Problem 32 - Algebra 2

Question

Solve each system of equations by using either substitution or elimination. \(2 p=7+q\) \(6 p-3 q=24\)

Step-by-Step Solution

Answer

The system of equations has no solution.

Step 1: Solve the First Equation for One Variable

The first equation is \(2p = 7 + q\). We will solve for \(q\) in terms of \(p\).\[ q = 2p - 7 \]

Step 2: Substitute the Expression for q into the Second Equation

Take the expression for \(q\) from Step 1 and substitute it into the second equation: \(6p - 3q = 24\).Substitute \(q = 2p - 7\):\[ 6p - 3(2p - 7) = 24 \]

Step 3: Simplify the Second Equation

Distribute the \(-3\) in the equation:\[ 6p - 6p + 21 = 24 \]

Step 4: Solve for p

Now simplify the equation:\[-6p + 6p\] cancels to \(0\), so the equation becomes:\[ 21 = 24 \]This creates a contradiction, so there is no value of \(p\) that satisfies this equation.

Step 5: Conclusion

Since the previous step resulted in a contradiction, this means that the system of equations has no solution. The lines represented by these equations are parallel and do not intersect.

Key Concepts

Substitution Method

The substitution method is one of the ways to tackle a system of equations. It involves expressing one variable in terms of another, making it easier to deal with the equations.

Here’s how it generally works:

• Solve one of the equations for one variable.
• Substitute the resulting expression into the other equation.
• Solve the new equation for the remaining variable.
• Back-substitute to find the other variable, if needed.

In the original exercise, the first equation was solved for the variable \( q \), giving us \( q = 2p - 7 \).

Substituting this expression for \( q \) into the second equation allowed to deduce information about how the two equations interact.

The power of substitution is in transforming a system of equations into an easily solvable single-variable equation.

Elimination Method

The elimination method, also known as the addition or subtraction method, is another effective technique to solve systems of equations.

It works by adding or subtracting equations to eliminate one variable, which allows us to solve for the other.

Here are the steps:

• Align the equations similarly by moving terms around.
• Multiply one or both equations so a variable's coefficients are equal or additive inverses.
• Add or subtract the equations to eliminate one variable.
• Solve for the remaining variable and back-substitute to find the other variable.

Though the exercise focused on substitution, elimination is a powerful alternative for systems with easily comparable coefficients, making it a versatile choice for various problems.

No Solution

A system of equations has no solution when the equations represent parallel lines that never meet, leading to a contradiction.

In practical terms, after using substitution or elimination, you might end up with a statement that's mathematically inaccurate, like \( 21 = 24 \).

This tells us something is amiss in terms of solution existence:

• The lines are parallel; hence, they do not intersect.
• Variables cancel each other out completely.
• The remaining numerical equation has no consistent solution.

Recognizing no solution is crucial, as trying to force a result could lead to conceptual misunderstandings.

Parallel Lines

In the realm of algebra, parallel lines signal that a system of equations has no solution.

This occurs when the lines represented by two equations have the same slope but different y-intercepts, causing them to never intersect.

The conditions for parallel lines are:

• The coefficients of \( p \) (or the relevant variable) are proportional.
• The constant terms don't form a matching combination when simplified.

In the exercise's context, the simplification led to \( 21 = 24 \), flagging that the lines from the equations do not crossover.

Understanding when lines are parallel aids in quickly identifying when multi-equation setups will not yield a viable numerical solution.