Exercise 51 Page 267

For simplicity, we will solve the given system of equations using the Substitution Method. x^2+y^2=1 & (I) y= 12x^2-1 & (II) Although the y-variable is isolated in Equation (II), we will isolate the x^2-variable in Equation (II). Then we will substitute its value for x^2 in Equation (I). In this way, we will not have to expand the power of a binomial. Let's do it! Now, let's substitute 2y+2 for x^2 in Equation (I). Notice that in Equation (I) we have a quadratic equation in terms of only the y-variable. Notice that there are many ways to solve the obtained equation. We will use the Quadratic Formula. Let's determine a, b, and c. y^2+2y+1=0 ⇔ 1y^2+ 2y+ 1=0 We see above that a= 1, b= 2, and c= 1. Let's substitute these values into the Quadratic Formula to solve the equation.

y=- b±sqrt(b^2-4ac)/2a

SubstituteValues

Substitute values

y=- 2±sqrt(2^2-4( 1)( 1))/2( 1)

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Simplify right-hand side

CalcPow

Calculate power

y=- 2±sqrt(4-4(1)(1))/2(1)

Multiply

Multiply

y=- 2±sqrt(4-4)/2

SubTerms

Subtract terms

y=- 2±sqrt(0)/2

CalcRoot

Calculate root

y=- 2± 0/2

Simplify numerator

y=- 2/2

CalcQuot

Calculate quotient

y=- 1

Now consider Equation (II). x^2=2y+2 We will substitute y=- 1 into the above equation to find the value for x.

x^2=2y+2

Substitute

y= - 1

x^2=2( - 1)+2

Multiply

Multiply

x^2=- 2+2

AddTerms

Add terms

x^2=0

SqrtEqn

sqrt(LHS)=sqrt(RHS)

x=0

We found that x=0, when y=- 1. Therefore, the solution of the system is (0,- 1).

Graphing

Note that Equation (I) represents a circle and Equation (II) represents a parabola. If you need explanations on graphing circles, please refer to this example. If you need explanations on graphing parabolas, please refer to this other example. Let's graph them and mark the point (0,- 1).

Both the circle and the parabola intersect at the point (0,- 1), which confirms our answer.