Notice that the y-variable is already isolated in Equation (II). Therefore, the most convenient method to solve the given system of equations is the Substitution Method.
y^2=x-3 & (I) y=x-3 & (II)
Let's substitute x-3 for y in Equation (I).
Notice that in Equation (I) we have a quadratic equation in terms of only the x-variable. There are many ways to solve this type of equation. We will use the Quadratic Formula. Let's determine a, b, and c.
x^2-7x+12=0 ⇔ 1x^2+( - 7)x+ 12=0
We see above that a= 1, b= - 7, and c= 12. Let's substitute these values into the Quadratic Formula to solve the equation.
We found that y=0, when x=3. Therefore, our second solution is (3,0).
Graphing
Notice that Equation (I) represents a parabola and Equation (II) represents a line. If you need explanations on graphing parabolas, please see this example. If you need explanations on graphing lines, please see this other example. Let's graph them and mark the points (4,1) and (3,0)!
The parabola and the line intersect at the points (4,1) and (3,0), which confirms our answer.
