Problem 57 - Intermediate Algebra

Question

Solve each system by any method, if possible. If a system is inconsistent or if the equations are dependent, state this. $$ \left\\{\begin{array}{l} y=-2.2 x+3.5 \\ y=-1.8 x+2.4 \end{array}\right. $$

Step-by-Step Solution

Answer

The solution is \((x, y) = (2.75, -2.55)\).

Step 1: Identify the Method

To solve the system of linear equations, we first need to choose a method. We can use the substitution method, the elimination method, or graph both equations. For this problem, let's choose the substitution method because both equations are already solved for \(y\).

Step 2: Set Equations Equal to Each Other

Since both equations are equal to \(y\), we can set the right sides of the equations equal to each other:\[-2.2x + 3.5 = -1.8x + 2.4\]

Step 3: Solve for x

To solve for \(x\), we first get all \(x\) terms on one side of the equation:\[-2.2x + 1.8x = 2.4 - 3.5\]Simplifying, we get:\[-0.4x = -1.1\]Now, solving for \(x\), we divide both sides by -0.4:\[x = \frac{-1.1}{-0.4} = 2.75\]

Step 4: Substitute x Back Into One Equation

Now that we have \(x = 2.75\), substitute it back into one of the original equations to solve for \(y\). Let's use the first equation:\[y = -2.2(2.75) + 3.5\]Calculating, we find:\[y = -6.05 + 3.5 = -2.55\]

Step 5: Check the Solution in the Other Equation

To ensure our solution is correct, substitute \(x = 2.75\) back into the second equation and see if it results in the same \(y\) value:\[y = -1.8(2.75) + 2.4\]Calculating, we get:\[y = -4.95 + 2.4 = -2.55\]Since it matches, our solution is confirmed.

Key Concepts

Substitution Method

The substitution method is a widely used technique for solving systems of linear equations. It involves expressing one of the variables in terms of another, and then substituting this expression into the other equation(s). This method is particularly useful when an equation is already solved for one variable, as it allows for straightforward substitution.

In our system of equations, both were already solved for \(y\):

• \( y = -2.2x + 3.5 \)
• \( y = -1.8x + 2.4 \)

Since both equations are equal to \(y\), our task is simply to set the right-hand sides equal to each other: \(-2.2x + 3.5 = -1.8x + 2.4\).

The substitution method is efficient for equations in this form, allowing us to solve for the unknown variable directly. Once one variable is found, it is substituted back into one of the original equations to find the other variable.

Solving for Variables

Solving for variables involves isolating the unknowns in the equation to find their numeric values. In our given system, we first solved for \(x\) after setting the equations equal to each other.

The process began by simplifying terms involving \(x\):

• Combine like terms: \(-2.2x + 1.8x = 2.4 - 3.5\)
• Resulting in \(-0.4x = -1.1\)

Next, isolating \(x\) by dividing both sides by \(-0.4\) provided the solution: \(x = 2.75\).

After finding \(x\), it was substituted back into one of the original equations to find \(y\). Using the first equation:

• Plug in \(x = 2.75\): \(y = -2.2(2.75) + 3.5\)
• Simplify to find \(y = -2.55\)

Verification was done by ensuring this \(y\) value satisfied the second equation as well.

Consistency of Equations

When solving systems of linear equations, determining the system's consistency is crucial. A consistent system has at least one solution, while an inconsistent system has none.

In our problem, the solution for \(x\) and \(y\) that satisfies both equations indicates the system is consistent. Inconsistent systems occur when equations represent parallel lines that never intersect, meaning no common solution exists.

In practice, rewriting terms and verifying that equations reduce to a valid solution ensures consistency. If during manipulation of the equations, a statement like "false equals true," such as \(0 = 1\), appears, it signals inconsistency.

Dependent and Independent Equations

Understanding the nature of equations as dependent or independent is fundamental in solving linear systems. Independent equations imply that the lines they represent intersect at exactly one point. This is the case in our example where we found a single solution \((x, y)\).

Dependent equations, on the other hand, represent the same line, thereby having infinitely many solutions. This occurs when one equation is a scalar multiple of the other, resulting in lines that coincide completely.

In this exercise, checking if original equations are scalar multiples or yield the same ratios for terms would help identify them as dependent. Our case with unique solutions confirms the equations are independent.