Question

problem 6
consider this system of equations:
x - 6y = 4
3x - 18y = 4
change one number to make a new system with an infinite number of solutions.
equation 1: x - 6y = 4
equation 2: 3x - 18y = 4

Explanation:

Step1: Recall the condition for infinite solutions

For a system of linear equations \(a_1x + b_1y = c_1\) and \(a_2x + b_2y = c_2\) to have infinite solutions, the two equations must be scalar multiples of each other, i.e., \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\).

Given Equation 1: \(x - 6y = 4\) (so \(a_1 = 1\), \(b_1=-6\), \(c_1 = 4\))
Equation 2: \(3x - 18y = 4\) (so \(a_2 = 3\), \(b_2=-18\), \(c_2 = 4\))

Check the ratios of coefficients: \(\frac{a_1}{a_2}=\frac{1}{3}\), \(\frac{b_1}{b_2}=\frac{-6}{-18}=\frac{1}{3}\), but \(\frac{c_1}{c_2}=\frac{4}{4} = 1\) (wait, no, \(c_2\) is 4? Wait, no, original Equation 2 is \(3x - 18y = 4\), but if we want \(\frac{c_1}{c_2}=\frac{1}{3}\), since \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{1}{3}\), then \(c_2\) should be \(3\times c_1=3\times4 = 12\). Or we can adjust Equation 2's constant term or Equation 1's coefficients. Alternatively, let's see: If we multiply Equation 1 by 3, we get \(3x-18y = 12\). So currently, Equation 2 is \(3x - 18y = 4\). So to make them scalar multiples, we can change the constant term of Equation 2 from 4 to 12, or change the constant term of Equation 1, or change the coefficients. But the problem says "change one number". Let's take Equation 2: \(3x - 18y = 4\). If we change the 4 (constant term) to 12, then Equation 2 becomes \(3x - 18y = 12\), which is 3 times Equation 1 (\(x - 6y = 4\) multiplied by 3 is \(3x - 18y = 12\)). So now the two equations are scalar multiples, so they represent the same line, hence infinite solutions.

Step2: Implement the change

So we can change the constant term of the second equation from 4 to 12. So the new system is:
Equation 1: \(x - 6y = 4\)
Equation 2: \(3x - 18y = 12\)

Now, check the ratios: \(\frac{1}{3}=\frac{-6}{-18}=\frac{4}{12}=\frac{1}{3}\), so they are scalar multiples, hence infinite solutions.

Answer:

Change the constant term of the second equation \(3x - 18y = 4\) to 12, so the new second equation is \(3x - 18y = 12\) (or other valid changes, like changing the coefficient of \(x\) in Equation 1 to 3, etc., but changing the constant term of Equation 2 to 12 is a simple one - number change).