to isolate the variable.
• This process often includes combining like terms for simplification.
• In steps, remember to perform equal operations on both sides to maintain equality.

To isolate \(y\), move the constant term and divide by the coefficient of \(y\). We find \(y = \frac{-a}{b-a}\), ensuring to note the condition \(a eq b\) to avoid division by zero.

Algebraic Manipulation

Algebraic manipulation is the art and science of reshaping equations in order to solve them. It includes operations such as expansion, simplification, and rearrangement. In this task, we have several opportunities to practice these skills. When we expand \(a(1-y) + by = 0\) to \(a - ay + by = 0\) and then simplify by combining like terms, we apply algebraic techniques to clean and straightforward formulas.

• Consistency in manipulation maintains equation integrity.
• Understanding properties of operations allows efficient solving steps.
• Remember: each manipulation should bring you closer to isolating variables or simplifying expressions.

Finally, when substituting \(y = \frac{-a}{b-a}\) back to find \(x\), further algebraic manipulation consolidates our understanding and solution. After finding \(y\), ride these skills to verify the values align with both original equations, reinforcing the correctness of the solution.