7. Solving Linear-Nonlinear Systems - 9. Conic Sections - McGraw Hill Glencoe Algebra 2, 2012

We want to solve the given system of equations. 8y = - 10x & (I) y^2 = 2x^2 - 7 & (II) Let's isolate the y-variable in the linear equation so that we can use the Substitution Method.

8y = - 10x y^2 = 2x^2 - 7

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(I): Solve for y

DivEqn

(I): .LHS /8.=.RHS /8.

y = - 10x8 y^2 = 2x^2 - 7

MoveNegNumToFrac

Put minus sign in front of fraction

y = - 10x8 y^2 = 2x^2 - 7

ReduceFrac

a/b=.a /2./.b /2.

y =- 5x4 y^2 = 2x^2 - 7

MovePartNumRight

a* b/c=a/c* b

y =- 54x y^2 = 2x^2 - 7

The y-variable is isolated in Equation (I). This allows us to substitute - 54x for y in Equation (II).

y =- 54x & (I) y^2 = 2x^2 - 7 & (II)

Substitute

(II): y= - 5/4x

y =- 54x ( - 54x)^2 = 2x^2 - 7

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(II): Simplify

NegBaseToPosBase

(II): (- a)^2=a^2

y =- 54x ( 54x)^2 = 2x^2 - 7

PowProdII

(II): (a b)^m=a^m b^m

y =- 54x ( 54)^2x^2 = 2x^2 - 7

PowQuot

(II): (a/b)^m=a^m/b^m

y =- 54x 2516x^2 = 2x^2 - 7

SubEqn

(II): LHS-2x^2=RHS-2x^2

y =- 54x 2516x^2 - 2x^2 = - 7

NumberToFrac

(II): a = 16* a/16

y =- 54x 2516x^2 - 3216x^2 = - 7

SubTerms

(II): Subtract terms

y =- 54x - 716x^2 = - 7

MultEqn

(II): LHS * 16=RHS* 16

y =- 54x - 7x^2 = - 7(16)

DivEqn

(II): .LHS /(- 7).=.RHS /(- 7).

y =- 54x x^2 = 16

To solve Equation (II) we take square roots on each side. Do not forget to consider both the positive and the negative results!

x^2 = 16

SqrtEqn

sqrt(LHS)=sqrt(RHS)

sqrt(x^2) = sqrt(16)

sqrt(a^2)=± a

x = ± sqrt(16)

CalcRoot

Calculate root

x = ± 4

This result tells us that we have two solutions for x. One solution is positive and the other negative. Now, consider Equation (I). y=- 5/4x We can substitute x = 4 and x = - 4 into the above equation to find the values for y. Let's start with x = 4.

y = - 5/4x

Substitute

x= 4

y = - 5/4( 4)

FracMultDenomToNumber

a/4* 4 = a

y = - 5

We found that y=- 5 when x=4. Therefore, one solution to the system is the point (4,- 5). To find the other solution, we will substitute - 4 for x in Equation (I) again.

y = - 5/4x

Substitute

x= - 4

y = - 5/4( - 4)

MultNegNegOnePar

- a(- b)=a* b

y = 5/4(4)

FracMultDenomToNumber

a/4* 4 = a

y = 5

We found that y=5 when x=- 4. Our second solution is the point (- 4, 5)

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 (4,- 5). We will substitute 4 and - 5 for x and y, respectively, in Equation (I) and Equation (II).

8y = - 10x & (I) y^2 = 2x^2 - 7 & (II)

(I), (II): x= 4, y= - 5

8( - 5)? =- 10( 4) ( - 5)^2? =2( 4)^2-7

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Evaluate

MultPosNeg

(I): a(- b)=- a * b

- 40 ? = - 10(4) (- 5)^2? =2(4)^2 - 7

MultNegPos

(I): (- a)b = - ab

- 40 = - 40 ✓ (- 5)^2? =2(4)^2 - 7

CalcPow