If the point (2k – 3, k + 2) lies on the line 2x + 3y + 15 = 0, find the value of k.
k=−157
k=−137
k=157
k=137
Since the point (2k - 3, k + 2) lies on the line 2x + 3y + 15 = 0, 2(2k−3)+3(k+2)+15=0 ⇒4k−6+3k+6+15=0 ⇒7k=−15⇒k=−157
If the points (2k-3,k+2) lie on the graph of the equation 2x+3y+15=0, find the value of k.