Question

1 how many solutions does each system of equations have? explain how you can tell.
a. ( x + 6y = 19 )
( x - 6y = 13 )

Explanation:

Step1: Analyze the system type

This is a system of linear equations. For a system of the form \(a_1x + b_1y = c_1\) and \(a_2x + b_2y = c_2\), we can check the slopes (or use elimination/substitution). Here, let's use elimination.

Step2: Add the two equations

Add \(x + 6y = 19\) and \(x - 6y = 13\).
\((x + 6y)+(x - 6y)=19 + 13\)
Simplify the left side: \(x+6y+x - 6y = 2x\)
Simplify the right side: \(32\)
So we get \(2x=32\).

Step3: Solve for x

Divide both sides by 2: \(x=\frac{32}{2}=16\).

Step4: Substitute x back to find y

Substitute \(x = 16\) into the first equation \(x + 6y = 19\):
\(16+6y = 19\)
Subtract 16 from both sides: \(6y=19 - 16=3\)
Divide by 6: \(y=\frac{3}{6}=\frac{1}{2}\).
Since we found a unique solution \((x,y)=(16,\frac{1}{2})\), the system has one solution. Also, we can check the slopes: rewrite the equations in slope - intercept form (\(y = mx + b\)).
For \(x + 6y = 19\), \(6y=-x + 19\), \(y=-\frac{1}{6}x+\frac{19}{6}\), slope \(m_1=-\frac{1}{6}\).
For \(x - 6y = 13\), \(-6y=-x + 13\), \(y=\frac{1}{6}x-\frac{13}{6}\), slope \(m_2=\frac{1}{6}\). Since \(m_1
eq m_2\), the lines intersect at one point, so one solution.

Answer:

The system of equations \(x + 6y = 19\) and \(x - 6y = 13\) has one solution. We can tell this by either solving the system (and finding a unique \((x,y)\) pair) or by noting that the slopes of the two lines (when written in slope - intercept form) are different, meaning the lines intersect at exactly one point.