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Core Connections Algebra 2, 2013

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Core Connections Algebra 2, 2013 View details

2. Section 4.2

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Exercise 87 Page 194

Practice makes perfect

a To solve the equation for y, we have to simplify the left-hand side and then perform inverse operations until y is isolated.

x-3(y+2)=6

Distribute -3

x-3y-6=6

LHS+3y=RHS+3y

x-6=6+3y

LHS-6=RHS-6

x-12=3y

Rearrange equation

3y=x-12

.LHS /3.=.RHS /3.

y=x-12/3

Write as a difference of fractions

y=x/3-12/3

Calculate quotient

y=x/3-4

a/b=1/b* a

y=1/3x-4

b Like in Part A, we have to perform inverse operations until y is by itself.

6x-1/y-3=2

LHS+3=RHS+3

6x-1/y=5

LHS * y=RHS* y

6x-1=5y

Rearrange equation

5y=6x-1

.LHS /5.=.RHS /5.

y=6x-1/5

Write as a difference of fractions

y=6x/5-1/5

MovePartNumRight

a* b/c=a/c* b

y=6/5x-1/5

Since the equation started out with y in the denominator, y cannot equal 0.

c To isolate y, we need to remove the radical sign. If we raise both sides to the power of 2, the radical and the power will cancel each other.

sqrt(y-4)=x+1

LHS^2=RHS^2

y-4=(x+1)^2

LHS+4=RHS+4

y=(x+1)^2+4

We have now isolated y and it's not necessary to expand the parentheses, because this form is possible to enter in the calculator. Note that in the original equation y cannot be less than -4.

d Like in Part C, we will raise both sides to the power of 2 to eliminate the radical sign.

sqrt(y+4)=x+2

LHS^2=RHS^2

y+4=(x+2)^2

ExpandPosPerfectSquare

(a+b)^2=a^2+2ab+b^2

y+4=x^2+4x+4

LHS-4=RHS-4

y=x^2+4x

Note that In the original equation y must be greater or equal to -4. Otherwise, we will end up with a negative argument under the square root which is undefined.

Subchapter links

4.2.1 Solving Inequalities with One or Two Variables p.188-190

4.2.2 Using Systems to Solve a Problem p.193-194

4.2.3 Application of Systems of Linear Inequalities p.197-198

4.2.4 Using Graphs to Find Solutions p.200-201