Problem 92 - Algebra 1: Concepts and Skills

Question

Use substitution to solve the system. $$ \begin{aligned} &x+4 y=30\\\ &x-2 y=0 \end{aligned} $$

Step-by-Step Solution

Answer

The solutions to the system of equations are \(x = 10\) and \(y = 5\).

Step 1: Isolating variable x

From the second equation in our system, which is \(x - 2y = 0\), we can easily isolate \(x\). In order to do this, we can rewrite the equation in the form \(x = 2y\).

Step 2: Substitution of x into the first equation

Now, we can substitute the value of \(x\) found from the second equation into the first one. It will result in the equation: \(2y + 4y = 30.\)

Step 3: Solving for y

Then, after combining like terms, we can solve for \(y\) through division: \(6y = 30\) simplifies to \(y = 5\).

Step 4: Substituting y into the second equation to find x

Next, we substitute \(y = 5\) back into our rearranged second equation from Step 1 to find \(x\). This gives us \(x = 2*5 = 10\).

Key Concepts

Understanding the Substitution Method

When it comes to solving systems of equations, the substitution method is one of the most straightforward and intuitive techniques. Essentially, this method involves replacing one variable with an expression derived from one of the equations. Here's the step-by-step process:

• First, you choose one of the equations and solve it for one of the variables in terms of the others. In our example, we used the equation \(x - 2y = 0\) to express \(x\) in terms of \(y\), which gives us \(x = 2y\).
• Next, substitute this expression into the other equation. This replaces one variable, reducing the system to an equation with a single variable. We replaced \(x\) in \(x + 4y = 30\) with \(2y\), leading to the equation \(2y + 4y = 30\).
• Finally, solve this new single-variable equation to find the value of one variable. Here, by solving \(6y = 30\), we found \(y = 5\).
• Once you have solved for one variable, backtrack to find the other variable using one of the original equations. Substituting \(y = 5\) into \(x = 2y\) results in \(x = 10\).

This method is particularly useful when one equation is easily manipulated to express one variable in terms of the other.

Diving into Linear Equations

Linear equations, as their name suggests, graphically represent straight lines in two dimensions. They are characterized by variables raised only to the power of one, forming a degree-1 polynomial. In our exercise, both equations \(x + 4y = 30\) and \(x - 2y = 0\) are linear.

• The standard form of a linear equation in two variables is \(ax + by = c\). Here, \(a\), \(b\), and \(c\) are constants.
• Linear equations in two variables typically describe a line in a coordinate plane. Solving such systems finds the intersection point, where both lines cross.
• Systems of linear equations may have a single solution (consistent system), no solution (parallel lines), or infinitely many solutions (the same line).

Linear equations form the foundation of algebra, being a simple yet powerful tool to describe relationships between quantities.

Solving Equations Step-by-Step

Solving equations, whether linear or nonlinear, involves finding the values of variables that make the equation true. This process is all about preserving equality and employing mathematical operations correctly. Let's revisit the steps used to solve our system:

• Isolate a Variable: Choose one equation and rearrange it to express one variable in terms of the others. In our case, we isolated \(x\) to get \(x = 2y\).
• Substitute and Simplify: Replace this isolated variable in the other equation and simplify. We substituted \(x = 2y\) into \(x + 4y = 30\), resulting in the simplified equation \(6y = 30\).
• Solve for One Variable: Simplify and solve the equation for one variable. Here, dividing both sides of \(6y = 30\) by 6 gave us \(y = 5\).
• Back-Substitute to Find the Other Variable: Once one variable is known, use it in one of the original equations to find the other variable. Substituting \(y = 5\) into our expression for \(x\) provided \(x = 10\).

By following methodical steps, solving equations becomes a process of logic and precision, ensuring the correct solution is reached every time.