Problem 9 - A Graphical Approach to Precalculus with Limits

Question

Solve each system by substitution. $$\begin{aligned}&y=2 x+3\\\&3 x+4 y=78\end{aligned}$$

Step-by-Step Solution

Answer

The solution is \((x, y) = (6, 15)\).

Step 1: Substitute y in Second Equation

Start with the equation for \( y \): \( y = 2x + 3 \). Replace \( y \) in the second equation \( 3x + 4y = 78 \) with \( 2x + 3 \). This gives us: \( 3x + 4(2x + 3) = 78 \).

Step 2: Simplify the Equation

Distribute 4 in the equation: \( 4(2x + 3) = 8x + 12 \). Substitute back into the equation, obtaining \( 3x + 8x + 12 = 78 \). Simplify this to get \( 11x + 12 = 78 \).

Step 3: Solve for x

Subtract 12 from both sides to find: \( 11x = 66 \). Divide both sides by 11 to isolate \( x \): \( x = 6 \).

Step 4: Substitute x back into y Equation

Now that we have \( x = 6 \), substitute it back into the equation for \( y \): \( y = 2(6) + 3 \).

Step 5: Solve for y

Calculate \( 2(6) + 3 = 12 + 3 = 15 \). Thus, \( y = 15 \).

Step 6: Final Solution

The solution of the system is \( x = 6 \) and \( y = 15 \). This means the intersection point of the lines is \((6, 15)\).

Key Concepts

Substitution Method

The substitution method is a reliable technique for solving systems of equations. Here's how it works:

You pick one of the equations from the system and solve it for one variable. In our case, we picked the equation \( y = 2x + 3 \). With this equation, we already have \( y \) expressed in terms of \( x \).

Next, you substitute this expression back into the other equation. This allows you to reduce the system into a single equation with one variable. So, we substituted \( 2x + 3 \) for \( y \) in the equation \( 3x + 4y = 78 \). Remember, always replace the variable consistently without missing any parts of the equation.

By using substitution, you aim to find a value for one of the variables, which then helps you solve the entire system.

Solving Linear Equations

Solving linear equations means finding the values of variables that satisfy the equation. It's often straightforward but needs careful attention to details.

For our problem, after substituting \( y \) into the second equation, we ended up with \( 11x + 12 = 78 \).
Breaking down these steps:

• Distribute wherever necessary - like we distributed \( 4 \) into \( 2x + 3 \), resulting in \( 8x + 12 \).
• Combine like terms - that means adding or subtracting similar terms (e.g., \( 3x + 8x \) gives \( 11x \)).

Once simplified, isolate the variable on one side of the equation. We subtracted 12 from 78 to simplify it further, leaving \( 11x = 66 \).
Finally, divide both sides by the coefficient of x, yielding \( x = 6 \). This is your solution for one of the variables.

Intersection Point

The intersection point is where two lines, representing equations, cross each other on a graph. When solving systems of equations, this point will indicate the values of the variables that satisfy both equations simultaneously.

In our context, we solved for \( x = 6 \) and then used this value to find \( y = 15 \).
We substitute \( x = 6 \) back into the expression for \( y \), i.e., \( y = 2(6) + 3 \). Thus, \( y \) equals 15.

When plotted, both equations \( y = 2x + 3 \) and \( 3x + 4y = 78 \) will intersect at \((6, 15)\). This point is significant because it represents a common solution to both equations.

Understanding intersection points is key in different math applications like geometry, physics, and engineering for determining where different conditions or equations meet.