Exercise 50 Page 267

We will solve the given system of equations using the Substitution Method. x^2+y^2=25 & (I) y=- 34x+ 254 & (II) The y-variable is isolated in Equation (II). This allows us to substitute its value - 34x+ 254 for y in Equation (I). Notice that in Equation (I) we have a quadratic equation in terms of only the x-variable. Notice that there are many ways to solve a quadratic equation. We will use Quadratic Formula. Let's determine a, b, and c. x^2-6x+9=0 ⇔ 1x^2+( - 6)x+ 9=0We see above that a= 1, b= - 6, and c= 9. Let's substitute these values into the Quadratic Formula to solve the equation.

x=- b±sqrt(b^2-4ac)/2a

SubstituteValues

Substitute values

x=- ( - 6)±sqrt(( - 6)^2-4( 1)( 9))/2( 1)

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Simplify right-hand side

CalcPow

Calculate power

x=- (- 6)±sqrt(36-4(1)(9))/2(1)

Multiply

Multiply

x=6±sqrt(36-36)/2

SubTerms

Subtract terms

x=6±sqrt(0)/2

CalcRoot

Calculate root

x=6± 0/2

Simplify numerator

x=6/2

CalcQuot

Calculate quotient

x=3

Now consider Equation (II) in the given nonlinear system. y=-3/4x+25/4 We will substitute x=3 into the above equation to find the value for y.

y=-3/4x+25/4

Substitute

x= 3

y=-3/4( 3)+25/4

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Simplify

Multiply

Multiply

y=-9/4+25/4

AddFrac

Add fractions

y=16/4

CalcQuot

Calculate quotient

y=4

We found that y=4 when x=3. Therefore, the solution of the system is (3,4).

Graphing

Let's notice that Equation (I) represents a circle and Equation (II) represents a line. If you need explanations on graphing circles, please refer to this site. If you need explanations on graphing lines, please refer to this site. Let's graph them and mark the point (3,4).

Notice that the circle and the line intersect about at the point (3,4), which confirms our answer.