Question

another system of equations
discuss: can you identify a point that balances just hanger a? just hanger b? both? neither?
hanger a: 2x = y
hanger b: y = 2x + 6
(chart: weight of circle (lb) on y - axis, weight of triangle (lb) on x - axis. lines for hanger a (2x = y) and hanger b (2x + 6 = y), point (2,10) marked on hanger bs line)

Explanation:

Step1: Analyze Hanger A's equation

Hanger A has the equation \(2x = y\). We can find points \((x,y)\) that satisfy this by choosing an \(x\) value and computing \(y\). For example, if \(x = 3\), then \(y=2\times3 = 6\), so \((3,6)\) balances Hanger A.

Step2: Analyze Hanger B's equation

Hanger B has the equation \(y = 2x + 6\). Let's check a point. If \(x = 4\), then \(y=2\times4 + 6=8 + 6 = 14\), so \((4,14)\) balances Hanger B.

Step3: Analyze the intersection (both hangers)

To find a point that balances both, we solve the system \(

$$\begin{cases}2x=y\\y = 2x+6\end{cases}$$

\). Substitute \(y = 2x\) into the second equation: \(2x=2x + 6\), which simplifies to \(0 = 6\), a contradiction. So no solution for both. But from the graph, we see the point \((2,10)\): check Hanger A: \(2\times2=4
eq10\), so not A. Check Hanger B: \(2\times2 + 6=4 + 6 = 10\), so \((2,10)\) balances Hanger B.

Answer:

• A point balancing Hanger A: e.g., \((3,6)\) (any \((x,2x)\) works).
• A point balancing Hanger B: e.g., \((2,10)\) (or \((4,14)\) etc., any \((x,2x + 6)\) works).
• Both: None (since the system is inconsistent).
• Neither: Points not on either line, e.g., \((1,5)\) (not on \(y = 2x\) or \(y=2x + 6\)).