Problem 17 - College Algebra

Question

Solve each system by substitution. $$ \begin{array}{l} 3 \quad x+5 y=9 \\ 30 x+50 y=-90 \end{array} $$

Step-by-Step Solution

Answer

The system has no solution as it leads to a false statement.

Step 1: Solve for x in the first equation

Start with the equation \(3x + 5y = 9\). Solve for \(x\) by rearranging the equation: \(3x = 9 - 5y\). Divide by 3 to isolate \(x\):\[ x = \frac{9 - 5y}{3} \]

Step 2: Substitute x into the second equation

Take the expression for \(x\) from Step 1 and substitute it into the second equation: \(30x + 50y = -90\). Replace \(x\) with \(\frac{9 - 5y}{3}\):\[ 30\left(\frac{9 - 5y}{3}\right) + 50y = -90 \]

Step 3: Simplify the equation

Simplify the expression by multiplying \(30\) with \(\frac{9 - 5y}{3}\):\[ 10(9 - 5y) + 50y = -90 \]Distribute the \(10\):\[ 90 - 50y + 50y = -90 \]

Step 4: Solve for y

Combine like terms in the simplified equation. The \(-50y\) and \(+50y\) cancel each other out:\[ 90 = -90 \]Since the statement is false, there is no solution to the system of equations.

Key Concepts

Substitution Method

The substitution method is an algebraic technique used to solve systems of linear equations. It is especially useful when dealing with two equations. The primary goal here is to express one variable in terms of the other in one of the equations. This expression is then substituted into the other equation, which allows us to solve for one of the variables.

In our given set of equations:

• First equation: \(3x + 5y = 9\)
• Second equation: \(30x + 50y = -90\)

Starting with the first equation, we rearranged to express \(x\) as \(x = \frac{9 - 5y}{3}\). Once we have this expression, it allows us to substitute \(x\) in the second equation. This leads to simplifying the second equation with only one variable, \(y\), which we can attempt to solve. The substitution method is a powerful technique, especially when one of the variables can be easily isolated in one of the equations.

No Solution

In the realm of systems of equations, the solution can fall into three categories: one solution, no solution, or infinitely many solutions. When we talk about 'no solution', it means that the equations are parallel lines on a graph. Parallel lines are lines that never intersect.

For our specific problem, once we substitute and simplify, we reach a point where the statement becomes \(90 = -90\). This is a contradiction, implying there is no value of \(x\) and \(y\) that will satisfy both equations simultaneously. Therefore, the system has no solution. When dealing with linear equations, recognizing a contradiction after substituting is a key indicator of parallel lines, or in other words, no solution.

Linear Equations

Linear equations are equations of the first order. This means that each term is either a constant or the product of a constant and a single variable. They graph as straight lines in a coordinate system.

Each linear equation in our system can be written in the standard form \(ax + by = c\). For our exercise, both equations:

• \(3x + 5y = 9\)
• \(30x + 50y = -90\)

are indeed linear. The simplicity of linear equations makes them suitable for solving using various methods like substitution, elimination, or graphing. The characteristics of these equations, such as slopes and y-intercepts, determine whether they intersect, are parallel, or are the same line. This explains why understanding the linear equations is crucial in predicting whether there will be a solution or not in a given system.