When you add 2+3 which equals 5 so it’ll be y=5x-7. And plug in y into the other equation. 5x-7-5x+8=0. -7-8=0, -7-8 =-1 therefore there is no solution because-7-8= -1 not 0.
Moving -8+5x to the other side of the equation, we have:
x²+3x-7+8-5x = 0
x²+3x-8x-7+8= 0
x²-5x+1 = 0
Since the resulting equation is quadratic, we will get two solutions as our x and substituting both value of x in equation 2 to get y will give us two solutions for our y variable making a total of 4 solutions.
Therefore in the system of equation given, there are exactly 4 solutions
