Question

consider this system of equations:
x - 6y = 4
3x - 18y = 4
change one number to make a new system with one solution.
equation 1 \t\t x - 6y = 4
equation 2 \t\t 3x - 18y = 4

Explanation:

Step1: Analyze the original system

The first equation is \(x - 6y = 4\). Multiply it by 3: \(3x - 18y = 12\). The second equation is \(3x - 18y = 4\). These are parallel lines (same slope, different intercepts) with no solution.

Step2: Modify a number for one solution

To have one solution, the two equations must represent non - parallel (intersecting) lines. Let's change the constant term of the second equation. Let's change the 4 in \(3x - 18y = 4\) to 12. Now the system is \(

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

\). But wait, that's the same line (infinite solutions). Oops, we need different slopes or different coefficients. Wait, the slope of \(ax+by = c\) is \(-\frac{a}{b}\). For \(x - 6y = 4\), slope is \(\frac{1}{6}\). For \(3x - 18y = c\), slope is \(\frac{3}{18}=\frac{1}{6}\). So to change the slope, we need to change the coefficient of \(x\) or \(y\) in the second equation. Let's change the coefficient of \(x\) in the second equation from 3 to 2. Now the system is \(

$$\begin{cases}x - 6y = 4\\2x - 18y = 4\end{cases}$$

\). The slopes are \(\frac{1}{6}\) and \(\frac{2}{18}=\frac{1}{9}\), which are different. So this system will have one solution.

Another way: Let's change the coefficient of \(y\) in the second equation. The first equation: \(x - 6y = 4\), slope \(m_1=\frac{1}{6}\). Second equation: \(3x - ay = 4\), slope \(m_2=\frac{3}{a}\). We want \(m_1
eq m_2\), so \(\frac{1}{6}
eq\frac{3}{a}\), \(a
eq18\). Let's set \(a = 17\). So the second equation becomes \(3x - 17y = 4\). Now the two lines have different slopes, so they will intersect at one point.

Let's take a simple change. Let's change the 3 in \(3x - 18y = 4\) to 2. So the new system is:
Equation 1: \(x - 6y = 4\)
Equation 2: \(2x - 18y = 4\)

To check, solve the first equation for \(x\): \(x=6y + 4\). Substitute into the second equation: \(2(6y + 4)-18y = 4\), \(12y+8 - 18y = 4\), \(- 6y=4 - 8=-4\), \(y=\frac{2}{3}\). Then \(x=6\times\frac{2}{3}+4 = 4 + 4=8\). So there is one solution \((8,\frac{2}{3})\).

Answer:

One possible answer is to change the coefficient of \(x\) in the second equation from 3 to 2, resulting in the system \(

$$\begin{cases}x - 6y = 4\\2x - 18y = 4\end{cases}$$

\) (or other valid changes like changing the coefficient of \(y\) or the constant term in a way that makes the slopes different). For example, changing the second equation to \(2x - 18y = 4\) (by changing the 3 to 2 in \(3x - 18y = 4\)) will create a system with one solution.