Problem 37 - Elementary Algebra Within Reach

Question

In Exercises 35-38, solve the system by the method of elimination. $$ \left\\{\begin{aligned} 3(x+5)-7 &=2(3-2 y) \\ 2 x+1 &=4(y+2) \end{aligned}\right. $$

Step-by-Step Solution

Answer

The solution to the system of equations is \(x = 1\) and \(y = 5/2\).

Step 1 - Simplify the Equations

Begin by expanding and simplifying both equations. For the first equation, distribute the 3 to both \(x\) and 5, and distribute the 2 to both 3 and -2y. The first equation becomes: 3x + 15 - 7 = 6 - 4y, which simplifies further to 3x - 4y = -8. For the second equation, distribute the 4 to both \(y\) and 2. The second equation becomes: 2x - 4y = -7.

Step 2 - Use the Method of Elimination

Subtract the second equation from the first to eliminate \(x\). Subtracting gives: (3x - 4y) - (2x - 4y) = -8 - (-7) This simplifies to:x = 1.

Step 3 - Solve for the other variable

Substitute \(x = 1\) into the second equation from Step 1: 2(1) - 4y = -7 Simplify to find \(y = 5/2.\)

Key Concepts

Solving Systems of Equations

Solving systems of equations is a fundamental concept in algebra where we find values for variables that satisfy all equations in the system simultaneously. These systems can consist of two or more equations involving the same set of variables. A system of linear equations can graphically be represented as lines on a coordinate plane. When we solve these equations, we're essentially looking for the point where these lines intersect, known as the solution to the system.

• If the lines intersect at a single point, there is a unique solution.
• If the lines are parallel and do not intersect, there are no solutions.
• If the lines are the same (coincident), there are infinitely many solutions.

The elimination method is a popular technique used to find solutions. It involves adding or subtracting equations to eliminate one variable, simplifying the problem to a single equation with one unknown. This is particularly useful when the system contains two equations with two variables, as shown in the given problem. By eliminating a variable, you can solve for the remaining variable first, then back-substitute to find the value of the eliminated variable.

Algebraic Simplification

Algebraic simplification is an essential step in solving systems of equations, as it involves transforming equations into simpler forms. This process makes it easier to solve for variables. Simplification often includes distributing, combining like terms, and rearranging terms to isolate variables or simplify expressions.In the original exercise, both equations are simplified by distribution: for example, distributing 3 in the first equation results in the terms \(3x + 15\). Similarly, distributing 2 across \(3-2y\) results in \(6 - 4y\). Ensuring that each equation is simplified correctly prepares you to use methods like elimination or substitution effectively. Simplification can also be about combining like terms. In our example, terms involving the same variable are combined, leading to streamlined equations and straightforward solutions. This type of simplification not only reduces the risk of computational errors but also makes the application of further algebraic methods more manageable.

Linear Equations

Linear equations are equations of the first degree, meaning that the variables in a linear equation are raised only to the power of one. These equations form straight lines when graphed on a Cartesian coordinate system.A general form of a linear equation in two variables is \(ax + by = c\), where \(a\), \(b\), and \(c\) are constants. In the system provided, both simplified equations, \(3x - 4y = -8\) and \(2x - 4y = -7\), are linear equations. Each represents a line on the graph, and their intersection point provides the solution to the system.Linear equations are foundational in algebra because they are simple and they bridge to more complex polynomial equations. Understanding how to work with them – including finding their slopes, intercepts, and intersections – is crucial not only in mathematics but also in real-world applications where relationships are represented linearly. They offer a straightforward way for evaluating situations with consistent rates of change.