Problem 41 - Algebra 1

Question

Solve the linear system. $$ \begin{aligned} &0.5 x+2.2 y=9\\\ &6 x+0.4 y=-22 \end{aligned} $$

Step-by-Step Solution

Answer

The solution to the system of equations is \(x = -4\) and \(y = 5\).

Step 1: Multiply the equations

First, we can simplify the equations by getting rid of decimal points. Multiply the first equation by 10 and the second one by 5: \[5x + 22y = 90\] and \[30x + 2y = -110\]

Step 2: Isolate one variable

Now, we want to isolate one variable so we can substitute it into the other equation. For example, isolate \(y\) in the second equation: \[2y = -30x - 110\] which simplifies to \[y = -15x - 55\]

Step 3: Substitute isolated variable

Substitute \(y = -15x - 55\) into the first equation: \[5x + 22(-15x - 55) = 90\]

Step 4: Solve for x

Solve for \(x\) from the equation: \[5x - 330x - 1210 = 90\] which simplifies to \[-325x = 1300\] so hence, \(x = -4\)

Step 5: Find y

Substitute \(x=-4\) into one of the original equations, for instance the first one: \[0.5(-4) + 2.2y = 9\] which simplifies to \[-2 + 2.2y = 9\] so hence, \(y = 5\)

Key Concepts

Linear Equations

Linear equations are foundational to algebra and represent relationships where variables change at a constant rate. As the term 'linear' suggests, the graph of a linear equation forms a straight line. A linear equation in two variables, such as x and y, can be written in the form ax + by = c, where a, b, and c are constants. The key to solving a linear system is finding the values of the unknown variables that satisfy all the equations at once.

For example, in the given exercise, we have two linear equations with two variables, x and y. The goal is to determine the specific values of x and y that make both equations true simultaneously. This is central to understanding algebra, as it demonstrates the concept of variables and how they can interact within mathematical models.

Substitution Method

The substitution method is a powerful algebraic technique used to solve systems of linear equations. It involves rearranging one equation to express one variable in terms of another and then substituting this expression into the second equation. This results in an equation with only one variable, making it easier to solve.

In this particular exercise, we began by isolating y in one of the equations, resulting in y = -15x - 55. By replacing y with this expression in the other equation, we merged the two equations into a single equation with just one variable, x. This greatly simplifies the process of finding the solution, as it reduces the problem to a simple one-variable equation. Once x is found, we can substitute its value back into the new expression for y to find the value of the second variable. This method not only simplifies the process but also ensures accuracy in the solution.

Algebraic Manipulation

Algebraic manipulation is the process of restructuring equations to make them easier to understand or solve. It involves operations such as adding, subtracting, multiplying, or dividing both sides of the equation by the same number, as well as distributing multiplication over addition and using inverse operations to isolate variables.

In our exercise, algebraic manipulation began with multiplying the original equations to eliminate decimal points, a common first step in simplifying equations. Then, to isolate variables effectively, we performed operations that transformed the equation into a more solvable form. For instance, distributing the multiplication through the parentheses in the equation 5x + 22(-15x - 55) = 90 and then combining like terms to arrive at -325x = 1300. By these manipulations, we made the path to find x straightforward. When using algebraic manipulation, it's essential to perform valid operations that maintain the equality and lead to the solution in consistent, logical steps.