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Chapter 5: Problem 33

Solve the given systems of equations by either method of this section. $$\begin{aligned} &44 A=1-15 B\\\ &5 B=22+7 A \end{aligned}$$

Short Answer

Expert verified
The solution to the system is \( A = -1 \) and \( B = 3 \).

Step by step solution

01

Express A in terms of B

Starting with the first equation, \( 44A = 1 - 15B \), we can solve for \( A \) by isolating it on one side of the equation. Thus, we get \( A = \frac{1 - 15B}{44} \).
02

Substitute A in the second equation

Next, substitute \( A = \frac{1 - 15B}{44} \) into the second equation \( 5B = 22 + 7A \). This gives us:\[5B = 22 + 7\left(\frac{1 - 15B}{44}\right)\]
03

Simplify the substituted equation

Multiply through by 44 to eliminate the fraction in the substituted equation:\[44 \times 5B = 44 \times 22 + 7 \times (1 - 15B)\]This becomes:\[220B = 968 + 7 - 105B\]Simplify it to: \[220B + 105B = 975\]Which results in:\[325B = 975\]
04

Solve for B

Divide both sides by 325 to find \( B \):\[B = \frac{975}{325}\]Simplify the fraction to get \( B = 3 \).
05

Solve for A using the value of B

Substitute \( B = 3 \) back into the expression for \( A \):\[A = \frac{1 - 15 \times 3}{44}\]Simplify the expression:\[A = \frac{1 - 45}{44} = \frac{-44}{44} = -1\]
06

Verify the solutions

Substitute \( A = -1 \) and \( B = 3 \) back into the original equations to verify:1. \( 44(-1) = 1 - 15(3) \rightarrow -44 = 1 - 45 \rightarrow -44 = -44 \) - Correct.2. \( 5(3) = 22 + 7(-1) \rightarrow 15 = 22 - 7 \rightarrow 15 = 15 \) - Correct.

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Key Concepts

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

Linear Equations
Linear equations are equations that involve variables raised to the power of one. They can directly be represented in the form of a straight line when plotted on a graph. In our system of equations: - The equation \( 44A = 1 - 15B \) is linear since both the terms \( A \) and \( B \) have a degree of one. - Similarly, \( 5B = 22 + 7A \) is another linear equation.
Each equation illustrates a relationship between two variables. When you have a set of two or more linear equations, it forms a system of equations. The goal in solving these systems is to find the values of the variables that satisfy all given equations simultaneously.
Substitution Method
The substitution method is one of the ways to solve a system of equations. It involves solving one of the equations for one variable and then substituting this expression into the other equation. Here's how it works in our problem:
  • First, we solved the equation \( 44A = 1 - 15B \) for \( A \), giving us an expression for \( A \) in terms of \( B \): \( A = \frac{1 - 15B}{44} \).
  • Next, substitute this expression into the other equation \( 5B = 22 + 7A \). This step replaces \( A \) with its equivalent expression involving \( B \), allowing us to solve for \( B \).
By taking these steps, you're converting the system of equations into a single equation in one variable, making it easier to solve for that variable first.
Mathematical Verification
Once you've calculated the values of the variables from a system of equations, mathematical verification ensures they are correct. This step is crucial because it's easy to make errors in calculations or substitutions.
In the provided solution, verification was done by plugging the found values \( A = -1 \) and \( B = 3 \) back into the original equations:
  • For the first equation: Substitute to confirm \( 44(-1) = 1 - 15(3) \), which checks out as both sides equal \(-44\).
  • For the second equation: Substitute to confirm \( 5(3) = 22 + 7(-1) \), which checks out as both sides equal \(15\).
By verifying, any mistakes in earlier steps can be caught and corrected.
Algebraic Solutions
Algebraic solutions involve using algebraic operations and manipulations to find the values of variables that satisfy all equations in a given system. In our example, several algebraic steps were employed:
  • Rearranging equations to isolate variables, as seen when \( A \) was expressed in terms of \( B \).
  • Substituting expressions into other equations to reduce the number of variables, showcased when the expression for \( A \) was substituted into the second equation.
  • Simplifying complex algebraic expressions, by multiplying through to eliminate fractions, and then rearranging terms to gather like terms.
  • Finally, solving for each variable through simple operations like division once the equation is reduced to a single variable form.
Each step of this process requires careful handling of algebraic operations to ensure accuracy and to ultimately solve the system of equations.

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Most popular questions from this chapter

Solve the given systems of equations by determinants. $$\begin{aligned} &x+2 y+z=2\\\ &3 x-6 y+2 z=2\\\ &2 x-z=8 \end{aligned}$$

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