Problem 18 - 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. $$ \left\\{\begin{array}{r}{2 r+4 s=5} \\ {16 r+50 s=55}\end{array}\right. $$

Step-by-Step Solution

Answer

The solution to the system of equations is \(r = 5/2\) and \(s = -5/6\).

Step 1: Scale the equations

To perform the elimination, it is often necessary to scale (multiply) the equations by appropriate numbers. In this case, multiplying the first equation by 8 yields \(16r + 32s = 40\). The system of equations now looks like this: \[ \left\{\begin{array}{r}{16 r+32 s=40} \ {16 r+50 s=55}\end{array}\right. \]

Step 2: Subtract the equations

By subtracting the second equation from the first, the variable \(r\) is eliminated, and an equation with only \(s\) is left. This yields: \(18s = -15\).

Step 3: Solve for variable s

Divide by 18 on both sides to solve for \(s\), which results in \(s = -15/18 = -5/6\).

Step 4: Substitute and solve for r

Substitute \(s = -5/6\) into the first original equation and solve for \(r\): \[2r + 4*(-5/6) = 5\] After simplifying, we obtain \(r = 5/2\).

Step 5: Check the Solution

Substitute \(r = 5/2\) and \(s = -5/6\) into the original system of equations and check if both are true. Checking the equations confirms that the solution is correct.

Key Concepts

Understanding the Elimination Method

The elimination method is a strategic approach used to solve systems of linear equations. It involves aligning and manipulating equations to eliminate one of the variables. This helps to simplify the problem into a single-variable equation, making it easier to solve.

In the elimination method, you often need to scale one or both equations by multiplying all terms by suitable numbers. This adjustment allows you to cancel out a variable when the equations are subtracted from one another.

• Start by aligning equations vertically, ensuring that similar variables and constants are in the same column.
• Decide on a variable to eliminate. Make coefficients of this variable equal and opposite in both equations by scaling.
• Subtract or add the equations, which effectively removes the chosen variable.

Once a variable is eliminated, you can solve the resulting equation for the remaining variable. After finding its value, substitute it back into one of the original equations to find the second variable. This method is efficient and often leads to neat and clear algebraic solutions.

Achieving an Algebraic Solution

To find an algebraic solution is to determine the values of variables that satisfy all the given equations in a system. Algebra involves using operations and rules to simplify and solve equations systematically.

During algebraic solutions, you will:

• Use arithmetic operations like addition, subtraction, multiplication, and division to manipulate equations.
• Transform equations to isolate variables, which means getting the variable terms on one side and constants on the other.
• Substitute found values of variables back into the equations to verify their correctness and to solve for the remaining variables.

Algebraic solutions provide exact answers and are therefore a powerful tool in mathematics. It is essential to verify your solutions by substituting them back into the original equations. This ensures that all the given conditions are satisfied.

Grasping Linear Equations

Linear equations are fundamental in algebra and represent a straight line when graphed on a coordinate plane. They can have one or multiple variables, but each variable is only raised to the first power.

A typical form of a linear equation in two variables is represented as: \[ ax + by = c \]Here,

• \( a \) and \( b \) are coefficients of the variables
• \( x \) and \( y \) are the variables
• \( c \) is the constant term

Linear equations are straightforward because they do not contain powers or roots of variables, and their solutions can be found using methods like substitution, elimination, or graphing. They can represent many real-life situations, making understanding them crucial. Solving systems of linear equations like the one in our exercise typically involves finding a point where the equations "intersect", representing a common solution to both equations.