Problem 23 - Algebra A Combined Function

Question

Solve each system of equations by the substitution method. $$ \left\\{\begin{array}{l} 6 x-3 y=5 \\ x+2 y=0 \end{array}\right. $$

Step-by-Step Solution

Answer

\(x = \frac{2}{3}\), \(y = -\frac{1}{3}\)

Step 1: Solve One Equation for One Variable

Start by solving the second equation for \(x\). From the equation \(x + 2y = 0\), isolate \(x\) by subtracting \(2y\) from both sides:\[ x = -2y \]

Step 2: Substitute into the Other Equation

Now take the expression for \(x\) from Step 1 and substitute it into the first equation. Replace \(x\) in the first equation, \(6x - 3y = 5\), with \(-2y\):\[ 6(-2y) - 3y = 5 \]

Step 3: Simplify and Solve for the Remaining Variable

Simplify the equation from Step 2:\[ -12y - 3y = 5 \]Combine like terms:\[ -15y = 5 \]Now solve for \(y\) by dividing both sides by \(-15\):\[ y = -\frac{1}{3} \]

Step 4: Find the Value of the Other Variable

Using the value of \(y\) found in Step 3, substitute back into the expression for \(x\) from Step 1:\[ x = -2(-\frac{1}{3}) \]Simplify the expression:\[ x = \frac{2}{3} \]

Step 5: Verify the Solution

To verify, substitute \(x = \frac{2}{3}\) and \(y = -\frac{1}{3}\) into both original equations:1. Check the first equation: \[ 6(\frac{2}{3}) - 3(-\frac{1}{3}) = 4 + 1 = 5 \] which is true.2. Check the second equation: \[ \frac{2}{3} + 2(-\frac{1}{3}) = \frac{2}{3} - \frac{2}{3} = 0 \] which is also true.Since both equations are satisfied, the solution is correct.

Key Concepts

System of Equations

When we talk about a "system of equations," we mean a set of two or more equations that share a common set of variables. In our exercise, the system is made up of two equations:

• \(6x - 3y = 5\)
• \(x + 2y = 0\)

A solution to the system is a pair (or set) of values for the variables that makes each equation in the system true. Solving these types of problems can tell you where the graphs of the equations intersect, if they do at all.

There are different methods to solve a system of equations, such as:

• Graphing, where you find the point of intersection on a graph
• Substitution, where you solve one equation for one variable and substitute that into the other equation
• Elimination, which involves adding or subtracting the equations to eliminate one of the variables

In this exercise, we use the substitution method.

Linear Equations

Linear equations are equations of the first degree, which means that the variables are raised only to the power of one. In simpler terms, the graph of a linear equation is always a straight line. The standard form of a linear equation with two variables, say \(x\) and \(y\), is \(ax + by = c\), where \(a\), \(b\), and \(c\) are constants.

In our set of equations:

• \(6x - 3y = 5\)
• \(x + 2y = 0\)

both are linear equations. This means when we solve them together, we are essentially finding the point at which the lines represented by these equations cross or intersect.

Linear equations are foundational in algebra and are the building blocks to understanding more complex mathematical concepts.

Solving Algebraic Equations

Solving algebraic equations essentially means finding the value(s) of the variable(s) that make the equation true. When we solve systems of algebraic equations, our goal is to find a set of values that satisfies all the equations simultaneously.

In the substitution method—as used in this exercise—we start by isolating one variable in one of the equations. Here, we solved for \(x\) in the second equation: \(x + 2y = 0\), resulting in \(x = -2y\).

Next, substitute \(-2y\) for \(x\) in the first equation. Solve the resulting expression to find the value of \(y\):

• Substitute to get \(-12y - 3y = 5\), which simplifies to \(-15y = 5\). This gives \(y = -\frac{1}{3}\).

Finally, use \(y = -\frac{1}{3}\) in our expression for \(x\) to find that \(x = \frac{2}{3}\).

Checking the solution is a critical step. Ensure both values satisfy the original equations. This verifies the accuracy, giving confidence in your solutions.