Question

slopes of non - parallel lines
the graph shows four linear functions.

• ( f(x)=2x )
• ( g(x)=2x - 10 )
• ( h(x)=-\frac{2}{3}x )
• ( j(x)=\frac{2}{3}x + 10 )

study the linear functions graphed to complete each statement.
consider the two lines represented by the functions ( h(x)=-\frac{2}{3}x ) and ( j(x)=\frac{2}{3}x + 10 ). to determine where the two lines intersect, you can set both functions equal to each other and solve.
( -\frac{2}{3}x=\frac{2}{3}x + 10 )
( cdotleft(\frac{2}{3}x + 10
ight) )
( x=-x - 15 )
( = - 15 )
( x=)
the equation ( -\frac{2}{3}x=\frac{2}{3}x + 10 )
the two lines described by the functions ( h(x)=-\frac{2}{3}x ) and ( j(x)=\frac{2}{3}x + 10 )
what patterns have you noticed? the slopes of parallel lines are

Explanation:

Step1: Start with the equation

We have the equation \(-\frac{2}{3}x=\frac{2}{3}x + 10\). To solve for \(x\), we first get all \(x\) terms on one side. Add \(\frac{2}{3}x\) to both sides:
\(-\frac{2}{3}x+\frac{2}{3}x=\frac{2}{3}x+\frac{2}{3}x + 10\)
Simplifying the left side gives \(0=\frac{4}{3}x + 10\).

Step2: Isolate the \(x\) term

Subtract 10 from both sides:
\(0 - 10=\frac{4}{3}x+10 - 10\)
Which simplifies to \(- 10=\frac{4}{3}x\).

Step3: Solve for \(x\)

Multiply both sides by \(\frac{3}{4}\) to solve for \(x\):
\(x=-10\times\frac{3}{4}=-\frac{30}{4}=-\frac{15}{2}=- 7.5\)

Wait, but maybe there was a miscalculation. Wait, let's re - do the first step. The original equation is \(-\frac{2}{3}x=\frac{2}{3}x + 10\). Let's move \(\frac{2}{3}x\) to the left:
\(-\frac{2}{3}x-\frac{2}{3}x=10\)
\(-\frac{4}{3}x = 10\)
Then \(x=10\times(-\frac{3}{4})=-\frac{30}{4}=-\frac{15}{2}=-7.5\)

Now, for the parallel lines part: The slope of \(h(x)=-\frac{2}{3}x\) is \(m_1 =-\frac{2}{3}\) and the slope of \(j(x)=\frac{2}{3}x + 10\) is \(m_2=\frac{2}{3}\). Wait, no, if two lines are parallel, their slopes are equal. But here \(m_1=-\frac{2}{3}\) and \(m_2 = \frac{2}{3}\), which are not equal. Wait, maybe the question about parallel lines is about other lines. Let's check the other lines: \(f(x)=2x\) and \(g(x)=2x - 10\) have the same slope (\(m = 2\)), so they are parallel. The pattern is that parallel lines have equal slopes.

But going back to the intersection of \(h(x)\) and \(j(x)\):

Starting over:
Equation: \(-\frac{2}{3}x=\frac{2}{3}x + 10\)
Add \(\frac{2}{3}x\) to both sides:
\(-\frac{2}{3}x+\frac{2}{3}x=\frac{2}{3}x+\frac{2}{3}x + 10\)
\(0=\frac{4}{3}x + 10\)
Subtract 10:
\(-10=\frac{4}{3}x\)
Multiply by \(\frac{3}{4}\):
\(x=-10\times\frac{3}{4}=-\frac{30}{4}=-\frac{15}{2}=-7.5\)

For the parallel lines: The functions \(f(x) = 2x\) and \(g(x)=2x - 10\) have the same slope (2), so parallel lines have equal slopes. The functions \(h(x)=-\frac{2}{3}x\) and \(j(x)=\frac{2}{3}x + 10\) have slopes that are negative reciprocals? No, \(-\frac{2}{3}\) and \(\frac{2}{3}\) are not negative reciprocals (negative reciprocal of \(-\frac{2}{3}\) is \(\frac{3}{2}\)). Wait, maybe the user made a typo, but assuming we are to solve \(-\frac{2}{3}x=\frac{2}{3}x + 10\):

Answer:

The solution for \(x\) when \(-\frac{2}{3}x=\frac{2}{3}x + 10\) is \(x =-\frac{15}{2}\) (or \(-7.5\)). The pattern for parallel lines is that parallel lines have equal slopes (e.g., \(f(x)=2x\) and \(g(x)=2x - 10\) are parallel as they have the same slope \(m = 2\)).