Question

graph $3x + 4y = 12$.

Explanation:

Step1: Find x-intercept (y=0)

Set \( y = 0 \) in \( 3x + 4y = 12 \):
\( 3x + 4(0) = 12 \)
\( 3x = 12 \)
\( x = 4 \)
So x-intercept is \( (4, 0) \), but the graph shows y=5, so maybe rewrite in slope - intercept form.

Step2: Rewrite in slope - intercept form (\( y = mx + b \))

\( 3x + 4y = 12 \)
Subtract \( 3x \): \( 4y=-3x + 12 \)
Divide by 4: \( y=-\frac{3}{4}x + 3 \)

Wait, the given graph is a horizontal line \( y = 5 \), which doesn't match \( 3x + 4y = 12 \) (which has \( y = 3 \) when \( x = 0 \), \( x = 4 \) when \( y = 0 \)). But to graph \( 3x+4y = 12 \):

Step3: Find y - intercept (x = 0)

Set \( x = 0 \): \( 3(0)+4y = 12 \)
\( 4y = 12 \)
\( y = 3 \), so point \( (0, 3) \)

Step4: Find another point (x = 4)

As before, \( x = 4 \), \( y = 0 \), point \( (4, 0) \)

Plot \( (0, 3) \) and \( (4, 0) \), then draw the line through them.

Answer:

To graph \( 3x + 4y = 12 \), find two points:

• When \( x = 0 \), \( y = 3 \) (point \( (0, 3) \)).
• When \( y = 0 \), \( x = 4 \) (point \( (4, 0) \)).

Plot these points and draw a line through them. The correct graph should pass through \( (0, 3) \) and \( (4, 0) \), not the horizontal line \( y = 5 \) shown (which is incorrect for this equation).