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Ch. 5 - Systems and Matrices
Lial - College Algebra 13th Edition
Lial13th EditionCollege AlgebraISBN: 9780136881063Not the one you use?Change textbook
Chapter 6, Problem 27

Solve each system by elimination. In systems with fractions, first clear denominators.
x/2+ y/3 = 4
3x/2+3y/2 = 15

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1
Identify the system of equations: \(\frac{x}{2} + \frac{y}{3} = 4\) and \(\frac{3x}{2} + \frac{3y}{2} = 15\).
Clear the denominators in each equation by multiplying both sides by the least common denominator (LCD). For the first equation, the LCD of 2 and 3 is 6, so multiply the entire equation by 6: \(6 \times \left( \frac{x}{2} + \frac{y}{3} \right) = 6 \times 4\).
Simplify the first equation after clearing denominators: \(6 \times \frac{x}{2} = 3x\) and \(6 \times \frac{y}{3} = 2y\), so the equation becomes \$3x + 2y = 24$.
For the second equation, the denominators are both 2, so multiply the entire equation by 2: \(2 \times \left( \frac{3x}{2} + \frac{3y}{2} \right) = 2 \times 15\).
Simplify the second equation after clearing denominators: \(2 \times \frac{3x}{2} = 3x\) and \(2 \times \frac{3y}{2} = 3y\), so the equation becomes \$3x + 3y = 30\(. Now use the elimination method on the system \)3x + 2y = 24\( and \)3x + 3y = 30$.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Clearing Denominators

Clearing denominators involves multiplying both sides of an equation by the least common denominator (LCD) to eliminate fractions. This simplifies the system into equations with integer coefficients, making them easier to manipulate and solve.
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System of Linear Equations

A system of linear equations consists of two or more linear equations with the same variables. The goal is to find values for the variables that satisfy all equations simultaneously, often by methods such as substitution, elimination, or graphing.
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Elimination Method

The elimination method solves systems by adding or subtracting equations to eliminate one variable. This reduces the system to a single-variable equation, which can be solved easily, and then substituted back to find the other variable.
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