Question

cepts. show your work!

1. 2x = 3y = 12

Explanation:

Step1: Find x-intercept (set y=0)

The equation is likely a typo, should be \(2x + 3y = 12\) (since \(2x = 3y = 12\) is inconsistent). For x - intercept, set \(y = 0\):
\(2x+3(0)=12\)
\(2x = 12\)

Step2: Solve for x

Divide both sides by 2:
\(x=\frac{12}{2}=6\)? Wait, no, maybe the original equation is \(2x=12\) and \(3y = 12\) (system of equations). Let's re - interpret: If it's two equations \(2x = 12\) and \(3y=12\).
For \(2x = 12\), solve for x: \(x=\frac{12}{2}=6\)? No, the graph has x from - 5 to 5. Wait, maybe the equation is \(2x+3y = 12\). Let's correct:
For x - intercept (\(y = 0\)):
\(2x+3(0)=12\)
\(2x=12\)
\(x = 6\) (but graph range is - 5 to 5, maybe the equation is \(2x+3y=12\) is wrong. Wait, maybe the equation is \(2x = 12\) (x - intercept when y = 0, but 2x=12 gives x = 6) and \(3y=12\) (y - intercept when x = 0, y = 4). But the graph has x up to 5. Maybe the original problem is \(2x+3y = 12\). Let's do that:
x - intercept (y = 0):
\(2x+3(0)=12\)
\(2x = 12\)
\(x = 6\) (but graph is small, maybe a typo, equation is \(2x+3y = 12\) or \(2x=12\) and \(3y = 12\) as two intercepts.
Wait, if we consider the equation as two separate equations: \(2x=12\) (so x - intercept is x = 6, but graph is - 5 to 5, maybe the equation is \(2x+3y = 12\). Let's recalculate:
x - intercept (\(y = 0\)):
\(2x+3(0)=12\Rightarrow2x = 12\Rightarrow x = 6\)
y - intercept (\(x = 0\)):
\(2(0)+3y=12\Rightarrow3y = 12\Rightarrow y = 4\)

But maybe the original equation is \(2x=12\) (x - intercept) and \(3y = 12\) (y - intercept). So x - intercept from \(2x = 12\) is \(x = 6\), y - intercept from \(3y=12\) is \(y = 4\).

Wait, the user's equation is written as \(2x = 3y = 12\), which is a bit ambiguous. It can be interpreted as a system where \(2x=12\) and \(3y = 12\). So:

For x - intercept (when we consider the x - value from \(2x = 12\)):

Step1: Solve \(2x = 12\)

Divide both sides by 2: \(x=\frac{12}{2}=6\) (but graph is - 5 to 5, maybe a typo, equation should be \(2x+3y = 12\))

For y - intercept (from \(3y = 12\)):

Step1: Solve \(3y = 12\)

Divide both sides by 3: \(y=\frac{12}{3}=4\)

Answer:

x - intercept: If \(2x = 12\), then \(x = 6\); y - intercept: If \(3y = 12\), then \(y = 4\). (Assuming the equation is a system \(2x = 12\) and \(3y = 12\))