Exercise 10 Page 151

We want to solve the given system of equations. Because the y-variable is already isolated in Equation II, we will be using the Substitution Method. 0=x^2+y^2-40 & (I) y=x+4 & (II) The y-variable is isolated in Equation (II). This allows us to substitute its value x+4 for y in Equation (I).

0=x^2+y^2-40 y=x+4

Substitute

(I): y= x^2-5x-2

0=x^2+( x+4)^2-40 y=x+4

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(I): Simplify

PowToProdTwoFac

(I): a^2=a* a

0=x^2+(x+4)(x+4)-40 y=x+4

Distr

(I): Distribute x+4

0=x^2+x(x+4)+4(x+4)-40 y=x+4

Distr

Distribute x

0=x^2+x^2+4x+4(x+4)-40 y=x+4

Distr

Distribute 4

0=x^2+x^2+4x+4x+16-40 y=x+4

AddTerms

Add terms

0=2x^2+8x-24 y=x+4

Notice that in Equation (I), we have a quadratic equation in terms of only the x-variable. 0=2x^2+8x-24 ⇔ 2x^2+ 8x+( - 24)=0Now, recall the Quadratic Formula. x=- b±sqrt(b^2-4ac)/2a We can substitute a= 2, b= 8, and c= - 24 into this formula to solve the quadratic equation.

x=- b±sqrt(b^2-4ac)/2a

SubstituteValues

Substitute values

x=- ( 8)±sqrt(( 8)^2-4( 2)( - 24))/2( 2)

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Solve for x

CalcPow

Calculate power

x=- 8±sqrt(64-4(2)(- 24))/2(2)

MultNegPos

(- a)b = - ab

x=- 8±sqrt(64-8(- 24))/2(2)

MultNegNegOnePar

- a(- b)=a* b

x=- 8±sqrt(64+192)/2(2)

AddTerms

Add terms

x=- 8±sqrt(256)/2(2)

Multiply

x=- 8±sqrt(256)/4

CalcRoot

Calculate root

x=- 8 ± 16/4

This result tells us that we have two solutions for x. One of them will use the positive sign and the other one will use the negative sign.

x=- 8 ± 16/4
x_x=- 8 + 16/4	x_2=- 8 - 16/4
x_1= 8/4	x_2=- 24/4
x_1= 2	x_2=- 6

Now, consider Equation (II). y=x+4 We can substitute x=2 and x=- 6 into the above equation to find the values for y. Let's start with x=2.

y=x+4

Substitute

x= 2

y= 2+4

AddTerms

Add terms

y=6

We found that y=6, when x=2. One solution of the system, which is a point of intersection of the two parabolas, is (2,6). To find the other solution, we will substitute - 6 for x in Equation (II) again.

y=x+4

Substitute

x= - 6

y= - 6+4

AddTerms

Add terms

y=- 2

We found that y=- 2, when x=- 6. Therefore, our second solution, which is the other point of intersection of the two parabolas, is (- 6, - 2).

Checking Our Answer

Checking the answer

We can check our answers by substituting the points into both equations. If they produce true statements, our solutions are correct. Let's start by checking (2,6). We will substitute 2 and 6 for x and y, respectively, in Equation (I) and Equation (II).

0=x^2+y^2-40 y=x+4

(I), (II): x= 2, y= 6

0=( 2)^2+( 6)^2-40 6= 2+4

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Simplify

CalcPow

(I): Calculate power

0=4+36-40 6=2+4

(I), (II): Add and subtract terms

0=0 ✓ 6=6 ✓

Since both equations produce true statements, the solution (2,6) is correct. Let's now check (- 6, -2).

0=x^2+y^2-40 y=x+4

(I), (II): x= - 6, y= - 2

0=( - 6)^2+( - 2)^2-40 - 2= - 6+4

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Simplify

CalcPow

(I): Calculate power

0=36+4-40 - 2=- 6+4

(I), (II): Add and subtract terms

0=0 ✓ - 2=- 2 ✓

Since again both equations produce true statements, the solution (- 6,- 2) is also correct.

Hints

Check the answer

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Checking Our Answer