Problem 24 - Algebra 1: Concepts and Skills

Question

Use linear combinations to solve the linear system. Then check your solution. \(3 b+2 c=46\) \(5 c+b=11\)

Step-by-Step Solution

Answer

The solution for the system of equations is \(b = 16\) and \(c = -1\).

Step 1: Arrange the Equations

Rearrange equation \(5c + b = 11\) to \(b = 11 -5c\). Now, we substituted for the variable \(b\) in the first equation: \(3b + 2c = 46\).

Step 2: Substitute

By substituting \(b = 11 - 5c\) into the equation \(3b + 2c= 46\) we get equation: \(3(11 - 5c) + 2c = 46\) by simplifying this equation, we get a new equation \(33 - 15c + 2c = 46\).

Step 3: Solve for \(c\)

Continue by simplifying the equation to solve for \(c\): \(33 - 13c =46\). After simplification, we get \(c = -1\).

Step 4: Solve for \(b\)

Now we use the value of \(c\) that we just found, and substitute it into the equation \(b = 11-5c\) to get \(b = 16\).

Step 5: Check the solution

Finally, we substitute these values of \(b\) and \(c\) into our original equations to confirm they satisfy both equations. Substituting \(b = 16\) and \(c = -1\) into \(3b + 2c = 46\) we get 46 = 46. Then substituting into \(5c + b = 11\) we get 11 = 11. Since both equations are satisfied, it means our solutions are correct.

Key Concepts

Linear Equations

Linear equations are the foundation of algebra, forming the basis for solving many types of mathematical problems. They are equations of the first degree, which means the highest power of the variable is one. The general form of a linear equation with two variables, such as in our exercise, is expressed as:

• \(ax + by = c\), where \(a\), \(b\), and \(c\) are constants.

Solving linear equations involves finding the values of the variables that make the equation true.
In the exercise example, the equations are linear because each term is either a constant or a constant times a single variable.
An important characteristic of linear equations is that they graph as straight lines. When linear equations contain more than one variable, they describe relationships between these variables.

System of Equations

A system of equations consists of two or more equations that share the same set of unknowns. In solving a system, the goal is to find values for the variables that satisfy all equations simultaneously.
Our exercise presents a system of linear equations:

• \(3b + 2c = 46\)
• \(5c + b = 11\)

There are several methods to solve systems of equations:

• Substitution - Solve one equation for one variable, then substitute into the other equations.
• Elimination (or linear combination) - Combine equations to eliminate one variable, making the system easier to solve.
• Graphical method - Graph each equation and find intersections.

In the solution, we used substitution and linear combination, replacing one variable to simplify solving for the others.
For linear equations, solutions are usually where the lines intersect on a graph.

Algebraic Solutions

Algebraic solutions to systems of equations involve using algebraic techniques to isolate variables and find their values.
These methods include:

• Simplifying the equations by combining like terms.
• Using substitution to replace a variable.
• Solving simple arithmetic operations.

In the given problem, we used algebraic manipulation to rearrange and solve the equations:

• Firstly, by expressing \(b\) in terms of \(c\) in the second equation.
• Substituting this expression in the first equation to get a single equation with one unknown.
• Finally solving for \(c\) and then using this value to find \(b\) with substitution.

By carefully performing these steps, the correct solution \(c = -1\) and \(b = 16\) was found.
This type of systematic approach is crucial in ensuring accuracy in algebraic solutions, and verifying the results helps in confirming they meet the condition of all initial equations.