Problem 29 - Algebra A Combined Function

Question

Solve each system of equations by the substitution method. $$ \left\\{\begin{array}{l} 5 x+10 y=20 \\ 2 x+6 y=10 \end{array}\right. $$

Step-by-Step Solution

Answer

Solution: \((x, y) = (2, 1)\).

Step 1: Solve One Equation for One Variable

We will solve the first equation for \( x \). The first equation is \( 5x + 10y = 20 \). To isolate \( x \), we can rewrite the equation as follows:\[5x = 20 - 10y\]Now, solve for \( x \):\[x = \frac{20 - 10y}{5}\]Simplify the expression:\[x = 4 - 2y\]

Step 2: Substitute the Expression into the Second Equation

Substitute \( x = 4 - 2y \) from Step 1 into the second equation, which is \( 2x + 6y = 10 \):\[2(4 - 2y) + 6y = 10\]Simplify by expanding and combining like terms:\[8 - 4y + 6y = 10\]Combine \(-4y\) and \(6y\):\[8 + 2y = 10\]

Step 3: Solve for the Remaining Variable

Now, solve for \( y \) in the equation \( 8 + 2y = 10 \):\[2y = 10 - 8\]\[2y = 2\]Divide both sides by 2:\[y = 1\]

Step 4: Substitute Back to Find the Other Variable

Now that we have \( y = 1 \), substitute it back into the expression for \( x \) from Step 1:\[x = 4 - 2(1)\]\[x = 4 - 2\]\[x = 2\]

Step 5: Write the Solution

The solution to the system of equations is \( (x, y) = (2, 1) \). This means that \( x = 2 \) and \( y = 1 \) satisfies both equations.

Key Concepts

System of Equations

A system of equations is a collection of two or more equations that involve the same set of variables. In the problem we're discussing, the system contains two linear equations with two variables, \(x\) and \(y\). Each equation represents a line, and solving the system involves finding the point(s) where these lines intersect. In our example, we're dealing with the following system:

• \(5x + 10y = 20\)
• \(2x + 6y = 10\)

These equations describe two straight lines on a coordinate plane. The solution to this system is the point (or points) where the lines meet, which signifies the set of values for \(x\) and \(y\) that satisfy both equations simultaneously. In this exercise, we'll use the substitution method to find that intersection and solve both equations.

Solving Equations

Solving an equation means finding the value(s) of the variable(s) that make the equation true. In the context of a system of equations, we not only solve one equation but find a common solution for all equations in the system.

With the substitution method, we start by solving one equation for a particular variable. In our example, we isolated \(x\) in the first equation,

• \(x = 4 - 2y\).

Afterwards, we substitute this expression into the other equation to find the value of the second variable.

Once we've solved for one variable, the next step is to substitute the found value into one of the original equations or expressions to find the value of the other variable. This approach allows us to reduce the system of equations to a simpler one-variable equation, making it easier to find the solution.

Algebra Step by Step

Approaching algebra problems step-by-step is crucial, especially when solving systems of equations. Let's look at how this is done with substitution:- **Step 1:** Substitute solution in one of the variables: First, we solve one of the equations for one variable. In this case, \(5x + 10y = 20\) is solved for \(x\), giving us \(x = 4 - 2y\).- **Step 2:** Substitute this solution into the other equation: Take the expression for \(x\) and insert it into the other equation. By doing so,\(2(4 - 2y) + 6y = 10\) simplifies to a single equation in terms of \(y\).- **Step 3:** Solve for one variable: The simplified equation \(2y = 2\) is solved to find \(y = 1\).- **Step 4:** Find the other variable: Finally, substitute \(y = 1\) back into \(x = 4 - 2y\) to get \(x = 2\).By breaking down the problem into manageable steps, solving a system becomes much less intimidating and more systematic. Each step builds upon the last, making it simpler to track errors and ensure solutions are correct.

Linear Equations

Linear equations form the foundation of many algebraic concepts and are always graphically represented as straight lines. In the exercise at hand, we are dealing with two such linear equations:

• \(5x + 10y = 20\)
• \(2x + 6y = 10\)

These equations are called linear because the highest power of the variable in each equation is one. The coefficients provide us the slope and intercepts to graph each line. For example, rewriting \(5x + 10y = 20\) in slope-intercept form \(y = mx + b\) makes it easier to understand graphically.

In linear equations, the relationship between the variables is consistent and unvarying, forming a straight line. When two such equations are written as a system, the task is to find the point where these lines intersect, which is the simultaneous solution for the system. This point corresponds to the specific values of \(x\) and \(y\) that satisfy both equations, confirming the found solution \((2, 1)\), as both equations are satisfied at this point.