Problem 44 - Algebra 2

Question

Solve each system of equations by using either substitution or elimination. $$ \begin{array}{l}{3 x-4 y=-2} \\ {5 x+2 y=40}\end{array} $$

Step-by-Step Solution

Answer

The solution to the system is \(x = 6\) and \(y = 5\).

Step 1: Choose the Method

Let's use the elimination method. Our goal will be to eliminate one of the variables, making it easier to solve the system of equations.

Step 2: Align the Equations for Elimination

The given system is: \(3x - 4y = -2\) and \(5x + 2y = 40\). To eliminate \(y\), we need to create equal and opposite coefficients for \(y\) in both equations.

Step 3: Make Coefficients of Y Equal

Multiply the first equation by 2 and the second equation by 4: \( 2(3x - 4y) = 2(-2) \) gives \( 6x - 8y = -4 \) \( 4(5x + 2y) = 4(40) \) gives \( 20x + 8y = 160 \).

Step 4: Add the Equations to Eliminate Y

Add the two new equations: \( (6x - 8y) + (20x + 8y) = -4 + 160 \). The \(y\) terms cancel out, leaving \(26x = 156\).

Step 5: Solve for X

Solve the resulting equation for \(x\):\[ x = \frac{156}{26} = 6 \].

Step 6: Substitute X into One Equation

Substitute \(x = 6\) back into the first original equation:\(3(6) - 4y = -2\) becomes \(18 - 4y = -2\).

Step 7: Solve for Y

Rearrange the equation to isolate \(y\):\( -4y = -2 - 18 \) which simplifies to \( -4y = -20 \). Divide both sides by -4 to find \( y = 5 \).

Step 8: Verify the Solution

Substitute \(x = 6\) and \(y = 5\) into the second equation:\(5(6) + 2(5) = 30 + 10 = 40\), which confirms that the values satisfy the system.

Key Concepts

Elimination Method

The elimination method is a powerful technique for solving systems of linear equations. The central idea is to either add or subtract the equations so that one variable cancels out, simplifying the system to a single equation with only one variable. This requires creating equal and opposite coefficients for one of the variables in the equations.

For instance, consider the equations \(3x - 4y = -2\) and \(5x + 2y = 40\). To eliminate the variable \(y\), we seek to make the \(y\) coefficients equal but opposite. By multiplying the first equation by 2 and the second by 4, we adjust coefficients to \(-8y\) and \(+8y\), respectively.

• Multiply the first equation: \(2(3x - 4y) = 6x - 8y = -4\).
• Multiply the second equation: \(4(5x + 2y) = 20x + 8y = 160\).

Adding these equations together, \(-8y + 8y\) cancels out, leaving \(26x = 156\), from which \(x\) can be easily determined. Eliminating a variable streamlines the solving process, reducing the system to a single-variable equation.

Substitution Method

The substitution method provides another way to tackle systems of linear equations by solving one equation for one variable and substituting that expression into the other equation.

This approach is straightforward when one of the equations makes it easy to isolate a variable, though this wasn't the chosen method for our problem. Let's imagine substituting:

• Solve for \(x\) in one equation, say \(x = \frac{y}{2} \)
• Substitute into the other equation to express it in terms of a single variable.
• Find the solution to the new single-variable equation.

After finding one variable, plug its value back into one of the original equations to get the other variable's value. This method can often be more intuitive and direct, especially when dealing with simpler equations or equations that are easily manipulated into one variable form.

Solving Linear Equations

To solve linear equations, we aim to find the values of the variables that satisfy all equations in the system. Each equation represents a straight line, and the solution reflects where these lines intersect in a graph.

When solving, it's essential to light upon the following strategies:

• First, choose between elimination or substitution based on the equations' structure.
• Apply the chosen method to simplify the system to a single equation in one variable.
• Solve the simplified equation to find one of the variable values.
• Substitute back to get the value of the remaining variable.

By using these systematic approaches, solving linear equations becomes clearer, making it easier to find the point of intersection, which provides the solution to the system. It’s important to check solutions by substituting back into the original equations to confirm their validity.

Algebra

Algebra is a foundational branch of mathematics that allows you to express relationships through variables and equations. It's not only about finding unknowns but about developing logical reasoning to manipulate symbols and expressions systematically.

Core principles of algebra include:

• Understanding variables as symbols representing numbers.
• Learning to manipulate expressions and equations to isolate variables.
• Knowing how to simplify and solve equations involving these variables.

One of algebra's applications is solving systems of equations. This involves finding common values for variables that satisfy multiple equations simultaneously. Concepts like the elimination and substitution methods exemplify algebra's power to break down complex problems into simpler, manageable steps through strategic manipulation of equations. With practice, algebra becomes a toolkit for solving diverse mathematical problems.