Problem 44
Question
Solve each equation for the indicated variable. \(\frac{x-a}{b}=\frac{y-a}{c}\) for \(y\)
Step-by-Step Solution
Verified Answer
The solution for \(y\) is \(y = \frac{cx + ba - ca}{b}\).
1Step 1: Clear the Fractions
To eliminate the fractions from the equation, cross-multiply. This gives us \[ c(x - a) = b(y - a) \]
2Step 2: Distribute and Expand
Distribute the coefficients \(c\) and \(b\) across the terms in the parentheses:\[ cx - ca = by - ba \]
3Step 3: Isolate Terms with y
To isolate the term containing \(y\), move \(-ba\) to the left side of the equation by adding \(ba\) to both sides:\[ cx - ca + ba = by \]
4Step 4: Solve for y
Divide both sides of the equation by \(b\) to solve for \(y\):\[ y = \frac{cx + ba - ca}{b} \]
Key Concepts
Cross-MultiplicationClearing FractionsIsolating VariablesLinear Equations
Cross-Multiplication
Cross-multiplication is a technique used to simplify equations involving fractions, particularly when there are two ratios equated against each other. In our exercise, we have the equation \(\frac{x-a}{b}=\frac{y-a}{c}\).
- Think of each side of the equation as a fraction. The goal here is to eliminate these fractions by multiplying across the equals sign.
- Multiply the numerator of one fraction by the denominator of the other, and vice versa, to form a new equation without fractions.
Clearing Fractions
Clearing fractions from an equation means manipulating the equation in such a way that no fractions remain. It allows for easier manipulation of the algebraic equation.
- Cross-multiplication, as described earlier, is one of the primary methods used to clear fractions.
- In our example, we eliminated the fractions in \(\frac{x-a}{b}=\frac{y-a}{c}\) by using cross-multiplication to generate \(c(x-a) = b(y-a)\).
Isolating Variables
Isolating a variable means rearranging an equation in order to represent one specific variable on one side of the equation, typically leaving that variable by itself.
- After clearing the fractions from our equation, we have \(cx - ca = by - ba\).
- Our goal now is to isolate \(y\). This involves moving all terms not containing \(y\) to the other side of the equation: \(cx - ca + ba = by\).
- Once \(y\) is isolated, you can solve for it by performing additional operations such as dividing or factoring.
Linear Equations
Linear equations are equations of the first degree, meaning they involve only terms that are constant or linear in nature; these terms do not have any exponents beyond one.
- In our exercise, the equation \(\frac{x-a}{b}=\frac{y-a}{c}\) is a linear equation once we apply cross-multiplication and simplify it to \(cx - ca = by - ba\).
- Linear equations are fundamental because they form a straight line when graphed, representing the simplest form of relationships in algebra.
- Solving a linear equation often involves isolating the variable of interest, a process we have applied to solve for \(y\) by following operations like addition, subtraction, and division.
Other exercises in this chapter
Problem 44
Solve each compound inequality and graph the solution sets. Express the solution sets in interval notation. \(5 x-22\)
View solution Problem 44
Solve each inequality and express the solution set using interval notation. 5-4 x>-2
View solution Problem 44
Solve each equation. A total of \(\$ 4000\) was invested, part of it at \(8 \%\) interest and the remainder at \(9 \%\). If the total yearly interest amounted t
View solution Problem 44
Use an algebraic approach to solve each problem. Suppose that the width of a rectangle is 3 centimeters less than two-thirds of its length. The perimeter of the
View solution