Problem 26 - Precalculus with Limits

Question

Solving a System by Elimination In Exercises \(13-30,\) solve the system by the method of elimination and check any solutions algebraically. $$ \left\\{\begin{array}{l}{0.05 x-0.03 y=0.21} \\ {0.07 x+0.02 y=0.16}\end{array}\right. $$

Step-by-Step Solution

Answer

The solution to the system of equations is \(x = 2.9\) and \(y = 2.17\).

Step 1: Multiply the equations to make elimination possible

In order to eliminate one of the variables, multiply each equation with suitable numbers such that the coefficients of y's or x's in both equations will cancel out when subtracted. Here, if we multiply the first equation by 0.02 and the second by 0.03, we wil get equations that have the same coefficient for y. After multiplication, the system becomes: \[\begin{cases}0.001x - 0.0006y = 0.0042 \0.0021x + 0.0006y = 0.0048\end{cases}\]

Step 2: Eliminate one variable

Add the two equations together. The y terms will cancel out:\(0.001x + 0.0021x = 0.0042 + 0.0048 0.0031x = 0.009\)

Step 3: Solve for x

Divide by the coefficient of x to find its value:\(x = 0.009 / 0.0031 = 2.9\)

Step 4: Substitute x in the original equation

Substitute \(x = 2.9\) into the first equation to solve for y:\(0.05 * 2.9 - 0.03y = 0.21 0.145 - 0.03y = 0.21 0.03y = 0.145 - 0.21 = -0.065\)

Step 5: Solve for y

Divide by the coefficient of \(y\) to find its value:\(y = -0.065 / -0.03 = 2.17\).

Key Concepts

The Elimination Method

When solving systems of equations, the elimination method is a powerful tool. This technique involves manipulating the equations in a system so that one of the variables cancels out, making it easier to solve for the remaining variables. Here's how it generally works:

• First, you aim to align the coefficients of one variable in both equations, so they are either equal or opposite. This often requires multiplying one or both of the equations by suitable constants.
• Next, the two equations are added or subtracted. This will eliminate that variable from the system, leaving you with a single equation with one unknown.
• With one less variable, you can solve this new equation using basic algebraic techniques.

The elimination method is beneficial when direct substitution is complex. By focusing on removing variables one at a time, it can simplify solving complex systems.

This method is systematically used to find solutions in many fields like economics, engineering, and physics where multiple variables need resolving.

Understanding Algebraic Solutions

An algebraic solution refers to the process of finding the values of unknown variables making the equation true. In our case, calculating a system's solution algebraically is crucial for verification.

• Algebra provides a structure to manipulate equations logically. By applying algebraic rules, you can simplify and solve equations systematically.
• Some steps typically involve rearranging equations to isolate variables, distributing constants across terms, or factoring expressions.

The importance of checking solutions algebraically cannot be overstated. After you have reached a conclusion, substitute the values back into the original equations to confirm correctness. If both equations are satisfied, then your solution is sound.

By consistently checking your work algebraically, you safeguard against mistakes and firmly grasp the dynamics between variables.

Variable Substitution for Solutions

Variable substitution is an essential part of solving systems of equations. Once a variable's value is determined, substituting back to find other unknowns is straightforward.

• This process involves taking the value found from one equation and substituting it into the original or another transformed equation to find remaining variables.
• It's crucial for variable substitution to be meticulously performed to avoid errors, especially sign or arithmetic mistakes.

By carefully substituting the value of one variable into another equation, you proceed to uncover the subsequent variable's value. This method is part of the broader algebraic strategy in tackling systems.

Additionally, substitution enriches your understanding of how each variable interplays to satisfy the system's constraints.