Problem 24 - Technical Mathematics with Calculus

Question

Solve for \(x\) and \(y\) in terms of the other literal quantities. $$\begin{aligned} &a x+b y=r\\\ &a x+c y=s \end{aligned}$$

Step-by-Step Solution

Answer

The solution for x is given by $$x=\frac{bs-cr}{ab-ac},$$ and the solution for y is $$y=\frac{ac r - abs}{b(ab-ac)}.$$

Step 1: Express one variable in terms of the other using the first equation

Isolate the variable y on one side in the first equation. This gives us the expression: $$y=\frac{r-ax}{b}.$$

Step 2: Substitute the expression for y into the second equation

Take the expression for y found from the first equation and substitute it into the second equation, resulting in:$$ax+c\left(\frac{r-ax}{b}\right)=s.$$

Step 3: Solve for x

Simplify the second equation to solve for x:$$ax+\frac{cr}{b}-\frac{ac}{b}x=s.$$ Combine like terms and solve for x: $$x \left(a - \frac{ac}{b} \right) = s - \frac{cr}{b}. $$Therefore: $$x= \frac{s-\frac{cr}{b}}{a-\frac{ac}{b}}.$$ Multiply numerator and denominator by b to clear the fraction: $$x=\frac{bs-c r}{ab-ac}.$$

Step 4: Substitute the value of x into y's expression

Now substitute the value of x back into the equation obtained for y in Step 1:$$y=\frac{r-a\left(\frac{bs-c r}{ab-ac}\right)}{b}.$$

Step 5: Simplify the expression for y

Simplify the above expression to solve for y: $$y=\frac{r( ab-ac)-abs+a cr}{b(ab-ac)}. $$Cancel out terms and simplify further: $$y=\frac{ac r - abs}{b(ab-ac)}.$$

Key Concepts

Literal Equation Solving

Literal equations are those that involve two or more variables or literal coefficients. They are common in algebra and are used to represent general relationships. In solving for a specific variable, it is important to isolate that variable on one side of the equation.

To achieve this, you may need to perform operations such as addition or subtraction to move terms from one side to the other, and multiplication or division to get the variable by itself. As seen in the given exercise, isolation of variable y was done in the first step. The literal equation was manipulated to express y in terms of x and other known quantities, making it a critical step to find the solution.

Substitution Method

The substitution method in solving systems of equations involves taking the expression for one variable from one equation and substituting it into the other equation. This method is particularly useful when one variable is easily isolated, as done in Step 1 of the provided exercise. Once the substitution has been made, as seen in Step 2, the resulting equation will have only one variable, enabling us to solve for that variable directly. It's important to ensure the substitution is done correctly to avoid any errors in calculation.

Algebraic Expressions

Algebraic expressions represent numerical relationships using variables and constants. They consist of terms that are variables, constants, or both, combined using arithmetic operations such as addition, subtraction, multiplication, and division. In our exercise, algebraic expressions were used to represent the relationships between the variables x, y, a, b, c, r, and s. Simplifying these expressions is an integral part of solving the system of equations.

Simplifying Equations

Simplifying equations is the process of reducing them to a simpler form. It involves combining like terms, distributing, and cancelling out factors to identify the essence of the equation. This makes solutions visually less complex and easier to understand. For instance, in our exercise, after substituting x back into y's expression, the final step involved simplifying to find the most reduced form of y. This systematic approach of simplifying ensures accuracy in solving the system of equations and provides a clearer view of the relationship between the variables.