TRANSFORM THE GIVEN EQUATION IN:- \[ 3 x+2 y-18=0 \] a) SLOPE-INTERCEPT FORM \( (y=m x+c) \) b) TWO INTERCEPTS FORM \( \left[\frac{x}{a}+\frac{y}{b}=1\right] \)

**Step 1. Write the given equation:** \[ 3x + 2y - 18 = 0 \] --- **(a) Slope-Intercept Form \(\left(y = mx + c\right)\):** 1. Isolate the \(y\)-term: \[ 2y = -3x + 18 \] 2. Divide by \(2\): \[ y = -\frac{3}{2}x + 9 \] Thus, the slope-intercept form is: \[ y = -\frac{3}{2}x + 9 \] --- **(b) Two Intercepts Form \(\left[\frac{x}{a} + \frac{y}{b} = 1\right]\):** 1. Find the \(x\)-intercept by letting \(y = 0\): \[ 3x - 18 = 0 \quad \Rightarrow \quad 3x = 18 \quad \Rightarrow \quad x = 6 \] So, \(a = 6\). 2. Find the \(y\)-intercept by letting \(x = 0\): \[ 2y - 18 = 0 \quad \Rightarrow \quad 2y = 18 \quad \Rightarrow \quad y = 9 \] So, \(b = 9\). 3. Substitute into the intercept form: \[ \frac{x}{6} + \frac{y}{9} = 1 \] --- **Final Answers:** - Slope-Intercept Form: \(\displaystyle y = -\frac{3}{2}x + 9\) - Two Intercepts Form: \(\displaystyle \frac{x}{6} + \frac{y}{9} = 1\)