/*! 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 57 Exercises \(55-57\) will help yo... [FREE SOLUTION] | 91Ó°ÊÓ

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Exercises \(55-57\) will help you prepare for the material covered in the next section. Solve: $$ \left\\{\begin{array}{r} A+B=3 \\ 2 A-2 B+C=17 \\ 4 A-2 C=14 \end{array}\right. $$

Short Answer

Expert verified
The solution to the system of equations is approximately \(A = 4.33\), \(B = -1.33\), and \(C = 4.33\).

Step by step solution

01

Simplify the equations if possible

The given equations are: \(A+B=3\), \(2A-2B+C=17\), and \(4A-2C=14\). The second equation can be simplified by dividing it by 2, resulting in : \(A - B + 0.5C = 8.5\). This simplifying step makes the further steps a bit easier.
02

Solve for A in terms of B from equation 1

Re-arrange the first equation to find \(A\) in terms of \(B\), i.e., \(A = 3 - B\).
03

Substitute A into Equation 2 & 3

Now that \(A\) has been expressed in terms of \(B\) from equation 1, it can be substituted into equations 2 and 3. Equation 2 becomes: \((3-B) - B + 0.5C = 8.5\). Simplifying this gives: \(2B - 0.5C = -5.5\). Equation 3 becomes: \(4(3-B) - 2C = 14 \). Simplifying this gives: \(4B +2C = 2\). We've now reduced the problem to a system of two equations with two variables (B and C).
04

Solve for B & C

Now, the system with B and C can be solved. Multiply the second equation by 0.5 to make addition/elimination possible. We now have: \(2B - 0.5C = -5.5\) and \(2B + C = 1\). Subtracting the first equation from the second gives: \(1.5C = 6.5\), so \( C = 6.5 / 1.5 \approx 4.33 \). Substituting \(C\) back into equation \(2B - 0.5 * 4.33 = -5.5\), we find \( B \approx -1.33 \).
05

Solve for A

Finally, substituting \(B\) back into the first equation \(A = 3 - (-1.33)\), we find \(A \approx 4.33 \).

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Key Concepts

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

Substitution Method
Solving a system of linear equations can be approached with different methods, one of which is the substitution method. This technique involves expressing one variable in terms of another from one equation and then substituting this expression into the other equations. This method helps in reducing the number of variables and equations step by step until you can solve for the unknowns.

Let's clarify this method with an example from the exercise given. Initially, equation 1 is manipulated to express variable A in terms of B, as shown in the solution. By substituting this expression into the other equations, we effectively eliminate A from those equations. The result is a simpler set of equations that can be solved sequentially to find the values of B and C, and eventually A. It is essential to execute the substitution correctly and maintain the balance of the equation throughout the process.
Simplifying Equations
Simplifying equations is a fundamental step when solving systems of linear equations. It involves breaking down complex expressions into simpler, more manageable components. This usually includes combining like terms, removing parentheses, and dividing or multiplying both sides of the equation by a common factor to reduce the coefficients.

In our exercise, the second equation is simplified by dividing all terms by 2. This step makes the equation easier to manipulate in later stages. Simplification helps to see the structure of the equations more clearly and can often reveal a direct path to the solution. When simplifying, it's crucial to do the same operation to both sides of the equation to maintain equality. Moreover, ensuring the simplified form is completely reduced primes the equation for efficient substitution or elimination.
Algebraic Expressions
Algebraic expressions are combinations of numbers, variables, and arithmetic operations. Understanding how to manipulate these expressions is instrumental in solving algebra problems, including systems of equations. In our example, algebraic expressions are simplified and rearranged to isolate variables, enabling us to utilize the substitution method effectively.

For instance, when we express A as 3 - B, we are crafting a new algebraic expression that can be used to replace A in other equations. Recognizing how terms in an expression relate and interact with one another is crucial, such as knowing that a term with A can be substituted with its equivalent expression from another equation. Mastering the manipulation of algebraic expressions not only aids in solving equations, but also in understanding the relationships between mathematical quantities.

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Most popular questions from this chapter

What does a dashed line mean in the graph of an inequality?

When is it easier to use the addition method rather than the substitution method to solve a system of equations?

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