Problem 8 - Algebra and Trigonometry with Analytic Geometry

Question

1-30: Use the method of substitution to solve the system. $$ \left\\{\begin{array}{l} 3 x-4 y+20=0 \\ 3 x+2 y+8=0 \end{array}\right. $$

Step-by-Step Solution

Answer

The solution to the system is \(x = -4\) and \(y = 2\).

Step 1: Solve the First Equation for x

Start with the first equation, \(3x - 4y + 20 = 0\). We need to express \(x\) in terms of \(y\). Move \(-4y + 20\) to the right side: \(3x = 4y - 20\). Now solve for \(x\) by dividing both sides by 3: \(x = \frac{4y - 20}{3}\).

Step 2: Substitute x in the Second Equation

Take the expression for \(x\) from Step 1 and substitute it into the second equation \(3x + 2y + 8 = 0\). Replace \(x\) with \(\frac{4y - 20}{3}\): \(3\left(\frac{4y - 20}{3}\right) + 2y + 8 = 0\).

Step 3: Simplify the Equation

Simplify the equation from Step 2 by distributing the 3: \(4y - 20 + 2y + 8 = 0\). Combine like terms: \(6y - 12 = 0\).

Step 4: Solve for y

From the simplified equation \(6y - 12 = 0\), add 12 to both sides: \(6y = 12\). Divide by 6 to solve for \(y\): \(y = 2\).

Step 5: Substitute y Back to Find x

Use the value of \(y\) from Step 4 in the expression for \(x\) from Step 1, \(x = \frac{4y - 20}{3}\). Substitute \(y = 2\) into this expression: \(x = \frac{4(2) - 20}{3}\). Calculate to find \(x = \frac{8 - 20}{3} = \frac{-12}{3} = -4\).

Step 6: Verify the Solution

Substitute \(x = -4\) and \(y = 2\) into the original equations to verify the solution. For the first equation, \(3(-4) - 4(2) + 20 = -12 - 8 + 20 = 0\) is true. For the second equation, \(3(-4) + 2(2) + 8 = -12 + 4 + 8 = 0\) is true. Both equations are satisfied.

Key Concepts

System of Equations

A system of equations is a set of two or more equations that share common variables. When working with a system of linear equations, the main objective is to find a set of values for the variables that solve all the equations simultaneously. This typically involves linear equations, where the variables are not raised to any power greater than one. In this exercise, we have two equations:

• \(3x - 4y + 20 = 0\)
• \(3x + 2y + 8 = 0\)

Each equation represents a line on a coordinate plane, and solving the system means finding the point where these lines intersect. This intersection point provides the solution for the values of the variables \(x\) and \(y\). There are a few methods to solve these systems, such as graphing, elimination, and substitution. Here, we will focus on the substitution method.

Algebra

Algebra involves working with variables to solve equations and analyze relationships. It provides a systematic approach to solving equations using operations like addition, subtraction, multiplication, and division. In this context, algebra enables us to rearrange equations and express one variable in terms of another. For instance:Starting with the equation \(3x - 4y + 20 = 0\), algebra helps us isolate the variable \(x\). By moving terms around, we find:

• \(3x = 4y - 20\)
• \(x = \frac{4y - 20}{3}\)

Through this process, algebra provides the transformation needed to express \(x\) in terms of \(y\), which is fundamental for using the substitution method. This clarity allows us to plug the expression into another equation, bridging the path to find solutions for all variables involved.

Solving Linear Equations

When solving linear equations, our goal is to determine the values of the variables that make the equation true. By using the substitution method, one variable is expressed in terms of another, allowing us to substitute it back into another equation. For example:With \(x = \frac{4y - 20}{3}\) derived from the first equation, we substitute this into the second equation:

• \(3\left(\frac{4y - 20}{3}\right) + 2y + 8 = 0\)
• \(4y - 20 + 2y + 8 = 0\)

By simplifying and combining like terms, we solve for \(y\):

• \(6y - 12 = 0\)
• \(y = 2\)

After finding \(y\), we substitute it back to find \(x\):

• \(x = -4\)

Checking the solutions in both original equations confirms that \(x = -4\) and \(y = 2\) satisfy the system, verifying our solutions are correct. This methodical approach ensures precision in finding the true intersection of the linear equations.