Problem 16 - Precalculus

Question

Solve each system of linear equations by substitution. $$\begin{aligned} &4 x-5 y=-7\\\ &3 x+8 y=30 \end{aligned}$$

Step-by-Step Solution

Answer

The solution is \( x = 2 \) and \( y = 3 \).

Step 1: Solve for One Variable

Let's solve the first equation for variable \( x \). Using the equation \( 4x - 5y = -7 \), we isolate \( x \) by adding \( 5y \) to both sides and then dividing by 4.\[ 4x = 5y - 7 \] Divide by 4: \[ x = \frac{5y - 7}{4} \].

Step 2: Substitute into the Second Equation

Now that we have \( x = \frac{5y - 7}{4} \), substitute this expression for \( x \) into the second equation \( 3x + 8y = 30 \), replacing \( x \):\[ 3\left(\frac{5y - 7}{4}\right) + 8y = 30 \].

Step 3: Clear the Fraction

To eliminate the fraction, multiply every term by 4, which is the denominator:\[ 3(5y - 7) + 8y imes 4 = 30 imes 4 \], resulting in:\[ 15y - 21 + 32y = 120 \].

Step 4: Combine Like Terms

Combine the \( y \) terms on the left side of the equation:\[ 47y - 21 = 120 \].

Step 5: Solve for \( y \)

Add 21 to both sides to isolate the terms with \( y \):\[ 47y = 141 \]. Divide both sides by 47 to solve for \( y \): \[ y = 3 \].

Step 6: Substitute Back to Find \( x \)

Use the expression \( x = \frac{5y - 7}{4} \) and substitute \( y = 3 \) into it:\[ x = \frac{5(3) - 7}{4} \]. This simplifies to \[ x = \frac{15 - 7}{4} = \frac{8}{4} = 2 \].

Step 7: Verify the Solution

Plug \( x = 2 \) and \( y = 3 \) into the original equations to check for correctness. First equation: \( 4(2) - 5(3) = 8 - 15 = -7 \), which is correct. Second equation: \( 3(2) + 8(3) = 6 + 24 = 30 \), which is also correct. Therefore, the solution is verified.

Key Concepts

Substitution Method

The substitution method is a technique used to solve a system of linear equations. It involves solving one of the equations for one variable and then substituting that expression into the other equation. This simplifies the problem as it reduces the number of variables in the second equation, making it easier to solve.
Here's how you can apply the substitution method:

• Solve one of the equations for one of the variables. This means isolating that variable on one side of the equation.
• Substitute this expression into the other equation. Replace the solved variable in this equation with its expression from the previous step.
• Solve the resulting single-variable equation.
• Substitute back to find the other variable. Once you have the value of one variable, substitute it back into the expression found in the first step to solve for the other variable.

This method is particularly useful when one of the variables can be easily isolated, leading to simpler arithmetic later.

Systems of Equations

A system of equations is a collection of two or more equations with the same set of unknowns. Solving a system of equations means finding the values of the variables that satisfy all the equations simultaneously.
In our exercise, we are given two equations:

• \( 4x - 5y = -7 \)
• \( 3x + 8y = 30 \)

These are linear equations, which graphically represent straight lines. The solution to the system is the point where the lines intersect, which means they share common values for \( x \) and \( y \).
There are different methods to solve systems of equations, such as graphing, substitution (used here), and elimination. Substitution and elimination are algebraic methods preferred for their precision, especially for more complex equations. Understanding how to analyze and manipulate these equations strengthens foundational precalculus skills.

Solution Verification

Verification is an essential part of solving a system of equations. It confirms the accuracy of the solution by substituting the found values back into the original equations. This step ensures that the calculated solution set actually satisfies these equations.
Let's recap the process:

• After determining values for all variables, go back to the original equations.
• Substitute the values of the variables into both equations.
• Check if the left-hand side equals the right-hand side in each equation.

If both original equations hold true with the substituted values, the solution is verified. This step not only confirms that you have the correct solution but also boosts confidence in your problem-solving skills.

Precalculus

Precalculus is a course that bridges the gap between algebra and calculus. It encompasses various mathematical topics necessary for understanding advanced concepts.
The skills learned in precalculus, such as solving systems of equations, are foundationally important as they form the basis for calculus problems.
Some key concepts within precalculus include:

• Functions and their properties
• Trigonometry
• Complex numbers
• Vectors and matrices

Grasping these concepts thoroughly eases the transition into calculus, where these skills are further built upon and expanded. In this context, linear equations and systems teach problem-solving and analytical skills crucial for higher-level mathematics.