Question

problems 5–6: consider this system of equations: \\(\

$$\begin{cases}x - 6y = 4 \\\\ 3x - 18y = 4\\end{cases}$$

\\)\

1. change one number to make a new system with one solution.\
2. change one number to make a new system with an infinite number of solutions.

Explanation:

Problem 5

Step 1: Analyze the original system

The original system is \(

$$\begin{cases}x - 6y = 4\\3x - 18y = 4\end{cases}$$

\). Notice that the second equation is \(3(x - 6y)=4\), but \(3\times4 = 12
eq4\), so the lines are parallel (no solution). To get one solution, we need the two lines to have different slopes. So we can change the coefficient of \(y\) in the second equation (or other coefficients/constants) to make the slopes different. Let's change the \(18\) in the second equation to a number not equal to \(18\) (since the slope of the first line is \(\frac{1}{6}\), slope of the second line is \(\frac{3}{18}=\frac{1}{6}\); we need a different slope). Let's change \(18\) to \(17\). Then the new system is \(

$$\begin{cases}x - 6y = 4\\3x - 17y = 4\end{cases}$$

\). Now, the slopes are \(\frac{1}{6}\) and \(\frac{3}{17}\) (different), so they will intersect at one point (one solution).

Step 2: Verify (optional)

We can solve the new system. From the first equation, \(x = 4 + 6y\). Substitute into the second equation: \(3(4 + 6y)-17y = 4\) → \(12 + 18y - 17y = 4\) → \(12 + y = 4\) → \(y=-8\), then \(x = 4+6\times(-8)=4 - 48=-44\). So there is one solution \((-44, -8)\).

Step 1: Recall the condition for infinite solutions

A system of linear equations \(

$$\begin{cases}a_1x + b_1y = c_1\\a_2x + b_2y = c_2\end{cases}$$

\) has infinite solutions if \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\). For the original system \(

$$\begin{cases}x - 6y = 4\\3x - 18y = 4\end{cases}$$

\), \(\frac{1}{3}=\frac{-6}{-18}\) (since \(\frac{1}{3}=\frac{-6}{-18}\)), but \(\frac{4}{4}=1
eq\frac{1}{3}\). So we need to make \(\frac{c_1}{c_2}=\frac{1}{3}\). So change the constant term of the second equation from \(4\) to \(12\) (since \(4\times3 = 12\)). Then the new system is \(

$$\begin{cases}x - 6y = 4\\3x - 18y = 12\end{cases}$$

\). Now, \(\frac{1}{3}=\frac{-6}{-18}=\frac{4}{12}\) (all equal to \(\frac{1}{3}\)), so the two equations are the same line (infinite solutions).

Step 2: Verify (optional)

The second equation \(3x - 18y = 12\) can be divided by \(3\) to get \(x - 6y = 4\), which is the same as the first equation. So every solution of the first equation is a solution of the second, and vice versa, so there are infinite solutions.

Answer:

One possible way is to change the \(18\) in \(3x - 18y = 4\) to \(17\), resulting in the system \(

$$\begin{cases}x - 6y = 4\\3x - 17y = 4\end{cases}$$

\) (other valid changes are also possible, like changing the constant term of the second equation, etc.).

Problem 6