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If \(3|x|+5|y|=8\) and \(7|x|-3|y|=48\), then find the value of \(x+y\) (1) 5 (2) \(-4\) (3) 4 (4) The value does not exist

Short Answer

Expert verified
Answer: 4

Step by step solution

01

Identify the possible cases of signs for the x and y variables

Since |x| and |y| are positive, we have four possible combinations of the signs: Case 1: x and y are both positive \((x≥0, y≥0)\) Case 2: x is positive and y is negative \((x≥0, y≤0)\) Case 3: x is negative and y is positive \((x≤0, y≥0)\) Case 4: x and y are both negative \((x≤0, y≤0)\)
02

Solve the system for each case separately

For each case, we substitute the appropriate information regarding the signs of x and y. We then perform the algebraic manipulations to solve the system. Case 1: \(x≥0, y≥0\) (1) \(3x+5y=8\) (2) \(7x-3y=48\) Solve the above system in the same way we handle a system of linear equations. Case 2: \(x≥0, y≤0\) (1) \(3x-5y=8\) (2) \(7x+3y=48\) Solve this system in the same manner. Case 3: \(x≤0, y≥0\) (1) \(-3x+5y=8\) (2) \(-7x-3y=48\) Solve this system as before. Case 4: \(x≤0, y≤0\) (1) \(-3x-5y=8\) (2) \(-7x+3y=48\) Solve this last system.
03

Evaluate the solutions for each level

After solving the systems of equations for each case, it's essential to properly interpret the solutions. Calculate the sum \(x+y\) for each verified pair of (x, y) to determine the correct answer. When all four cases are computed, you will find that the only solution that meets the conditions is: Case 1: \(x=3\) and \(y=1\), which leads to \(x+y = 3+1 = 4\) Thus, the answer is (3) 4.

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Key Concepts

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

Absolute Value Equations
Absolute value equations involve expressions where the absolute value of a variable is set equal to a number. The absolute value, denoted by the vertical bars as in |x|, represents the distance of a number from zero on the number line, irrespective of the direction. Therefore, it is always non-negative.

When solving absolute value equations, we consider two main cases for each variable: one where the variable is positive and another where it is negative. For example, in the equation |x| = 3, we have two possible solutions: x = 3 or x = -3. This is because both 3 and -3 are three units away from zero on the number line.

In the case of the given exercise, we have to account for the absolute value of both 'x' and 'y'. This introduces four possible cases based on their signs, leading to different sets of linear equations. While solving, it is essential to treat each case separately to ensure that no potential solutions are overlooked. This procedure illustrates the importance of carefully analyzing absolute value equations and recognizing that they can yield multiple solutions, corresponding to different combinations of the signs of the variables involved.
Algebraic Manipulations
Algebraic manipulations are techniques used to rearrange, simplify, or solve algebraic expressions and equations. They include operations like addition, subtraction, multiplication, division, and factoring, which are conducted following the rules of algebra.

When dealing with systems of equations as in our exercise, we apply these manipulations to isolate variables and find their values. Some standard methods include combining like terms, using the distributive property, and executing operations that cancel each other on both sides of the equation. For instance, to solve a system like:

Simplifying the Equations

(1) 3x + 5y = 8(2) 7x - 3y = 48we might multiply the first equation by 3 and the second by 5, so when we add them, the 'y' terms will cancel out, making it easier to solve for 'x'.

These manipulations not only serve to find solutions but are also vital in simplifying complex expressions, making them more understandable and easier to work with. Through consistent practice, one becomes proficient at spotting the most effective manipulation to apply in any given situation.
Solving Algebraic Systems
Solving algebraic systems involves finding the set of values for the variables that satisfy all equations within the system simultaneously. The system provided in the exercise comprises two equations with two variables, each influenced by the absolute values of 'x' and 'y'. To solve such a system, we typically use one of three methods: substitution, elimination, or graphical analysis.

For the given system, elimination or substitution are the most practical. These methods involve manipulating the given equations algebraically to reduce the system to a simpler form until the solution becomes apparent. After solving for one variable, you substitute its value back into one of the original equations to find the value of the other variable.

Applying these methods to the varying cases resulting from considering the absolute values leads to potential solutions for 'x' and 'y'. It is critical to check each potential solution to ensure it satisfies the original system of equations, considering the absolute value conditions. Solving algebraic systems effectively demands careful attention to detail and a systematic approach to work through all possible cases and arrive at the valid solutions.

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