Exercise 47 Page 267

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=0We see above that a= 1, b= - 7, and c= 12. Let's substitute these values into the Quadratic Formula to solve the equation.

x=- b±sqrt(b^2-4ac)/2a

SubstituteValues

Substitute values

x=-( - 7)±sqrt(( - 7)^2-4( 1)( 12))/2( 1)

▼

Simplify right-hand side

NegNeg

- (- a)=a

x=7±sqrt(49-4(1)(12))/2(1)

CalcPow

Calculate power

x=7±sqrt(49-4(1)(12))/2(1)

Multiply

Multiply

x=7±sqrt(49-48)/2

SubTerms

Subtract terms

x=7±sqrt(1)/2

CalcRoot

Calculate root

x=7± 1/2

We can calculate the first root using the positive sign and the second root using the negative sign.

x=7± 1/2
x=7+ 1/2	x=7- 1/2
x=4	x=3

Now consider Equation (II). y=x-3 We can substitute x=4 and x=3 into the above equation to find the values for y. Let's start with x=4.

y=x-3

Substitute

x= 4

y= 4-3

SubTerms

Subtract terms

y=1

We found that y=1 when x=4. One solution of the system is (4,1). To find the other solution, we will substitute 3 for x in Equation (II) again.

y=x-3

Substitute

x= 3

y= 3-3

SubTerms

Subtract terms

y=0

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.