Problem 44 - Intermediate Algebra

Question

For Problems 19-48, solve each system by using either the substitution or the elimination-by-addition method, whichever seems more appropriate. (Objective 2) $$ \left(\begin{array}{l} \frac{2}{3} x+\frac{1}{5} y=0 \\ \frac{3}{2} x-\frac{3}{10} y=-15 \end{array}\right) $$

Step-by-Step Solution

Answer

The solution is \( x = -6 \) and \( y = 20 \).

Step 1: Simplify the System of Equations

Given the system of equations: \[ \frac{2}{3}x + \frac{1}{5}y = 0 \] and \[ \frac{3}{2}x - \frac{3}{10}y = -15 \]It is beneficial to eliminate fractions by multiplying each equation by the least common multiple of denominators. For the first equation (that has denominators 3 and 5), this is 15, and for the second equation (that has denominators 2 and 10), this is 10. After simplifications:1. First Equation: Multiply by 15: \[ 10x + 3y = 0 \]2. Second Equation: Multiply by 10: \[ 15x - 3y = -150 \]

Step 2: Use Elimination Method

Add the two simplified equations to eliminate \(y\):\[ 10x + 3y = 0 \] \[ 15x - 3y = -150 \]Adding these gives:\[ 25x = -150 \]This allows us to solve directly for \(x\).

Step 3: Solve for x

From the equation \( 25x = -150 \), divide both sides by 25 to get:\[ x = \frac{-150}{25} = -6 \]

Step 4: Substitute to Find y

Substitute \( x = -6 \) back into the simplified form of either original equation. We use the first simplified equation: \[ 10x + 3y = 0 \]Substitute \(x = -6\):\[ 10(-6) + 3y = 0 \]\[ -60 + 3y = 0 \]Add 60 to both sides:\[ 3y = 60 \]Divide by 3:\[ y = 20 \]

Step 5: Conclusion: System Solution

After solving, the solution to the system of equations is \( x = -6 \) and \( y = 20 \).

Key Concepts

Elimination Method

The elimination method is a fundamental technique for solving systems of equations. It involves manipulating the equations in such a way that one of the variables gets "eliminated" when the equations are added or subtracted from each other. This method is particularly useful when the coefficients of one of the variables are opposites or can easily be made opposites through multiplication.
To use the elimination method effectively:

• First, ensure the coefficients of either variable are opposites. You may need to multiply one or both equations by a constant to achieve this.
• Add or subtract the equations to cancel out one variable. This leaves you with a single-variable equation that can be solved directly.
• Substitute the found value back into one of the original equations to find the value of the other variable.

The strength of elimination lies in its ability to simplify the system to just one equation if done correctly. This method works best when the equations are straightforward and numbers lend themselves to easy manipulation.

Substitution Method

The substitution method is another powerful tool for solving systems of linear equations. This approach is flexible and can be used when one of the equations is already solved for one variable or can be easily manipulated to solve for one. The method involves expressing one variable in terms of the other and substituting it into the second equation.

• Begin by solving one of the equations for one variable. Choose the equation and variable that simplifies most easily.
• Substitute the expression obtained for this variable into the other equation. This results in a single equation with one variable that you can solve.
• Once you find the value of the variable from the single equation, substitute it back into one of the original equations to find the value of the second variable.

The substitution method is particularly advantageous when one equation is already isolated or nearly isolated for one variable. It simplifies the system by reducing it to solving a simple linear equation.

Solving Linear Equations

Solving linear equations is a core aspect of handling systems of equations. A linear equation usually takes the form \( ax + by = c \), where \( a \), \( b \), and \( c \) are constants and \( x \) and \( y \) are variables. The goal is to find the values of \( x \) and \( y \) that satisfy both equations simultaneously.
To effectively solve linear equations:

• Use either the substitution or elimination method to isolate variables.
• Once an equation in one variable is established, solve for that variable by straightforward arithmetic operations such as addition, subtraction, multiplication, or division.
• Use the found value to determine the second variable by substituting it back into one of the original equations.

Understanding and solving linear equations is pivotal, as it forms the basis for solving more complex mathematical problems. Mastering these tools allows for efficient resolution of equations and development of algebraic reasoning skills.