We want to solve the given system of equations using the substitution method.
y=x^2-2x+3 & (I) y=x+1 & (II)
The y-variable is isolated in Equation (II). This allows us to substitute its value, x+1, for y in Equation (I).
Notice that in Equation (I), we have a quadratic equation in terms of only the x-variable.
x^2-3x+2=0 ⇔ 1x^2+( - 3)x+ 2=0Now, recall the Quadratic Formula.
x=- b±sqrt(b^2-4ac)/2a
We can substitute a= 1, b= - 3, and c= 2 into this formula to solve the quadratic equation.
We found that y=3 when x=2. One solution of the system, which is a point of intersection of a parabola and a line, is (2,3). To find the other solution, we will substitute 1 for x in Equation (II).
We found that y=2 when x=1. Therefore our second solution, which is the other point of intersection of the parabola and the line, is (1, 2).
Checking Our Answer
Checking the answer
We can check our answers by substituting the points into both equations. If they both produce a true statement, our solutions are correct. Let's start by checking (2,3). We will substitute 2 and 3 for x and y, respectively, in Equation (I) and Equation (II).
