Question

solve the system of equations graphed on the coordinate axes below.
$y = \frac{2}{3}x + 3$
$y = -\frac{1}{4}x + 3$

Explanation:

Step1: Set the equations equal

Since both equations equal \( y \), set \( \frac{2}{3}x + 3 = -\frac{1}{4}x + 3 \).

Step2: Subtract 3 from both sides

\( \frac{2}{3}x + 3 - 3 = -\frac{1}{4}x + 3 - 3 \) simplifies to \( \frac{2}{3}x = -\frac{1}{4}x \).

Step3: Add \( \frac{1}{4}x \) to both sides

\( \frac{2}{3}x + \frac{1}{4}x = -\frac{1}{4}x + \frac{1}{4}x \). Find a common denominator (12) for the left side: \( \frac{8}{12}x + \frac{3}{12}x = \frac{11}{12}x \), so \( \frac{11}{12}x = 0 \).

Step4: Solve for \( x \)

Multiply both sides by \( \frac{12}{11} \): \( x = 0 \times \frac{12}{11} = 0 \).

Step5: Find \( y \)

Substitute \( x = 0 \) into either equation, e.g., \( y = \frac{2}{3}(0) + 3 = 3 \).

Answer:

The solution is \( x = 0 \), \( y = 3 \) (or the ordered pair \( (0, 3) \)).