Problem 56 - Precalculus: Mathematics for Calculus

Question

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

Step-by-Step Solution

Answer

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

Step 1: Express y in terms of x

Start by solving the second equation for one variable. The given equation is \( x + y = 1 \), so we can express \( y \) as \( y = 1 - x \).

Step 2: Substitute y in the first equation

Substitute \( y = 1 - x \) in the first equation \( ax + by = 0 \). This gives \( ax + b(1-x) = 0 \).

Step 3: Simplify the equation

Distribute \( b \) in the equation: \( ax + b - bx = 0 \), which simplifies to \( (a - b)x + b = 0 \).

Step 4: Solve for x

Rearrange the equation \((a-b)x + b = 0\) to solve for \( x \). Subtract \( b \) from both sides to get \( (a-b)x = -b \). Then divide both sides by \( a-b \) to find \( x = \frac{-b}{a-b} \).

Step 5: Solve for y using x

Substitute \( x = \frac{-b}{a-b} \) back into \( y = 1 - x \). This gives \( y = 1 - \frac{-b}{a-b} = 1 + \frac{b}{a-b} \).

Step 6: Simplify the expression for y

Find a common denominator to combine the terms: \( y = \frac{a-b}{a-b} + \frac{b}{a-b} = \frac{a}{a-b} \).

Key Concepts

Linear Equations

Linear equations are fundamental to understanding algebra and mathematics in general. These equations are typically presented in the form of ax + b = c, where 'a', 'b', and 'c' are constants, and 'x' is the variable. They're called linear because they can be graphically represented as straight lines on a coordinate plane.

In the exercise we are focusing on (\( ax + by = 0 \) and \( x + y = 1 \)), each equation represents a line. The solution to these equations is the intersecting point of these lines on the graph. Linear equations are simple yet form the building blocks for more complex mathematical problems.

Solving these equations requires finding values of the variables that satisfy all equations involved. Although linear equations can appear in various forms, such as slope-intercept form or standard form, the method of solving remains consistent, relying primarily on algebraic manipulation.

System of Equations

A system of equations consists of two or more equations with the same variables. The goal is to find a solution that satisfies all equations simultaneously. Systems of equations can be classified based on the number of solutions they have:

• Consistent System - Has at least one solution.
• Inconsistent System - Has no solutions.
• Dependent System - Has infinitely many solutions.

For our exercise, the system consists of two linear equations:\( ax + by = 0 \) and \( x + y = 1 \). These form a consistent and independent system because they intersect at exactly one point, given that \( a eq b \). Solving systems of equations involves finding the intersection point in their graphical representation.

Different methods, such as graphing, substitution, and elimination, can help find the solution set. In the current problem, the substitution method provides an efficient path to the solution.

Substitution Method

The substitution method is a powerful technique for solving systems of equations, especially when one equation is easily solvable for one variable. This method involves three main steps:

• Solving one of the equations for one variable.
• Substituting the expression obtained into the other equation.
• Solving the resulting one-variable equation.

In the given problem, we first solve \( x + y = 1 \) for \( y \), obtaining \( y = 1 - x \).
We substitute this expression into the first equation \( ax + by = 0 \), resulting in an equation with a single variable, \( x \).
This simplifies to \((a-b)x + b = 0\). Solving for \( x \) results in \( x = \frac{-b}{a-b} \).

Finally, to find \( y \), we substitute this value of \( x \) into \( y = 1-x \), which simplifies to \( y = \frac{a}{a-b} \). The substitution method is invaluable as it breaks down complex systems into more manageable steps, facilitating an easier and more straightforward solution.