/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 25 Solve the system by the method o... [FREE SOLUTION] | 91Ó°ÊÓ

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Solve the system by the method of elimination and check any solutions algebraically. $$\left\\{\begin{array}{l} 3 x+\frac{1}{4} y=1 \\ 2 x-\frac{1}{3} y=0 \end{array}\right.$$

Short Answer

Expert verified
The solution to the system of equations is \(x = 3/17\) and \(y = 32/17\)

Step by step solution

01

Multiply the Equations to Make the Coefficients of y the Same

The coefficients of y in the two equations are 1/4 and -1/3. The least common multiple of 4 and 3 is 12. Therefore, multiply the first equation by 3 and the second equation by 4 \[9x + y = 3\], \[8x - 4/3 y = 0\].
02

Eliminate y Using these New Equations

Sum the two equations to eliminate y. This gives a new equation: \[17x = 3\]
03

Solve for x

Divide both sides of the equation by 17 to get the value for x: \[x = 3/17\]
04

Substitute x into One of the Original Equations to Solve for y

Substitute \(x = 3/17\) into the first original equation, which gives \[3(3/17) + 1/4y =1\]. Solving this yields \[y = 4 - \frac{36}{17}\]. Therefore, \(y = \frac{32}{17}\)
05

Checking the Solution Algebraically

This involves substituting \(x = 3/17\) and \(y = 32/17\) back into the original equations. After simplification, both equations result in true statements, validating the solution.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Solving Systems of Equations
When we talk about solving systems of equations in algebra, we're referring to finding the values for variables that make all the equations work together simultaneously. A 'system' just means a set of two or more equations that have the same variables.

There are multiple methods to solve these systems, including graphing, substitution, and elimination. The elimination method, which is highlighted in our exercise, involves adding or subtracting equations to cancel out one of the variables, enabling you to solve for the other. This method works well for linear equations, where variables are to the first power, and is particularly handy when the equations are too complex for simple substitution.
Algebraic Methods
The term algebraic methods refers to the various techniques used to solve equations and systems of equations. Some common methods include factoring, using the quadratic formula, substitution, and elimination. Each method has its own set of rules and applicable scenarios.

For example, factoring is often the go-to method for solving quadratic equations when they can be easily decomposed into binomial factors, whereas the quadratic formula can be used when factoring is not straightforward. Understanding which method to use and when is a crucial skill developed through practice and familiarity with different types of equations.
Least Common Multiple
The least common multiple (LCM) is a foundational concept in algebra, directly applicable when summing or subtracting fractions and when applying the elimination method in systems of equations. The LCM of two or more numbers is the smallest number that all of them divide into without leaving a remainder.

For instance, in our exercise, we need to find the LCM of 4 and 3 to standardize the coefficients of the variable y. Once found, it allows us to manipulate the equations so that when added or subtracted, one of the variables is eliminated. Calculating LCM can often be done through prime factorization or by listing multiples of the numbers until we find a common one.
Substitution Method
The substitution method is another powerful algebraic technique for solving systems of equations. This method involves isolating one variable in one equation and then 'substituting' this expression into the other equation. This helps to eliminate the isolated variable and find the value of the remaining variable.

In practice, you typically start with the simpler of the two equations or the one that's easiest to isolate a variable from. After finding the value of one variable, substitute it back into one of the original equations to solve for the second variable. Substitution is particularly effective when the coefficients of the variables are already set up to make isolating one straightforward.

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