Problem 40 - Precalculus

Question

Solve each system of linear equations by elimination. $$\begin{array}{rr} -0.04 x+1.12 y= & 9.815 \\ 2.79 x+1.19 y= & -0.165 \end{array}$$

Step-by-Step Solution

Answer

The solution to the system is \(x = -3.74\) and \(y = 8.63\).

Step 1: Arrange Equations

The given system of equations is: Equation 1: \(-0.04x + 1.12y = 9.815\)Equation 2: \(2.79x + 1.19y = -0.165\). Before proceeding to elimination, ensure both equations are aligned: coefficients of \(x\) and \(y\) should be in the same position for both equations.

Step 2: Eliminate One Variable

To eliminate one variable, make the coefficients of \(y\) equal in both equations. Multiply Equation 1 by 1.19 and Equation 2 by 1.12 to attempt to eliminate \(y\):Equation 1 becomes:\(-0.0476x + 1.328y = 11.67985\)Equation 2 becomes:\(3.1248x + 1.328y = -0.1848\) Now subtract the new Equation 1 from Equation 2 to eliminate \(y\).

Step 3: Solve for x

Subtract the modified Equation 1 from Equation 2:\(3.1248x + 1.328y - (-0.0476x + 1.328y) = -0.1848 - 11.67985\)This simplifies to:\(3.1724x = -11.86465\)Solve for \(x\) by dividing both sides by 3.1724.\(x = \frac{-11.86465}{3.1724} = -3.74\).

Step 4: Substitute to Solve for y

Substitute \(x = -3.74\) back into original Equation 1:\(-0.04(-3.74) + 1.12y = 9.815\)Calculate:\(0.1496 + 1.12y = 9.815\)Subtract 0.1496 from both sides:\(1.12y = 9.6654\)Solve for \(y\) by dividing both sides by 1.12:\(y = \frac{9.6654}{1.12} = 8.63\).

Step 5: Verify Solution

Substitute \(x = -3.74\) and \(y = 8.63\) back into Equation 2 to verify:\(2.79(-3.74) + 1.19(8.63) = -0.165\)Calculate:\(-10.4346 + 10.2697 = -0.1649\)Since this is approximately equal to \(-0.165\), the solution \(x = -3.74\) and \(y = 8.63\) satisfies both equations.

Key Concepts

Elimination Method

The elimination method is a systematic process used to solve systems of linear equations by removing one variable, allowing for the straightforward solution of the other variable. This approach is particularly useful when faced with equations that are easier to manipulate algebraically.

### How It Works

• Begin by aligning the system of equations, ensuring all variables and constants are positioned correctly.
• Decide which variable to eliminate first. It's often convenient to eliminate the variable that can lead to simpler calculations.
• To eliminate this variable, multiply each equation by a suitable number so that the coefficients of the chosen variable, in both equations, become equal.
• Next, subtract one equation from the other (or add them, if necessary) to eliminate the chosen variable. This operation should leave you with a single equation in one variable.

The process uses multiplication and subtraction to reduce and simplify equations strategically, leading to a single-variable equation that can be easily solved. Once you have one variable's value, substituting back into one of the original equations gives the solution for the other variable.

The power of this method lies in its ability to systematically and efficiently boil down complex systems of equations into something much more manageable.

Substitution

Substitution is another technique used to solve systems of linear equations, particularly when one of the equations is easy to solve for one of the variables. This method provides an alternative to elimination and often works better for equations that can be quickly rearranged to isolate a variable.

### Steps of Substitution

• First, choose one of the equations and solve it for one of the variables. This equation should be the simplest one to rearrange.
• Substitute this expression into the other equation. This step essentially reduces the system of equations into a single equation with only one variable.
• Solve the resulting equation to find the value of the isolated variable.
• Finally, substitute this value back into the expression obtained in the first step to find the value of the other variable.

Substitution is incredibly useful in its ability to methodically alternate focus onto the different variables. It highlights how functions and relationships between variables intertwine, ultimately resulting in clear solutions.

Linear Equations

Linear equations form the backbone of algebra and are algebraic expressions that represent a straight line when graphed. A linear equation in two variables can typically be written in the form \(ax + by = c\), where \(x\) and \(y\) are variables, and \(a\), \(b\), and \(c\) are constants.

### Key Characteristics of Linear Equations

• Each linear equation graphs as a straight line, demonstrating a consistent, proportional relationship between the variables.
• Changes in either variable result in predictable and proportionate changes in the other variable.
• They are characterized by constant coefficients and do not contain any exponents or powers higher than one.

Solving a system of linear equations involves finding a solution that satisfies all equations in the system simultaneously. Whether using elimination or substitution, the goal is to find the intersection point of the lines represented by the equations, which corresponds to the values of the variables that satisfy each equation. Mastery of linear equations is pivotal to progressing in algebra, as they appear repeatedly in more complex mathematical concepts and real-world applications.