Question

one equation in a system of equations with one - solution is 4x + 2y = 14. determine if each equation could be the second equation in the system.
a. 2x + y = 7
b. 3x - 6y=-12
c. 2x + 6y = 32
d. - 3x + 10y = 1
e. 2x + y = 5
a. the line represented by 2x + y = 7
the equation 2x + y = 7
the line represented by 4x + 2y = 14. this means they have
in common, and the system would have

Explanation:

Step1: Recall condition for one - solution system

A system of linear equations \(a_1x + b_1y=c_1\) and \(a_2x + b_2y = c_2\) has one solution when \(\frac{a_1}{a_2}
eq\frac{b_1}{b_2}\). The given equation is \(4x + 2y=14\) (which simplifies to \(2x + y = 7\)).

Step2: Check each option

For option a: The first equation is \(2x + y=7\) and the second is \(2x + y = 5\). Here \(\frac{2}{2}=\frac{1}{1}
eq\frac{7}{5}\), so it's inconsistent (no solution).
For option b: The first equation \(2x + y=7\) and the second \(x-6y=-12\). We have \(\frac{2}{1}
eq\frac{1}{-6}\), so the system has one - solution.
For option c: The first equation \(2x + y=7\) and the second \(2x+6y = 32\). Here \(\frac{2}{2}=\frac{1}{1}
eq\frac{7}{32}\), the lines are parallel (no solution).
For option d: The first equation \(2x + y=7\) and the second \(-3x + 10y=1\). We have \(\frac{2}{-3}
eq\frac{1}{10}\), so the system has one - solution.
For option e: The first equation \(2x + y=7\) and the second \(2x + y=5\) is inconsistent as \(\frac{2}{2}=\frac{1}{1}
eq\frac{7}{5}\).

Answer:

B. \(x - 6y=-12\), D. \(-3x + 10y = 1\)